Vectors and Three-Dimensional Space

Geometric Vectors and Magnitudes

In mathematics and physics, many quantities have both a size and a direction. For example, a wind blowing east at 20 km/h is not described fully by the number 20 alone. It also needs a direction. Quantities like this are called vectors.

A geometric vector is often drawn as an arrow. The arrow shows three important ideas:

• its starting point,
• its direction, and
• its length, which is called its magnitude.

This lesson explains how to represent vectors geometrically, how to calculate their magnitudes, and how to recognize when two vectors are equivalent.

1. What is a geometric vector?

A geometric vector is a directed line segment. If a vector goes from point \(A\) to point \(B\), we write it as \(\overrightarrow{AB}\).

The vector \(\overrightarrow{AB}\) tells us how to move from \(A\) to \(B\). It does not mainly describe the location of the arrow on the page. What matters most is:

• how far it goes, and
• which way it points.

For example, an arrow 5 units to the right represents the same vector no matter where it is drawn, as long as its length and direction stay the same.

2. Magnitude of a vector

The magnitude of a vector is its length. If \(\vec{v}\) is a vector, its magnitude is written as \(|\vec{v}|\) or sometimes \(\|\vec{v}\|\).

If a vector lies in a 2D plane and has horizontal change \(x\) and vertical change \(y\), then its magnitude is found using the Pythagorean theorem:

$$ |\vec{v}| = \sqrt{x^2 + y^2} $$

If a vector lies in 3D space and has components \((x,y,z)\), then its magnitude is:

$$ |\vec{v}| = \sqrt{x^2 + y^2 + z^2} $$

This formula comes from extending the Pythagorean theorem into three dimensions.

3. Representing vectors using coordinates

If a vector starts at \(A(x_1,y_1)\) and ends at \(B(x_2,y_2)\), then the vector from \(A\) to \(B\) is:

$$ \overrightarrow{AB} = (x_2-x_1,\; y_2-y_1) $$

In 3D, if \(A(x_1,y_1,z_1)\) and \(B(x_2,y_2,z_2)\), then:

$$ \overrightarrow{AB} = (x_2-x_1,\; y_2-y_1,\; z_2-z_1) $$

These coordinate differences tell us how much movement happens in each direction.

4. Equivalent vectors

Two vectors are equivalent if they have the same magnitude and the same direction, even if they begin at different points.

For instance, a vector 3 units right and 2 units up is equivalent to any other vector that also moves 3 units right and 2 units up.

This means vectors can be shifted around in space without changing what they represent, as long as their length and direction do not change.

5. Important difference: vector vs magnitude

• A vector has both size and direction.
• A magnitude is only the size of the vector.

For example, if a force is represented by the vector \((6,8)\), then the vector tells us the direction and movement in the plane, while its magnitude tells us only how strong or long it is.

6. Finding magnitude from coordinates

If a vector is written as \(\vec{v}=(a,b)\), then:

$$ |\vec{v}| = \sqrt{a^2+b^2} $$

If a vector is written as \(\vec{v}=(a,b,c)\), then:

$$ |\vec{v}| = \sqrt{a^2+b^2+c^2} $$

Always square each component first, add them, and then take the square root.

Worked Example 1: Magnitude of a 2D vector

Find the magnitude of \(\vec{v}=(3,4)\).

Step 1: Use the 2D magnitude formula.

$$ |\vec{v}| = \sqrt{3^2+4^2} $$

Step 2: Square each number.

$$ |\vec{v}| = \sqrt{9+16} $$

Step 3: Add and simplify.

$$ |\vec{v}| = \sqrt{25} = 5 $$

Answer: The magnitude is \(5\).

This is a very common vector because it forms a \(3\)-\(4\)-\(5\) right triangle.

Worked Example 2: Vector between two points

Find the vector and its magnitude from \(A(1,2)\) to \(B(7,10)\).

Step 1: Subtract coordinates.

$$ \overrightarrow{AB} = (7-1,\;10-2) = (6,8) $$

Step 2: Find the magnitude.

$$ |\overrightarrow{AB}| = \sqrt{6^2+8^2} = \sqrt{36+64} = \sqrt{100} = 10 $$

Answer: The vector is \((6,8)\) and its magnitude is \(10\).

This means that moving from \(A\) to \(B\) requires 6 units in the horizontal direction and 8 units in the vertical direction.

Worked Example 3: Equivalent vectors

Determine whether \(\overrightarrow{PQ}\) and \(\overrightarrow{RS}\) are equivalent, where:

• \(P(2,1)\), \(Q(6,4)\)
• \(R(-3,5)\), \(S(1,8)\)

Step 1: Find \(\overrightarrow{PQ}\).

$$ \overrightarrow{PQ} = (6-2,\;4-1) = (4,3) $$

Step 2: Find \(\overrightarrow{RS}\).

$$ \overrightarrow{RS} = (1-(-3),\;8-5) = (4,3) $$

Step 3: Compare the vectors.

Both vectors are \((4,3)\), so they have the same direction and the same magnitude.

Answer: Yes, they are equivalent vectors.

Worked Example 4: Magnitude in 3D

Find the magnitude of the vector \(\vec{u}=(2,-3,6)\).

Step 1: Use the 3D magnitude formula.

$$ |\vec{u}| = \sqrt{2^2+(-3)^2+6^2} $$

Step 2: Square each component.

$$ |\vec{u}| = \sqrt{4+9+36} $$

Step 3: Add and simplify.

$$ |\vec{u}| = \sqrt{49} = 7 $$

Answer: The magnitude is \(7\).

This tells us the vector has length 7 units in three-dimensional space.

7. Visual meaning of components

A vector such as \((a,b)\) can be thought of as:

• \(a\) units horizontally,
• \(b\) units vertically.

If \(a\) is positive, the vector moves right. If \(a\) is negative, it moves left.

If \(b\) is positive, the vector moves up. If \(b\) is negative, it moves down.

In 3D, the third component \(c\) describes movement along the third axis.

8. Common mistakes to avoid

• Mixing up points and vectors: A point gives a location, while a vector gives movement or direction.
• Subtracting in the wrong order: For \(\overrightarrow{AB}\), always do endpoint minus starting point.
• Forgetting to square negative numbers correctly: For example, \((-3)^2=9\), not \(-9\).
• Confusing equivalent vectors with equal starting points: Equivalent vectors do not need to begin at the same place.

9. Why this matters

Geometric vectors are the foundation for later ideas such as vector addition, dot product, cross product, displacement, velocity, and force in three-dimensional space.

If you understand how to read a vector, find its magnitude, and identify equivalent vectors, then you are building the core skills needed for more advanced vector work.

Summary

A geometric vector is a directed line segment that has both magnitude and direction. The magnitude of a vector is its length, found using the Pythagorean theorem. In 2D, use \(\sqrt{x^2+y^2}\), and in 3D, use \(\sqrt{x^2+y^2+z^2}\). Two vectors are equivalent if they have the same direction and the same magnitude, even if they are drawn in different places.

Vector Operations and Unit Vectors

Vectors are quantities that have both magnitude (size) and direction. They are used to describe things like displacement, velocity, force, and acceleration. In three-dimensional space, vectors help us quantify motion and forces in different directions at the same time.

In this lesson, you will learn how to represent vectors, perform basic vector operations, and use unit vectors to write vectors clearly in component form. These skills are essential before moving on to ideas like dot products, cross products, and 3D geometry.

1. Representing a Vector

A vector can be shown in several ways:

• As an arrow, where the length shows magnitude and the arrowhead shows direction.
• By coordinates, such as from point \(A(x_1,y_1,z_1)\) to point \(B(x_2,y_2,z_2)\).
• By components, such as \(\langle a,b,c \rangle\).
• Using unit vectors: \(a\mathbf{i}+b\mathbf{j}+c\mathbf{k}\).

In 3D space, the standard unit vectors are:

• \(\mathbf{i}\): one unit in the \(x\)-direction
• \(\mathbf{j}\): one unit in the \(y\)-direction
• \(\mathbf{k}\): one unit in the \(z\)-direction

So the vector \(\langle 3,-2,5 \rangle\) can also be written as:

$$3\mathbf{i}-2\mathbf{j}+5\mathbf{k}$$

2. Vector from Two Points

If a vector goes from point \(A(x_1,y_1,z_1)\) to point \(B(x_2,y_2,z_2)\), then:

$$\overrightarrow{AB}=\langle x_2-x_1,\; y_2-y_1,\; z_2-z_1 \rangle$$

In unit vector form, this becomes:

$$\overrightarrow{AB}=(x_2-x_1)\mathbf{i}+(y_2-y_1)\mathbf{j}+(z_2-z_1)\mathbf{k}$$

This tells us how far and in what direction we move in each coordinate direction to go from \(A\) to \(B\).

3. Magnitude of a Vector

The magnitude, or length, of a vector \(\mathbf{v}=\langle a,b,c \rangle\) is written as \(|\mathbf{v}|\) and found using:

$$|\mathbf{v}|=\sqrt{a^2+b^2+c^2}$$

For a 2D vector \(\langle a,b \rangle\), the formula is:

$$|\mathbf{v}|=\sqrt{a^2+b^2}$$

The magnitude helps us measure the actual size of a displacement, force, or velocity.

4. Vector Addition

To add vectors, add their corresponding components.

If:

$$\mathbf{u}=\langle a,b,c \rangle \quad \text{and} \quad \mathbf{v}=\langle d,e,f \rangle$$

then:

$$\mathbf{u}+\mathbf{v}=\langle a+d,\; b+e,\; c+f \rangle$$

In unit vector form:

$$ (a\mathbf{i}+b\mathbf{j}+c\mathbf{k})+(d\mathbf{i}+e\mathbf{j}+f\mathbf{k})=(a+d)\mathbf{i}+(b+e)\mathbf{j}+(c+f)\mathbf{k} $$

Graphical meaning: Vector addition can be shown using the tip-to-tail method. Place the tail of the second vector at the tip of the first vector. The resultant vector goes from the start of the first vector to the end of the second vector.

