Vector Algebra and 3D Geometry

Lesson: Scalars vs. Geometric Vectors

In maths and physics, we often describe quantities such as length, mass, speed, force, and displacement. Some of these quantities need only a size to be fully described. Others need both a size and a direction. This is the key idea behind scalars and vectors.

Understanding the difference is important because scalars and vectors are handled differently. In particular, vectors can be added and subtracted using geometric methods such