5. Vector Subtraction

To subtract vectors, subtract corresponding components:

$$\mathbf{u}-\mathbf{v}=\langle a-d,\; b-e,\; c-f \rangle$$

Subtraction can also be thought of as adding the opposite vector:

$$\mathbf{u}-\mathbf{v}=\mathbf{u}+(-\mathbf{v})$$

Graphically, \(-\mathbf{v}\) has the same magnitude as \(\mathbf{v}\) but points in the opposite direction.

6. Scalar Multiplication

A scalar is an ordinary number. When a vector is multiplied by a scalar, each component is multiplied by that number.

If \(k\) is a scalar and \(\mathbf{v}=\langle a,b,c \rangle\), then:

$$k\mathbf{v}=\langle ka,kb,kc \rangle$$

Scalar multiplication changes the magnitude of a vector, and it may also change its direction:

• If \(k>1\), the vector gets longer.
• If \(0<k<1\), the vector gets shorter.
• If \(k<0\), the vector also reverses direction.

7. Unit Vectors

A unit vector is a vector with magnitude 1. It is used to describe direction only.

The standard unit vectors in 3D are:

$$\mathbf{i}=\langle 1,0,0 \rangle, \quad \mathbf{j}=\langle 0,1,0 \rangle, \quad \mathbf{k}=\langle 0,0,1 \rangle$$

Each has magnitude 1:

$$|\mathbf{i}|=|\mathbf{j}|=|\mathbf{k}|=1$$

Any vector can be written as a combination of these three unit vectors. For example:

$$\langle 4,-3,2 \rangle=4\mathbf{i}-3\mathbf{j}+2\mathbf{k}$$

8. Finding a Unit Vector in the Direction of a Given Vector

To find a unit vector in the same direction as \(\mathbf{v}\), divide the vector by its magnitude:

$$\hat{\mathbf{v}}=\frac{\mathbf{v}}{|\mathbf{v}|}$$

If \(\mathbf{v}=\langle a,b,c \rangle\), then:

$$\hat{\mathbf{v}}=\left\langle \frac{a}{|\mathbf{v}|},\frac{b}{|\mathbf{v}|},\frac{c}{|\mathbf{v}|}\right\rangle$$

This is useful when you want direction without changing the vector’s orientation.

9. Important Properties of Vector Operations

• Commutative property of addition: \(\mathbf{u}+\mathbf{v}=\mathbf{v}+\mathbf{u}\)
• Associative property of addition: \((\mathbf{u}+\mathbf{v})+\mathbf{w}=\mathbf{u}+(\mathbf{v}+\mathbf{w})\)
• Distributive property: \(k(\mathbf{u}+\mathbf{v})=k\mathbf{u}+k\mathbf{v}\)
• Zero vector: \(\mathbf{0}=\langle 0,0,0 \rangle\)
• Opposite vector: if \(\mathbf{v}=\langle a,b,c \rangle\), then \(-\mathbf{v}=\langle -a,-b,-c \rangle\)

These properties help simplify expressions and check your work.

Worked Example 1: Adding and Subtracting Vectors

Let:

$$\mathbf{u}=\langle 3,-1,4 \rangle, \quad \mathbf{v}=\langle 2,5,-6 \rangle$$

Find \(\mathbf{u}+\mathbf{v}\) and \(\mathbf{u}-\mathbf{v}\).

Solution:

Add corresponding components:

$$\mathbf{u}+\mathbf{v}=\langle 3+2,\; -1+5,\; 4+(-6) \rangle=\langle 5,4,-2 \rangle$$

Subtract corresponding components:

$$\mathbf{u}-\mathbf{v}=\langle 3-2,\; -1-5,\; 4-(-6) \rangle=\langle 1,-6,10 \rangle$$

In unit vector form:

$$\mathbf{u}+\mathbf{v}=5\mathbf{i}+4\mathbf{j}-2\mathbf{k}$$

$$\mathbf{u}-\mathbf{v}=\mathbf{i}-6\mathbf{j}+10\mathbf{k}$$

Worked Example 2: Scalar Multiplication

Let \(\mathbf{a}=\langle -2,3,1 \rangle\). Find:

$$2\mathbf{a} \quad \text{and} \quad -3\mathbf{a}$$

Solution:

Multiply each component by 2:

$$2\mathbf{a}=\langle 2(-2),\; 2(3),\; 2(1) \rangle=\langle -4,6,2 \rangle$$

Multiply each component by \(-3\):

$$-3\mathbf{a}=\langle -3(-2),\; -3(3),\; -3(1) \rangle=\langle 6,-9,-3 \rangle$$

Notice that the negative scalar reverses the direction of the vector.

Worked Example 3: Vector from Two Points

Find the vector \(\overrightarrow{PQ}\) where \(P(1,-2,3)\) and \(Q(5,4,-1)\).

Solution:

Use:

$$\overrightarrow{PQ}=\langle x_2-x_1,\; y_2-y_1,\; z_2-z_1 \rangle$$

Substitute the coordinates:

$$\overrightarrow{PQ}=\langle 5-1,\; 4-(-2),\; -1-3 \rangle$$

$$\overrightarrow{PQ}=\langle 4,6,-4 \rangle$$

In unit vector form:

$$\overrightarrow{PQ}=4\mathbf{i}+6\mathbf{j}-4\mathbf{k}$$

Worked Example 4: Finding a Unit Vector

Find a unit vector in the direction of \(\mathbf{v}=\langle 2,-3,6 \rangle\).

Step 1: Find the magnitude of \(\mathbf{v}\).

$$|\mathbf{v}|=\sqrt{2^2+(-3)^2+6^2}$$

$$|\mathbf{v}|=\sqrt{4+9+36}=\sqrt{49}=7$$

Step 2: Divide the vector by its magnitude.

$$\hat{\mathbf{v}}=\frac{1}{7}\langle 2,-3,6 \rangle=\left\langle \frac{2}{7},-\frac{3}{7},\frac{6}{7} \right\rangle$$

In unit vector form:

$$\hat{\mathbf{v}}=\frac{2}{7}\mathbf{i}-\frac{3}{7}\mathbf{j}+\frac{6}{7}\mathbf{k}$$

This vector has the same direction as \(\mathbf{v}\), but its magnitude is 1.

10. Graphical Interpretation in 2D and 3D

Although drawing 3D vectors on paper can be difficult, the main ideas stay the same as in 2D.

• Addition: place vectors tip-to-tail.
• Subtraction: add the opposite vector.
• Scalar multiplication: stretch, shrink, or reverse the vector.

Even when you cannot sketch a 3D diagram perfectly, the component method gives an exact and reliable way to work.

11. Common Mistakes to Avoid

• Adding a component to the wrong coordinate position.
• Forgetting that subtracting a negative becomes addition.
• Mixing up a vector’s magnitude with the vector itself.
• Thinking a unit vector means “small vector.” A unit vector specifically means magnitude 1.
• Forgetting to divide every component when finding a unit vector.

12. Quick Check Questions

1. If \(\mathbf{u}=\langle 1,2,-3 \rangle\) and \(\mathbf{v}=\langle 4,-1,5 \rangle\), find \(\mathbf{u}+\mathbf{v}\).
2. Find \(3\langle -2,1,4 \rangle\).
3. Find the magnitude of \(\langle 6,8,0 \rangle\).
4. Write \(\langle -5,2,7 \rangle\) in unit vector form.
5. Find a unit vector in the direction of \(\langle 0,3,4 \rangle\).

Answers:

1. \(\langle 5,1,2 \rangle\)
2. \(\langle -6,3,12 \rangle\)
3. \(10\)
4. \(-5\mathbf{i}+2\mathbf{j}+7\mathbf{k}\)
5. \(\left\langle 0,\frac{3}{5},\frac{4}{5}\right\rangle\)

Summary

Vectors describe quantities with both magnitude and direction. You can add, subtract, and scale vectors by working directly with their components. Unit vectors \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) let us write vectors clearly in the \(x\)-, \(y\)-, and \(z\)-directions.

A unit vector in the direction of a given vector is found by dividing the vector by its magnitude. Once you are confident with these operations, you will be ready to study more advanced vector ideas such as dot products, cross products, and equations of lines and planes in 3D space.

Lesson: The Dot Product and Orthogonality

In vector mathematics, we often want to compare two vectors. For example, we may want to know whether two forces act in similar directions, whether two paths meet at a right angle, or what angle lies between two directions in space. The dot product helps us answer these questions.

The dot product is a way of combining two vectors to get a single number, called a scalar. That number tells us something important about how the two vectors are related in direction.

One of the most useful ideas connected to the dot product is orthogonality. Two vectors are orthogonal if they are perpendicular to each other. In 12th Grade Maths, the dot product gives us a quick test for perpendicular vectors.

1. What is the dot product?

Suppose we have two vectors in 2D or 3D:

$$ \vec{a} = \langle a_1, a_2, a_3 \rangle, \quad \vec{b} = \langle b_1, b_2, b_3 \rangle $$

The dot product of \(\vec{a}\) and \(\vec{b}\) is written as \(\vec{a} \cdot \vec{b}\) and is found by multiplying matching components and then adding:

$$ \vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3 $$

In 2D, the formula is similar:

$$ \langle a_1, a_2 \rangle \cdot \langle b_1, b_2 \rangle = a_1b_1 + a_2b_2 $$

Notice that the answer is not a vector. It is just a number.

2. Geometric meaning of the dot product

The dot product also has a geometric formula:

$$ \vec{a} \cdot \vec{b} = |\vec{a}|\,|\vec{b}|\cos\theta $$

Here:

• \(|\vec{a}|\) is the magnitude, or length, of \(\vec{a}\)
• \(|\vec{b}|\) is the magnitude, or length, of \(\vec{b}\)
• \(\theta\) is the angle between the two vectors

This formula shows that the dot product depends on both the lengths of the vectors and the angle between them.

3. Magnitude of a vector

To use the geometric formula, we need to know how to find a vector's magnitude.

For a 3D vector \(\vec{a} = \langle a_1, a_2, a_3 \rangle\),

$$ |\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} $$

For a 2D vector \(\langle a_1, a_2 \rangle\),

$$ |\vec{a}| = \sqrt{a_1^2 + a_2^2} $$

4. Using the dot product to find the angle between vectors

From the geometric formula, we can rearrange to get:

$$ \cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|\,|\vec{b}|} $$

Then we find the angle by taking the inverse cosine:

$$ \theta = \cos^{-1}\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|\,|\vec{b}|}\right) $$

This works as long as neither vector is the zero vector.

5. Orthogonality: when vectors are perpendicular

Two non-zero vectors are orthogonal if the angle between them is \(90^\circ\).

Since

$$ \cos 90^\circ = 0 $$

the geometric formula becomes

$$ \vec{a} \cdot \vec{b} = |\vec{a}|\,|\vec{b}|\cdot 0 = 0 $$

So we get an important rule:

If two non-zero vectors have dot product 0, then they are orthogonal.

This gives a quick test for perpendicular vectors:

• Find the dot product.
• If the result is 0, the vectors are orthogonal.
• If the result is not 0, they are not orthogonal.

6. What the sign of the dot product tells us

• If \(\vec{a} \cdot \vec{b} > 0\), the angle between the vectors is acute (less than \(90^\circ\)).
• If \(\vec{a} \cdot \vec{b} = 0\), the angle is right (exactly \(90^\circ\)).
• If \(\vec{a} \cdot \vec{b} < 0\), the angle is obtuse (greater than \(90^\circ\)).

This helps us understand the directional relationship between vectors even before finding the exact angle.

7. Step-by-step method

When working with dot products, follow these steps:

1. Write the vectors clearly.
2. Multiply corresponding components.
3. Add the products to get the dot product.
4. If needed, find magnitudes.
5. Use the angle formula or check whether the dot product equals 0.

Worked Example 1: Find a dot product in 2D

Find \(\vec{a} \cdot \vec{b}\) for

$$ \vec{a} = \langle 3, 4 \rangle, \quad \vec{b} = \langle 2, -1 \rangle $$

Step 1: Multiply matching components.

$$ 3(2) + 4(-1) $$

Step 2: Add.

$$ 6 - 4 = 2 $$

So,

$$ \vec{a} \cdot \vec{b} = 2 $$

The answer is a scalar, not a vector.

Worked Example 2: Test for orthogonality

Determine whether the vectors

$$ \vec{u} = \langle 2, 5 \rangle, \quad \vec{v} = \langle 5, -2 \rangle $$

are orthogonal.

Step 1: Compute the dot product.

$$ \vec{u} \cdot \vec{v} = 2(5) + 5(-2) $$ $$ = 10 - 10 = 0 $$

Step 2: Interpret the result.

Since the dot product is 0, the vectors are orthogonal.

That means they meet at a right angle.

Worked Example 3: Find the angle between two 3D vectors

Find the angle between

$$ \vec{a} = \langle 1, 2, 2 \rangle, \quad \vec{b} = \langle 2, 1, 2 \rangle $$

Step 1: Find the dot product.

$$ \vec{a} \cdot \vec{b} = 1(2) + 2(1) + 2(2) = 2 + 2 + 4 = 8 $$

Step 2: Find the magnitudes.

$$ |\vec{a}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 $$ $$ |\vec{b}| = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 $$

Step 3: Use the angle formula.

$$ \cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|\,|\vec{b}|} = \frac{8}{3 \cdot 3} = \frac{8}{9} $$

So,

$$ \theta = \cos^{-1}\left(\frac{8}{9}\right) $$

Using a calculator,

$$ \theta \approx 27.3^\circ $$

So the angle between the vectors is approximately \(27.3^\circ\).

Worked Example 4: Decide whether the angle is acute, right, or obtuse

Consider

$$ \vec{p} = \langle 4, -1, 2 \rangle, \quad \vec{q} = \langle -2, 3, 1 \rangle $$

Step 1: Find the dot product.

$$ \vec{p} \cdot \vec{q} = 4(-2) + (-1)(3) + 2(1) $$ $$ = -8 - 3 + 2 = -9 $$

Step 2: Use the sign.

Since the dot product is negative, the angle between the vectors is obtuse.

We do not need to calculate the exact angle unless asked.

8. Common mistakes to avoid

• Do not add the vectors. The dot product uses multiplication of matching components, then addition.
• Do not give a vector as the answer. A dot product always gives a scalar.
• Be careful with negative signs. A small sign error can change the result completely.
• Do not forget magnitudes when finding the angle.
• Remember the zero vector issue. The angle formula is not used with the zero vector because its magnitude is 0.

9. Why this matters in vectors and 3D space

In physics and geometry, vectors are used to describe forces, motion, velocity, displacement, and directions in space. The dot product helps us measure how much one vector points in the same direction as another.

It is especially useful for:

• checking whether lines or directions are perpendicular,
• finding angles in 2D and 3D,
• understanding how strongly one force acts in a particular direction.

10. Quick review of key formulas

For vectors \(\vec{a} = \langle a_1, a_2, a_3 \rangle\) and \(\vec{b} = \langle b_1, b_2, b_3 \rangle\):

$$ \vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3 $$ $$ |\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} $$ $$ \vec{a} \cdot \vec{b} = |\vec{a}|\,|\vec{b}|\cos\theta $$ $$ \cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|\,|\vec{b}|} $$

Orthogonality test:

$$ \vec{a} \cdot \vec{b} = 0 \quad \Rightarrow \quad \text{vectors are orthogonal (if both are non-zero)} $$

Brief Summary

The dot product is a way of multiplying two vectors to get a scalar. You calculate it by multiplying corresponding components and adding the results. It can also be used to find the angle between vectors using \(\vec{a} \cdot \vec{b} = |\vec{a}|\,|\vec{b}|\cos\theta\).

Most importantly, the dot product helps us test for orthogonality: if two non-zero vectors have a dot product of 0, then they are perpendicular. This makes the dot product one of the most important tools for working with vectors in 2D and 3D space.

Vector Projections help us break a vector into parts: one part that points along a chosen direction, and another part that is perpendicular to that direction.

This idea is very useful in physics and geometry. For example, if a force acts at an angle, projection helps us find how much of that force acts in the direction of motion. That is exactly the part that matters for calculating work.

In 12th Grade Maths, vector projections are closely connected to the dot product. If you understand dot product, then projection becomes much easier.

Goal of this lesson: learn what scalar and vector projection mean, how to calculate them, and how to use them to split a vector into parallel and orthogonal components.

1. Review: the dot product

For vectors \(\mathbf{a}\) and \(\mathbf{b}\), the dot product is

$$\mathbf{a}\cdot\mathbf{b} = |\mathbf{a}|\,|\mathbf{b}|\cos\theta$$

where \(\theta\) is the angle between the vectors.

This tells us something important: \(|\mathbf{a}|\cos\theta\) is the amount of \(\mathbf{a}\) in the direction of \(\mathbf{b}\). That is the basic idea of projection.

2. Scalar projection

The scalar projection of \(\mathbf{a}\) onto \(\mathbf{b}\) is the signed length of the shadow of \(\mathbf{a}\) in the direction of \(\mathbf{b}\).

It is written as

$$\operatorname{comp}_{\mathbf{b}}\mathbf{a} = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}$$

This is a number, not a vector.

If the result is positive, then \(\mathbf{a}\) points partly in the same direction as \(\mathbf{b}\). If it is negative, then \(\mathbf{a}\) points partly in the opposite direction.

3. Vector projection

The vector projection of \(\mathbf{a}\) onto \(\mathbf{b}\) is the actual vector that lies along \(\mathbf{b}\).

It is written as

$$\operatorname{proj}_{\mathbf{b}}\mathbf{a} = \left(\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^2}\right)\mathbf{b}$$

This gives a vector parallel to \(\mathbf{b}\).

Another way to think about it is:

$$\operatorname{proj}_{\mathbf{b}}\mathbf{a} = \left(\operatorname{comp}_{\mathbf{b}}\mathbf{a}\right)\frac{\mathbf{b}}{|\mathbf{b}|}$$

So we take the scalar projection and multiply by the unit vector in the direction of \(\mathbf{b}\).

4. Parallel and orthogonal components

Any vector \(\mathbf{a}\) can be split into two parts relative to a direction \(\mathbf{b}\):

• a parallel component, which is the projection onto \(\mathbf{b}\)
• an orthogonal component, which is perpendicular to \(\mathbf{b}\)

This is written as

$$\mathbf{a} = \mathbf{a}_{\parallel} + \mathbf{a}_{\perp}$$

where

$$\mathbf{a}_{\parallel} = \operatorname{proj}_{\mathbf{b}}\mathbf{a}$$

and

$$\mathbf{a}_{\perp} = \mathbf{a} - \mathbf{a}_{\parallel}$$

The key fact is that

$$\mathbf{a}_{\perp}\cdot\mathbf{b} = 0$$

That means the orthogonal component is perpendicular to \(\mathbf{b}\).

5. Why projection matters in forces and motion

Suppose a force \(\mathbf{F}\) acts on an object that moves in direction \(\mathbf{d}\). Only the part of the force in the direction of motion contributes to the work done.

The amount of force in that direction is the scalar projection of \(\mathbf{F}\) onto \(\mathbf{d}\):

$$\operatorname{comp}_{\mathbf{d}}\mathbf{F} = \frac{\mathbf{F}\cdot\mathbf{d}}{|\mathbf{d}|}$$

The work done is

$$W = \mathbf{F}\cdot\mathbf{d} = |\mathbf{F}|\,|\mathbf{d}|\cos\theta$$

So projection helps us measure the effective force in the direction of motion.

6. Step-by-step method for finding a projection

1. Find the dot product \(\mathbf{a}\cdot\mathbf{b}\).
2. Find \(|\mathbf{b}|\) or \(|\mathbf{b}|^2\).
3. Use the correct formula:
   • Scalar projection: \(\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}\)
   • Vector projection: \(\left(\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^2}\right)\mathbf{b}\)
4. If needed, subtract the projection from the original vector to get the perpendicular component.

7. Worked Example 1: scalar projection in 2D

Find the scalar projection of \(\mathbf{a} = \langle 3,4 \rangle\) onto \(\mathbf{b} = \langle 1,2 \rangle\).

Step 1: Find the dot product.

$$\mathbf{a}\cdot\mathbf{b} = 3(1) + 4(2) = 3 + 8 = 11$$

Step 2: Find the magnitude of \(\mathbf{b}\).

$$|\mathbf{b}| = \sqrt{1^2 + 2^2} = \sqrt{5}$$

Step 3: Use the scalar projection formula.

$$\operatorname{comp}_{\mathbf{b}}\mathbf{a} = \frac{11}{\sqrt{5}}$$

Answer: the scalar projection is \(\frac{11}{\sqrt{5}}\).

This means the signed length of the component of \(\mathbf{a}\) in the direction of \(\mathbf{b}\) is \(\frac{11}{\sqrt{5}}\).

8. Worked Example 2: vector projection in 2D

Find the vector projection of \(\mathbf{a} = \langle 3,4 \rangle\) onto \(\mathbf{b} = \langle 1,2 \rangle\).

From the previous example, we already know

$$\mathbf{a}\cdot\mathbf{b} = 11$$

Now find \(|\mathbf{b}|^2\):

$$|\mathbf{b}|^2 = 1^2 + 2^2 = 5$$

Use the vector projection formula:

$$\operatorname{proj}_{\mathbf{b}}\mathbf{a} = \left(\frac{11}{5}\right)\langle 1,2 \rangle$$ $$\operatorname{proj}_{\mathbf{b}}\mathbf{a} = \left\langle \frac{11}{5}, \frac{22}{5} \right\rangle$$

Answer:

$$\operatorname{proj}_{\mathbf{b}}\mathbf{a} = \left\langle \frac{11}{5}, \frac{22}{5} \right\rangle$$

This vector lies along \(\mathbf{b}\), so it is the parallel part of \(\mathbf{a}\) relative to \(\mathbf{b}\).

9. Worked Example 3: splitting into parallel and perpendicular parts

Let \(\mathbf{a} = \langle 2,5,-1 \rangle\) and \(\mathbf{b} = \langle 1,0,2 \rangle\). Write \(\mathbf{a}\) as the sum of a component parallel to \(\mathbf{b}\) and a component perpendicular to \(\mathbf{b}\).

Step 1: Find the projection of \(\mathbf{a}\) onto \(\mathbf{b}\).

$$\mathbf{a}\cdot\mathbf{b} = 2(1) + 5(0) + (-1)(2) = 2 + 0 - 2 = 0$$

Also,

$$|\mathbf{b}|^2 = 1^2 + 0^2 + 2^2 = 5$$

So

$$\operatorname{proj}_{\mathbf{b}}\mathbf{a} = \left(\frac{0}{5}\right)\mathbf{b} = \langle 0,0,0 \rangle$$

This means \(\mathbf{a}\) has no component in the direction of \(\mathbf{b}\).

Step 2: Find the perpendicular component.

$$\mathbf{a}_{\perp} = \mathbf{a} - \mathbf{a}_{\parallel} = \langle 2,5,-1 \rangle - \langle 0,0,0 \rangle = \langle 2,5,-1 \rangle$$

Answer:

$$\mathbf{a}_{\parallel} = \langle 0,0,0 \rangle$$ $$\mathbf{a}_{\perp} = \langle 2,5,-1 \rangle$$

So \(\mathbf{a}\) is already perpendicular to \(\mathbf{b}\).

10. Worked Example 4: force and work

A force \(\mathbf{F} = \langle 6,2 \rangle\) newtons moves an object through displacement \(\mathbf{d} = \langle 3,0 \rangle\) metres. Find:

• the component of the force in the direction of motion
• the work done

Step 1: Find the scalar projection of \(\mathbf{F}\) onto \(\mathbf{d}\).

$$\mathbf{F}\cdot\mathbf{d} = 6(3) + 2(0) = 18$$ $$|\mathbf{d}| = \sqrt{3^2 + 0^2} = 3$$ $$\operatorname{comp}_{\mathbf{d}}\mathbf{F} = \frac{18}{3} = 6$$

So the force in the direction of motion is 6 N.

Step 2: Find the work done.

$$W = \mathbf{F}\cdot\mathbf{d} = 18$$

Answer:

• Component of force in the direction of motion: \(6\text{ N}\)
• Work done: \(18\text{ J}\)

This example shows why projection matters: even though the total force is not purely horizontal, only the horizontal part contributes to the work because the motion is horizontal.

11. Common mistakes to avoid

• Mixing up scalar and vector projection. Scalar projection is a number. Vector projection is a vector.
• Projecting onto the wrong vector. In \(\operatorname{proj}_{\mathbf{b}}\mathbf{a}\), the vector lies in the direction of \(\mathbf{b}\), not \(\mathbf{a}\).
• Forgetting the square in the denominator for vector projection: use \(|\mathbf{b}|^2\), not just \(|\mathbf{b}|\).
• Ignoring signs. A negative scalar projection means the component points opposite to the chosen direction.
• Using the zero vector as the direction vector. You cannot project onto \(\langle 0,0,0 \rangle\) because it has no direction.

12. Quick check questions

Try these on your own:

1. Find \(\operatorname{comp}_{\langle 2,0 \rangle}\langle 5,1 \rangle\).
2. Find \(\operatorname{proj}_{\langle 1,1 \rangle}\langle 4,2 \rangle\).
3. If \(\mathbf{a}_{\parallel} = \operatorname{proj}_{\mathbf{b}}\mathbf{a}\), how do you find \(\mathbf{a}_{\perp}\)?
4. Why does the perpendicular component not contribute to work done in a given direction?

13. Final summary

Vector projection tells us how much of one vector acts in the direction of another vector. The scalar projection gives the signed length, while the vector projection gives the actual vector along that direction.

Using projection, we can write a vector as the sum of a parallel component and a perpendicular component. This is especially useful in problems involving forces, motion, and work.

The main formulas to remember are:

$$\operatorname{comp}_{\mathbf{b}}\mathbf{a} = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}$$ $$\operatorname{proj}_{\mathbf{b}}\mathbf{a} = \left(\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^2}\right)\mathbf{b}$$ $$\mathbf{a}_{\perp} = \mathbf{a} - \operatorname{proj}_{\mathbf{b}}\mathbf{a}$$

If you can identify the direction vector correctly and apply these formulas carefully, you can solve most vector projection questions with confidence.

Three-Dimensional Coordinate Systems help us describe positions in space using three numbers instead of two. In a 2D coordinate plane, a point is written as \((x,y)\). In 3D space, a point is written as \((x,y,z)\), where the extra coordinate tells us how far the point is above or below a flat plane.

This idea is very important in mathematics, physics, engineering, and computer graphics. It allows us to describe locations, motion, forces, and shapes in real space.

In this lesson, you will learn how to:

• understand the three axes in 3D space,
• plot and interpret points \((x,y,z)\),
• find the distance between two points in 3D,
• find the midpoint of a line segment in 3D,
• use these ideas to solve geometry problems in space.

1. The three axes in 3D space

A three-dimensional coordinate system has three number lines that meet at one point called the origin. The origin is \((0,0,0)\).

• The x-axis usually shows left-right movement.
• The y-axis usually shows forward-backward movement.
• The z-axis usually shows up-down movement.

These three axes are perpendicular to each other. Together, they let us locate any point in space.

If a point is \((x,y,z)\):

• \(x\) tells how far to move along the x-axis,
• \(y\) tells how far to move along the y-axis,
• \(z\) tells how far to move parallel to the z-axis.

For example, the point \((3,-2,5)\) means:

• move 3 units in the positive x-direction,
• move 2 units in the negative y-direction,
• move 5 units in the positive z-direction.

2. Coordinate planes in 3D

In 3D space, there are three important coordinate planes:

• xy-plane: all points where \(z=0\)
• yz-plane: all points where \(x=0\)
• xz-plane: all points where \(y=0\)

These planes help us picture where a point lies in space. For example, if a point has \(z=0\), then it lies directly on the xy-plane.

3. How to interpret a point in 3D

You can think of plotting \((x,y,z)\) in stages:

1. Start at the origin \((0,0,0)\).
2. Move \(x\) units along the x-axis.
3. From there, move parallel to the y-axis by \(y\) units.
4. Then move parallel to the z-axis by \(z\) units.

This step-by-step view helps when drawing or imagining a 3D point.

4. Distance between two points in 3D

In 2D, the distance formula comes from the Pythagorean theorem. In 3D, we extend the same idea by including the change in the z-coordinate.

If the points are \(A(x_1,y_1,z_1)\) and \(B(x_2,y_2,z_2)\), then the distance between them is:

$$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$$

This formula measures the straight-line distance through space.

Why this formula makes sense:

The differences \((x_2-x_1)\), \((y_2-y_1)\), and \((z_2-z_1)\) tell how far apart the points are in each direction. Squaring and adding them, then taking the square root, is a 3D version of the Pythagorean theorem.

5. Midpoint of a line segment in 3D

The midpoint is the point exactly halfway between two endpoints. In 3D, we find it by averaging each coordinate separately.

If the endpoints are \(A(x_1,y_1,z_1)\) and \(B(x_2,y_2,z_2)\), then the midpoint is:

$$M\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\right)$$

This works because the midpoint must be halfway in the x-direction, halfway in the y-direction, and halfway in the z-direction.

6. Special cases to notice

• If two points have the same \(z\)-coordinate, they lie in a plane parallel to the xy-plane.
• If two points differ only in the z-coordinate, then the segment between them is vertical.
• If one coordinate is zero, the point lies on one of the coordinate planes.
• If two coordinates are zero, the point lies on an axis.

Recognizing these patterns makes problems easier to visualize.

Worked Example 1: Interpreting and identifying location

Describe the location of the point \(P(-4,2,0)\).

Step 1: Read each coordinate.

• \(x=-4\): move 4 units in the negative x-direction.
• \(y=2\): move 2 units in the positive y-direction.
• \(z=0\): there is no vertical movement.

Step 2: Interpret the result.

Because \(z=0\), the point lies on the xy-plane.

Answer: \(P(-4,2,0)\) is 4 units left, 2 units forward, and lies on the xy-plane.

Worked Example 2: Distance between two points in 3D

Find the distance between \(A(1,2,3)\) and \(B(5,-1,7)\).

Use the distance formula:

$$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$$

Substitute the coordinates:

$$d=\sqrt{(5-1)^2+(-1-2)^2+(7-3)^2}$$ $$d=\sqrt{4^2+(-3)^2+4^2}$$ $$d=\sqrt{16+9+16}$$ $$d=\sqrt{41}$$

Answer: The distance between the points is \(\sqrt{41}\) units.

Worked Example 3: Midpoint in 3D

Find the midpoint of the segment joining \(C(6,-2,4)\) and \(D(2,8,-6)\).

Use the midpoint formula:

$$M\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\right)$$

Substitute the values:

$$M\left(\frac{6+2}{2},\frac{-2+8}{2},\frac{4+(-6)}{2}\right)$$ $$M\left(\frac{8}{2},\frac{6}{2},\frac{-2}{2}\right)$$ $$M(4,3,-1)$$

Answer: The midpoint is \((4,3,-1)\).

Worked Example 4: Using distance in a spatial geometry problem

A drone moves from point \(A(0,0,0)\) to point \(B(3,4,12)\). How far did it travel in a straight line?

Use the distance formula:

$$d=\sqrt{(3-0)^2+(4-0)^2+(12-0)^2}$$ $$d=\sqrt{3^2+4^2+12^2}$$ $$d=\sqrt{9+16+144}$$ $$d=\sqrt{169}$$ $$d=13$$

Answer: The drone traveled 13 units.

This example shows how 3D coordinates can model real movement through space.

7. Connecting 2D and 3D formulas

It helps to compare the formulas you already know from 2D with the new 3D versions.

• 2D distance: \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
• 3D distance: \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\)

• 2D midpoint: \(\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\)
• 3D midpoint: \(\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\right)\)

The pattern is simple: in 3D, you do the same thing as in 2D, but include the z-coordinate as well.

8. Common mistakes to avoid

• Mixing up coordinates: Always keep the order \((x,y,z)\).
• Forgetting negative signs: Be careful when subtracting coordinates, especially if one value is negative.
• Using the midpoint formula incorrectly: Add corresponding coordinates first, then divide each sum by 2.
• Leaving out the z-part in the distance formula: In 3D, all three coordinate differences matter.

9. Quick check for understanding

Try these on your own:

1. Which coordinate plane contains the point \((5,-1,0)\)?
2. Find the distance between \((2,1,3)\) and \((2,1,9)\).
3. Find the midpoint of \((0,4,-2)\) and \((8,-2,6)\).

Answers:

1. The xy-plane, because \(z=0\).
2. $$d=\sqrt{(2-2)^2+(1-1)^2+(9-3)^2}=\sqrt{36}=6$$
3. $$M\left(\frac{0+8}{2},\frac{4+(-2)}{2},\frac{-2+6}{2}\right)=(4,1,2)$$

Summary

A three-dimensional coordinate system uses the coordinates \((x,y,z)\) to locate points in space. The three axes and the three coordinate planes help us visualize where points are placed.

To find the straight-line distance between two points in space, use

$$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$$

To find the midpoint of a segment in space, use

$$M\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\right)$$

Once you are comfortable reading points, finding distances, and finding midpoints in 3D, you have a strong foundation for studying vectors, geometry in space, and real-world motion problems.

The Cross Product

In three-dimensional space, vectors can describe movement, force, direction, and position. Sometimes we do not just want to compare two vectors, but also create a new vector from them. The cross product does exactly that.

The cross product takes two vectors in 3D and produces a new vector that is perpendicular to both of them. This makes it very useful in geometry, physics, and engineering.

It is also closely connected to area. The magnitude of the cross product gives the area of the parallelogram formed by the two vectors.

1. What is the cross product?

If \(\vec{a}\) and \(\vec{b}\) are vectors in 3D, then their cross product is written as \(\vec{a} \times \vec{b}\).

The result is another vector, not a number. This is different from the dot product, which gives a scalar.

If the angle between \(\vec{a}\) and \(\vec{b}\) is \(\theta\), then:

$$ |\vec{a} \times \vec{b}| = |\vec{a}|\,|\vec{b}|\sin\theta $$

This tells us two important things:

• The size of the cross product depends on the lengths of the two vectors.
• It also depends on the sine of the angle between them.

2. Geometric meaning

The vector \(\vec{a} \times \vec{b}\) points in a direction that is orthogonal, or perpendicular, to both \(\vec{a}\) and \(\vec{b}\).

Its magnitude equals the area of the parallelogram formed by the two vectors:

$$ \text{Area of parallelogram} = |\vec{a} \times \vec{b}| $$

Since a triangle is half of a parallelogram, the area of the triangle formed by the same two vectors is:

$$ \text{Area of triangle} = \frac{1}{2}|\vec{a} \times \vec{b}| $$

This means the cross product is a powerful geometric tool for finding area in 3D space.

3. When is the cross product zero?

From the formula

$$ |\vec{a} \times \vec{b}| = |\vec{a}|\,|\vec{b}|\sin\theta $$

we can see that the cross product is zero when:

• one of the vectors is the zero vector, or
• \(\sin\theta = 0\), which happens when \(\theta = 0^\circ\) or \(180^\circ\).

So if two non-zero vectors are parallel or anti-parallel, their cross product is the zero vector.

4. Formula in component form

Suppose

$$ \vec{a} = \langle a_1, a_2, a_3 \rangle \quad \text{and} \quad \vec{b} = \langle b_1, b_2, b_3 \rangle $$

Then the cross product is:

$$ \vec{a} \times \vec{b} = \left\langle a_2b_3 - a_3b_2,\; a_3b_1 - a_1b_3,\; a_1b_2 - a_2b_1 \right\rangle $$

This is the formula you will often use in calculations.

You may also see it written using a determinant-style layout:

$$ \vec{a} \times \vec{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} $$

Expanding gives:

$$ \vec{a} \times \vec{b} = \mathbf{i}(a_2b_3-a_3b_2) - \mathbf{j}(a_1b_3-a_3b_1) + \mathbf{k}(a_1b_2-a_2b_1) $$

Be careful with the middle term: it has a minus sign.

5. Direction: the right-hand rule

There are two directions perpendicular to a plane, so how do we know which direction the cross product points?

We use the right-hand rule:

• Point the fingers of your right hand in the direction of \(\vec{a}\).
• Curve them toward \(\vec{b}\).
• Your thumb points in the direction of \(\vec{a} \times \vec{b}\).

This direction matters because:

$$ \vec{a} \times \vec{b} = -(\vec{b} \times \vec{a}) $$

So reversing the order changes the sign of the result.

6. Important properties

• Anti-commutative: \(\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})\)
• Zero with itself: \(\vec{a} \times \vec{a} = \vec{0}\)
• Distributive: \(\vec{a} \times (\vec{b}+\vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}\)
• Scalar multiplication: \((k\vec{a}) \times \vec{b} = k(\vec{a} \times \vec{b})\)

The cross product is only defined in the standard way for 3D vectors at this level.

7. Standard unit vector cross products

The standard basis vectors satisfy:

$$ \mathbf{i} \times \mathbf{j} = \mathbf{k}, \quad \mathbf{j} \times \mathbf{k} = \mathbf{i}, \quad \mathbf{k} \times \mathbf{i} = \mathbf{j} $$

Reversing the order changes the sign:

$$ \mathbf{j} \times \mathbf{i} = -\mathbf{k}, \quad \mathbf{k} \times \mathbf{j} = -\mathbf{i}, \quad \mathbf{i} \times \mathbf{k} = -\mathbf{j} $$

Also,

$$ \mathbf{i} \times \mathbf{i} = \mathbf{j} \times \mathbf{j} = \mathbf{k} \times \mathbf{k} = \vec{0} $$

These facts can help you check answers quickly.

8. Worked Example 1: Find a cross product

Find \(\vec{a} \times \vec{b}\) for

$$ \vec{a} = \langle 1,2,3 \rangle, \quad \vec{b} = \langle 4,5,6 \rangle $$

Step 1: Use the component formula

$$ \vec{a} \times \vec{b} = \langle a_2b_3-a_3b_2,\; a_3b_1-a_1b_3,\; a_1b_2-a_2b_1 \rangle $$

Substitute the values:

$$ \vec{a} \times \vec{b} = \langle 2(6)-3(5),\; 3(4)-1(6),\; 1(5)-2(4) \rangle $$ $$ = \langle 12-15,\; 12-6,\; 5-8 \rangle $$ $$ = \langle -3,6,-3 \rangle $$

Answer:

$$ \vec{a} \times \vec{b} = \langle -3,6,-3 \rangle $$

This new vector is perpendicular to both \(\vec{a}\) and \(\vec{b}\).

9. Worked Example 2: Show the result is perpendicular

Using the result from Example 1, check that \(\langle -3,6,-3 \rangle\) is perpendicular to both original vectors.

To check perpendicular vectors, use the dot product. If the dot product is zero, the vectors are perpendicular.

First with \(\vec{a} = \langle 1,2,3 \rangle\):

$$ \langle -3,6,-3 \rangle \cdot \langle 1,2,3 \rangle = (-3)(1) + (6)(2) + (-3)(3) $$ $$ = -3 + 12 - 9 = 0 $$

Now with \(\vec{b} = \langle 4,5,6 \rangle\):

$$ \langle -3,6,-3 \rangle \cdot \langle 4,5,6 \rangle = (-3)(4) + (6)(5) + (-3)(6) $$ $$ = -12 + 30 - 18 = 0 $$

Since both dot products are zero, the cross product is perpendicular to both original vectors.

10. Worked Example 3: Area of a parallelogram

Find the area of the parallelogram formed by

$$ \vec{u} = \langle 2,-1,1 \rangle, \quad \vec{v} = \langle 1,3,2 \rangle $$

Step 1: Find the cross product

$$ \vec{u} \times \vec{v} = \langle (-1)(2) - (1)(3),\; (1)(1) - (2)(2),\; (2)(3) - (-1)(1) \rangle $$ $$ = \langle -2-3,\; 1-4,\; 6+1 \rangle $$ $$ = \langle -5,-3,7 \rangle $$

Step 2: Find its magnitude

$$ |\vec{u} \times \vec{v}| = \sqrt{(-5)^2 + (-3)^2 + 7^2} $$ $$ = \sqrt{25+9+49} = \sqrt{83} $$

Answer: The area of the parallelogram is

$$ \sqrt{83} $$

square units.

11. Worked Example 4: Area of a triangle and order of vectors

Let

$$ \vec{p} = \langle 0,2,1 \rangle, \quad \vec{q} = \langle 3,1,-1 \rangle $$

Find:

1. \(\vec{p} \times \vec{q}\)
2. the area of the triangle formed by these vectors
3. \(\vec{q} \times \vec{p}\)

Step 1: Compute \(\vec{p} \times \vec{q}\)

$$ \vec{p} \times \vec{q} = \langle 2(-1)-1(1),\; 1(3)-0(-1),\; 0(1)-2(3) \rangle $$ $$ = \langle -2-1,\; 3-0,\; 0-6 \rangle $$ $$ = \langle -3,3,-6 \rangle $$

Step 2: Find the area of the triangle

$$ |\vec{p} \times \vec{q}| = \sqrt{(-3)^2 + 3^2 + (-6)^2} $$ $$ = \sqrt{9+9+36} = \sqrt{54} = 3\sqrt{6} $$

So the area of the parallelogram is \(3\sqrt{6}\). The area of the triangle is half of that:

$$ \text{Area of triangle} = \frac{1}{2}(3\sqrt{6}) = \frac{3\sqrt{6}}{2} $$

Step 3: Reverse the order

$$ \vec{q} \times \vec{p} = -(\vec{p} \times \vec{q}) = \langle 3,-3,6 \rangle $$

Notice that reversing the order changes the direction, but not the magnitude. That means the area stays the same.

12. Common mistakes to avoid

• Mixing up dot product and cross product: dot product gives a number; cross product gives a vector.
• Using the wrong order: \(\vec{a} \times \vec{b}\) is not the same as \(\vec{b} \times \vec{a}\).
• Sign error in the middle component: when using the determinant method, the middle term must be handled carefully.
• Forgetting magnitude for area: area is \(|\vec{a} \times \vec{b}|\), not the vector itself.
• For triangle area: divide by 2.

13. How to approach cross product questions

When solving a problem, this step-by-step method helps:

1. Write the two vectors clearly in component form.
2. Use the cross product formula carefully.
3. Simplify to get the resulting vector.
4. If the question asks for area, find the magnitude.
5. If it asks for a triangle, divide by 2.
6. If needed, check perpendicularity using the dot product.

14. Why the cross product matters

The cross product helps us understand shape and direction in 3D space. It gives a normal vector to a plane, helps find areas, and is used in physical situations involving turning effects and forces.

At this level, the most important ideas are:

• the cross product creates a vector perpendicular to two given vectors,
• its magnitude gives the area of a parallelogram, and
• the order of the vectors affects the direction of the result.

Summary

The cross product of two 3D vectors, \(\vec{a} \times \vec{b}\), gives a new vector perpendicular to both \(\vec{a}\) and \(\vec{b}\). Its magnitude is

$$ |\vec{a} \times \vec{b}| = |\vec{a}|\,|\vec{b}|\sin\theta $$

which also equals the area of the parallelogram formed by the vectors. To find it from components, use

$$ \vec{a} \times \vec{b} = \langle a_2b_3-a_3b_2,\; a_3b_1-a_1b_3,\; a_1b_2-a_2b_1 \rangle $$

Remember that reversing the order changes the sign:

$$ \vec{a} \times \vec{b} = -(\vec{b} \times \vec{a}) $$

and that triangle area is half the magnitude of the cross product.

Lines in 3D Space help us describe paths, directions, and motion when points are located using three coordinates: \(x\), \(y\), and \(z\).

In two dimensions, a line can often be described using slope. In three dimensions, slope alone is not enough, because a line can tilt in many different ways. Instead, we describe a line using a point on the line and a direction vector.

This lesson will show you how to write the vector equation and parametric equations of a line in 3D space, and how to move between these forms.

1. The key idea: point + direction

A line in 3D is determined if we know:

• a fixed point on the line, and
• a vector parallel to the line, called a direction vector.

Suppose a line passes through the point

$$A(x_0,y_0,z_0)$$

and has direction vector

$$\vec d = \langle a,b,c \rangle.$$

Then every point \(P(x,y,z)\) on the line can be reached by starting at \(A\) and moving some amount along the direction vector.

That idea gives the vector equation of the line:

$$\vec r = \langle x_0,y_0,z_0 \rangle + t\langle a,b,c \rangle,$$

where \(t\) is a real number called the parameter.

The parameter \(t\) tells us how far and in which direction we move along the line:

• If \(t=0\), we are at the given point.
• If \(t>0\), we move in the direction of \(\vec d\).
• If \(t<0\), we move in the opposite direction.

2. From vector form to parametric form

If

$$\vec r = \langle x_0,y_0,z_0 \rangle + t\langle a,b,c \rangle,$$

then matching components gives the parametric equations:

$$x = x_0 + at,$$

$$y = y_0 + bt,$$

$$z = z_0 + ct.$$

These equations describe how each coordinate changes as \(t\) changes.

3. Why parametric equations are useful

Parametric equations are useful because they let us:

• generate points on the line,
• check whether a point lies on the line,
• describe motion in space,
• connect geometry with vectors.

For example, if \(t=2\), then we can substitute into the equations for \(x\), \(y\), and \(z\) to get one specific point on the line.

4. Understanding the direction vector

The direction vector tells us the line's orientation in space. If the direction vector is

$$\langle a,b,c \rangle,$$

then for every increase of 1 in \(t\):

• \(x\) changes by \(a\),
• \(y\) changes by \(b\),
• \(z\) changes by \(c\).

Any nonzero scalar multiple of a direction vector gives the same line. For example, \(\langle 2,4,-6\rangle\) and \(\langle 1,2,-3\rangle\) point in the same direction, so either can be used.

5. Writing the line through two points

If you are given two points instead of a point and a direction vector, you can find the direction vector by subtracting coordinates.

Suppose the line passes through

$$A(x_1,y_1,z_1) \quad \text{and} \quad B(x_2,y_2,z_2).$$

Then a direction vector is

$$\overrightarrow{AB} = \langle x_2-x_1,\; y_2-y_1,\; z_2-z_1 \rangle.$$

So the vector equation of the line can be written as

$$\vec r = \langle x_1,y_1,z_1 \rangle + t\langle x_2-x_1,\; y_2-y_1,\; z_2-z_1 \rangle.$$

6. Symmetric form

Sometimes a line is also written in symmetric form. Starting from

$$x=x_0+at, \quad y=y_0+bt, \quad z=z_0+ct,$$

we solve each for \(t\):

$$t=\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}.$$

So the symmetric form is

$$\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}.$$

This form is useful, but be careful: it only works directly when \(a\), \(b\), and \(c\) are all nonzero. If one component is 0, then that coordinate is constant instead.

For example, if the direction vector is \(\langle 2,0,-1\rangle\), then:

• \(x=x_0+2t\)
• \(y=y_0\)
• \(z=z_0-t\)

So \(y\) stays fixed.

7. Worked Example 1: Write vector and parametric equations from a point and direction

Find the vector and parametric equations of the line through the point \((1,-2,3)\) with direction vector \(\langle 4,1,-2\rangle\).

Step 1: Use the standard form

Point: \((1,-2,3)\)

Direction vector: \(\langle 4,1,-2\rangle\)

So the vector equation is

$$\vec r = \langle 1,-2,3 \rangle + t\langle 4,1,-2 \rangle.$$

Step 2: Write the parametric equations

Match components:

$$x=1+4t,$$

$$y=-2+t,$$

$$z=3-2t.$$

Answer:

Vector equation:

$$\vec r = \langle 1,-2,3 \rangle + t\langle 4,1,-2 \rangle$$

Parametric equations:

$$x=1+4t, \quad y=-2+t, \quad z=3-2t$$

8. Worked Example 2: Find the equation of a line through two points

Find the equation of the line through \(A(2,1,-1)\) and \(B(5,7,3)\).

Step 1: Find a direction vector

$$\overrightarrow{AB} = \langle 5-2,\; 7-1,\; 3-(-1) \rangle = \langle 3,6,4 \rangle.$$

Step 2: Use point + direction

Using point \(A(2,1,-1)\), the vector equation is

$$\vec r = \langle 2,1,-1 \rangle + t\langle 3,6,4 \rangle.$$

Step 3: Write parametric equations

$$x=2+3t,$$

$$y=1+6t,$$

$$z=-1+4t.$$

Check: If \(t=1\), we get

$$x=5, \quad y=7, \quad z=3,$$

which is point \(B\). So the equation is correct.

9. Worked Example 3: Check whether a point lies on a line

Does the point \((7,0,1)\) lie on the line

$$x=1+2t, \quad y=3-t, \quad z=-1+t?$$

Step 1: Use one equation to find \(t\)

From \(x=1+2t\):

$$7=1+2t$$

$$2t=6$$

$$t=3.$$

Step 2: Check the other coordinates with \(t=3\)

For \(y\):

$$y=3-3=0$$

For \(z\):

$$z=-1+3=2$$

The given point has \(z=1\), but the line gives \(z=2\).

Conclusion: The point \((7,0,1)\) does not lie on the line.

10. Worked Example 4: Write symmetric form

Write the symmetric form of the line

$$\vec r = \langle -2,4,1 \rangle + t\langle 3,-5,2 \rangle.$$

Step 1: Write parametric equations

$$x=-2+3t,$$

$$y=4-5t,$$

$$z=1+2t.$$

Step 2: Solve each for \(t\)

$$t=\frac{x+2}{3}, \quad t=\frac{4-y}{5}, \quad t=\frac{z-1}{2}.$$

It is common to keep the middle expression in denominator form matching the direction component:

$$\frac{x+2}{3}=\frac{y-4}{-5}=\frac{z-1}{2}.$$

Answer:

$$\frac{x+2}{3}=\frac{y-4}{-5}=\frac{z-1}{2}.$$

11. Common mistakes to avoid

• Using a point as the direction vector. A point tells where the line is; a direction vector tells where it goes.
• Subtracting points in the wrong way carelessly. If you use \(\overrightarrow{BA}\) instead of \(\overrightarrow{AB}\), that is fine, but the direction vector will point the opposite way. It still gives the same line.
• Forgetting the parameter. A line in 3D needs a variable like \(t\) to represent all points on the line.
• Checking only one coordinate when testing a point. A point lies on the line only if the same value of \(t\) works in all three equations.
• Using symmetric form when a direction component is zero. In that case, keep that coordinate as a constant.

12. Quick method summary

If you know a point and a direction vector:

1. Write the point as \(\langle x_0,y_0,z_0\rangle\).
2. Write the direction vector as \(\langle a,b,c\rangle\).
3. Use

$$\vec r=\langle x_0,y_0,z_0\rangle + t\langle a,b,c\rangle.$$

If you know two points:

1. Subtract to get a direction vector.
2. Use one of the points with that direction vector.

13. Final summary

A line in 3D space is best described using a point on the line and a direction vector parallel to the line. The vector equation is

$$\vec r = \langle x_0,y_0,z_0 \rangle + t\langle a,b,c \rangle,$$

and the matching parametric equations are

$$x=x_0+at, \quad y=y_0+bt, \quad z=z_0+ct.$$

When two points are given, subtract them to find a direction vector. Once you understand that a line is created by starting at one point and moving along a direction, writing equations of lines in 3D becomes much more manageable.

Planes in 3D Space

In three-dimensional geometry, a plane is a flat surface that extends forever in all directions within 3D space. You can think of it like an infinitely large sheet of paper. In coordinate geometry, we describe a plane using an equation involving the variables \(x\), \(y\), and \(z\).

One of the most important ways to define a plane is by using:

• a point on the plane, and
• a normal vector, which is a vector perpendicular to the plane.

This lesson explains how to construct the scalar equation of a plane from that information, how to recognize its meaning, and how to use it in problems.

1. The key idea: the normal vector

A normal vector tells us the orientation of the plane. If a vector is perpendicular to the plane, then every direction that lies along the plane must be perpendicular to that normal vector.

Suppose a plane contains the point \(P_0(x_0,y_0,z_0)\) and has normal vector

$$\vec{n}=\langle a,b,c \rangle.$$

Now take any other point \(P(x,y,z)\) on the plane. Then the vector from \(P_0\) to \(P\) is

$$\overrightarrow{P_0P}=\langle x-x_0,\; y-y_0,\; z-z_0 \rangle.$$

Because this vector lies in the plane, it must be perpendicular to the normal vector. Two perpendicular vectors have dot product zero, so:

$$\vec{n}\cdot \overrightarrow{P_0P}=0.$$

Substituting the components gives:

$$\langle a,b,c\rangle \cdot \langle x-x_0, y-y_0, z-z_0\rangle = 0.$$

So the equation of the plane is:

$$a(x-x_0)+b(y-y_0)+c(z-z_0)=0.$$

This is called the point-normal form of the equation of a plane.

2. Scalar equation of a plane

If we expand the point-normal form, we get a scalar equation in the form

$$ax+by+cz+d=0,$$

where \(a\), \(b\), and \(c\) come from the normal vector, and \(d\) is a constant.

This means:

• The vector \(\langle a,b,c\rangle\) is normal to the plane.
• Any point \((x,y,z)\) on the plane must satisfy the equation.

So if you know a plane equation such as

$$2x-3y+z-7=0,$$

then the normal vector is

$$\langle 2,-3,1\rangle.$$

3. How to construct the equation of a plane

To find the scalar equation of a plane from a point and a normal vector, follow these steps:

1. Write down the point \((x_0,y_0,z_0)\).
2. Write down the normal vector \(\langle a,b,c\rangle\).
3. Substitute into

$$a(x-x_0)+b(y-y_0)+c(z-z_0)=0.$$

4. Expand and simplify.

4. Why this works

The equation works because every point on the plane creates a vector from the known point in a direction that stays inside the plane. Since the normal vector is perpendicular to all directions in the plane, their dot product must be zero.

This connects plane equations directly to vector ideas from 3D geometry.

Worked Example 1: Find a plane from a point and a normal vector

Find the equation of the plane passing through \((1,2,3)\) with normal vector \(\langle 2,-1,4\rangle\).

Step 1: Use point-normal form

$$2(x-1)-1(y-2)+4(z-3)=0.$$

Step 2: Expand

$$2x-2-y+2+4z-12=0.$$

Step 3: Simplify

$$2x-y+4z-12=0.$$

So the scalar equation of the plane is

$$\boxed{2x-y+4z-12=0}.$$

Check: Substitute the point \((1,2,3)\):

$$2(1)-2+4(3)-12=2-2+12-12=0.$$

The point satisfies the equation, so the result is correct.

Worked Example 2: Write the plane equation and identify the normal vector

A plane passes through \((4,-1,2)\) and has normal vector \(\langle 3,5,-2\rangle\). Find its scalar equation.

Start with

$$3(x-4)+5(y+1)-2(z-2)=0.$$

Expand:

$$3x-12+5y+5-2z+4=0.$$

Simplify:

$$3x+5y-2z-3=0.$$

Therefore, the plane equation is

$$\boxed{3x+5y-2z-3=0}.$$

The normal vector is still

$$\boxed{\langle 3,5,-2\rangle}.$$

Worked Example 3: Determine whether a point lies on a plane

Does the point \((2,1,3)\) lie on the plane

$$x-2y+z-3=0?$$

Substitute \(x=2\), \(y=1\), and \(z=3\):

$$2-2(1)+3-3=2-2+3-3=0.$$

Because the result is 0, the point lies on the plane.

Now test the point \((1,0,1)\):

$$1-2(0)+1-3=1+1-3=-1.$$

Because the result is not 0, this point does not lie on the plane.

Worked Example 4: Find the equation from a point and simplify carefully

Find the plane through \((-2,3,1)\) with normal vector \(\langle -1,2,5\rangle\).

Use point-normal form:

$$-1(x+2)+2(y-3)+5(z-1)=0.$$

Expand:

$$-x-2+2y-6+5z-5=0.$$

Simplify:

$$-x+2y+5z-13=0.$$

You may leave it like this, or multiply every term by \(-1\):

$$x-2y-5z+13=0.$$

Both equations represent the same plane. Multiplying an equation by a nonzero constant does not change the set of points that satisfy it.

5. Important observations

• If two plane equations are multiples of each other, they describe the same plane.
• The coefficients of \(x\), \(y\), and \(z\) form a normal vector.
• To check whether a point lies on a plane, substitute its coordinates into the equation.
• A plane in scalar form is usually written as \(ax+by+cz+d=0\).

6. Common mistakes to avoid

• Using the point incorrectly: In \(x-x_0\), \(y-y_0\), and \(z-z_0\), be careful with negative coordinates. For example, if \(x_0=-2\), then \(x-x_0=x-(-2)=x+2\).
• Mixing up the normal vector and a point: The normal vector gives the coefficients \(a\), \(b\), and \(c\). The point gives \(x_0\), \(y_0\), and \(z_0\).
• Sign errors when expanding: Always distribute carefully.
• Forgetting the equation equals 0: A plane equation is written so that all terms are on one side.

7. Quick method to remember

If you know:

• point \((x_0,y_0,z_0)\)
• normal vector \(\langle a,b,c\rangle\)

then write:

$$\boxed{a(x-x_0)+b(y-y_0)+c(z-z_0)=0}.$$

Then expand to get:

$$\boxed{ax+by+cz+d=0}. $$

8. Brief summary

A plane in 3D can be defined by a point on the plane and a normal vector perpendicular to it. Using the dot product idea, the point-normal form is

$$a(x-x_0)+b(y-y_0)+c(z-z_0)=0.$$

After expanding, we get the scalar form

$$ax+by+cz+d=0.$$

The coefficients \(a\), \(b\), and \(c\) give a normal vector, and any point on the plane must satisfy the equation.

Intersections and Angles in 3D is about describing how lines and planes meet in space, and how to measure the angles between them. In three-dimensional geometry, objects can intersect in different ways: two lines may cross, two planes may meet in a line, and a line may meet a plane at a point. Understanding these situations is important in vector geometry, physics, engineering, and any problem involving motion or forces in space.

In this lesson, you will learn how to:

• decide whether lines and planes intersect,
• find intersection points or lines of intersection,
• calculate the angle between two lines,
• calculate the angle between two planes,
• calculate the angle between a line and a plane.

The key tools are vector equations, Cartesian equations, the dot product, and the idea of a normal vector to a plane.

1. Lines in 3D

A line in 3D is often written in vector form as

$$\mathbf{r} = \mathbf{a} + t\mathbf{d}$$

where:

• \(\mathbf{a}\) is the position vector of a point on the line,
• \(\mathbf{d}\) is the direction vector of the line,
• \(t\) is a real parameter.

For example,

$$\mathbf{r} = \begin{pmatrix}1\\2\\3\end{pmatrix} + t\begin{pmatrix}2\\-1\\4\end{pmatrix}$$

represents a line through \((1,2,3)\) with direction vector \((2,-1,4)\).

2. Planes in 3D

A plane is often written in Cartesian form as

$$ax + by + cz = d$$

The vector

$$\mathbf{n} = \begin{pmatrix}a\\b\\c\end{pmatrix}$$

is called a normal vector because it is perpendicular to the plane.

For example, the plane

$$2x - y + 3z = 7$$

has normal vector

$$\mathbf{n} = \begin{pmatrix}2\\-1\\3\end{pmatrix}.$$

Normal vectors are very important because angles involving planes are usually found using their normal vectors.

3. Intersection of two lines

Two lines intersect if they share a common point. To find out whether they intersect, write both lines in parametric form and set their coordinates equal.

If line 1 is

$$\mathbf{r} = \mathbf{a} + t\mathbf{d}_1$$

and line 2 is

$$\mathbf{r} = \mathbf{b} + s\mathbf{d}_2,$$

then compare the \(x\)-, \(y\)-, and \(z\)-coordinates. Solve for \(t\) and \(s\).

There are three possible outcomes:

• one solution: the lines intersect,
• no solution: the lines do not meet,
• infinitely many solutions: the lines are the same line.

In 3D, two non-parallel lines may still fail to intersect. Such lines are called skew lines.

4. Angle between two lines

The angle between two lines is found using their direction vectors. If the direction vectors are \(\mathbf{d}_1\) and \(\mathbf{d}_2\), then

$$\cos \theta = \frac{\mathbf{d}_1 \cdot \mathbf{d}_2}{|\mathbf{d}_1||\mathbf{d}_2|}$$

where \(\theta\) is the angle between the vectors.

When talking about the angle between lines, we usually take the acute angle. So if the calculation gives an obtuse angle, use the smaller angle instead.

5. Intersection of two planes

Two planes can:

• intersect in a line,
• be parallel and separate,
• be the same plane.

If the normal vectors are not parallel, the planes intersect in a line.

To find the line of intersection:

1. Write the equations of both planes.
2. Solve the two equations simultaneously.
3. Let one variable be a parameter.
4. Express the result as a vector equation of a line.

6. Angle between two planes

The angle between two planes is the angle between their normal vectors.

If the planes have normal vectors \(\mathbf{n}_1\) and \(\mathbf{n}_2\), then

$$\cos \theta = \frac{\mathbf{n}_1 \cdot \mathbf{n}_2}{|\mathbf{n}_1||\mathbf{n}_2|}$$

Again, the angle between planes is usually taken to be the acute angle.

7. Intersection of a line and a plane

A line intersects a plane if some point on the line satisfies the plane equation.

To find the intersection point:

1. Write the line in parametric form.
2. Substitute the expressions for \(x\), \(y\), and \(z\) into the plane equation.
3. Solve for the parameter.
4. Substitute back into the line equation.

If no value of the parameter works, then the line is parallel to the plane. If every value works, then the line lies in the plane.

8. Angle between a line and a plane

The angle between a line and a plane is not found directly from the plane equation. Instead, first use the angle between the line's direction vector and the plane's normal vector.

If \(\alpha\) is the angle between the line and the plane, and \(\theta\) is the angle between the line's direction vector \(\mathbf{d}\) and the plane's normal vector \(\mathbf{n}\), then

$$\alpha = 90^\circ - \theta$$

or equivalently

$$\sin \alpha = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}.$$

This works because the normal vector is perpendicular to the plane.

Worked Example 1: Find whether two lines intersect

Consider the lines

$$L_1: \mathbf{r} = \begin{pmatrix}1\\2\\3\end{pmatrix} + t\begin{pmatrix}1\\-1\\2\end{pmatrix}$$

and

$$L_2: \mathbf{r} = \begin{pmatrix}3\\0\\7\end{pmatrix} + s\begin{pmatrix}-1\\1\\-2\end{pmatrix}.$$

Write each line in coordinates:

For \(L_1\):

$$x = 1+t, \quad y = 2-t, \quad z = 3+2t$$

For \(L_2\):

$$x = 3-s, \quad y = s, \quad z = 7-2s$$

Set coordinates equal:

$$1+t = 3-s$$

$$2-t = s$$

$$3+2t = 7-2s$$

From \(2-t=s\), substitute into the first equation:

$$1+t = 3-(2-t)=1+t$$

This is always true.

Now check the third equation:

$$3+2t = 7-2(2-t)=7-4+2t=3+2t$$

This is also always true.

So the system has infinitely many solutions. The two lines are actually the same line.

Worked Example 2: Find the angle between two lines

Find the angle between the lines with direction vectors

$$\mathbf{d}_1 = \begin{pmatrix}2\\1\\2\end{pmatrix}, \quad \mathbf{d}_2 = \begin{pmatrix}1\\-2\\2\end{pmatrix}.$$

Use the dot product formula:

$$\cos \theta = \frac{\mathbf{d}_1 \cdot \mathbf{d}_2}{|\mathbf{d}_1||\mathbf{d}_2|}$$

First find the dot product:

$$\mathbf{d}_1 \cdot \mathbf{d}_2 = (2)(1) + (1)(-2) + (2)(2) = 2-2+4 = 4$$

Now find the magnitudes:

$$|\mathbf{d}_1| = \sqrt{2^2+1^2+2^2} = \sqrt{9} = 3$$

$$|\mathbf{d}_2| = \sqrt{1^2+(-2)^2+2^2} = \sqrt{9} = 3$$

So

$$\cos \theta = \frac{4}{3\cdot 3} = \frac{4}{9}$$

Therefore,

$$\theta = \cos^{-1}\left(\frac{4}{9}\right)$$

$$\theta \approx 63.6^\circ$$

So the angle between the lines is approximately \(63.6^\circ\).

Worked Example 3: Find the line of intersection of two planes

Find the line where the planes

$$x+y+z=4$$

and

$$x-y+2z=5$$

intersect.

We solve the equations together. Let \(z = t\).

Then the equations become

$$x+y = 4-t$$

$$x-y = 5-2t$$

Add the equations:

$$2x = 9-3t$$

$$x = \frac{9-3t}{2}$$

Subtract the second from the first:

$$2y = (4-t) - (5-2t) = -1+t$$

$$y = \frac{-1+t}{2}$$

Since \(z=t\), the line of intersection is

$$\mathbf{r} = \begin{pmatrix}\frac{9}{2}\\-\frac{1}{2}\\0\end{pmatrix} + t\begin{pmatrix}-\frac{3}{2}\\\frac{1}{2}\\1\end{pmatrix}$$

We can also multiply the direction vector by 2 to make it simpler:

$$\mathbf{r} = \begin{pmatrix}\frac{9}{2}\\-\frac{1}{2}\\0\end{pmatrix} + \lambda\begin{pmatrix}-3\\1\\2\end{pmatrix}$$

for parameter \(\lambda\).

This is the line where the two planes meet.

Worked Example 4: Find the angle between a line and a plane

Find the angle between the line

$$\mathbf{r} = \begin{pmatrix}1\\0\\2\end{pmatrix} + t\begin{pmatrix}2\\1\\2\end{pmatrix}$$

and the plane

$$x - 2y + 2z = 5.$$

The line has direction vector

$$\mathbf{d} = \begin{pmatrix}2\\1\\2\end{pmatrix}$$

and the plane has normal vector

$$\mathbf{n} = \begin{pmatrix}1\\-2\\2\end{pmatrix}.$$

Use

$$\sin \alpha = \frac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}$$

First find the dot product:

$$\mathbf{d}\cdot\mathbf{n} = (2)(1) + (1)(-2) + (2)(2) = 2-2+4 = 4$$

Find the magnitudes:

$$|\mathbf{d}| = \sqrt{2^2+1^2+2^2} = 3$$

$$|\mathbf{n}| = \sqrt{1^2+(-2)^2+2^2} = 3$$

So

$$\sin \alpha = \frac{4}{9}$$

Therefore,

$$\alpha = \sin^{-1}\left(\frac{4}{9}\right)$$

$$\alpha \approx 26.4^\circ$$

So the angle between the line and the plane is approximately \(26.4^\circ\).

Important ideas to remember

• The angle between lines comes from their direction vectors.
• The angle between planes comes from their normal vectors.
• The angle between a line and a plane is connected to the angle between the line's direction vector and the plane's normal vector.
• To find intersections, write equations clearly and solve them systematically.
• In 3D, lines do not have to be parallel to fail to intersect; they may be skew.

Common mistakes

• Using a point on the line instead of the direction vector when finding angles.
• Using the plane equation directly for angles instead of the normal vector.
• Forgetting that the angle between a line and a plane is the complement of the angle between the line and the normal.
• Assuming that two non-parallel lines in 3D must intersect.
• Not checking all three coordinate equations when testing whether lines intersect.

Brief Summary

In 3D geometry, intersections are found by solving equations for lines and planes together. Angles are usually found with the dot product: use direction vectors for lines, normal vectors for planes, and a direction vector with a normal vector for a line and a plane. Once you know which vectors represent the geometry, the calculations become much more manageable.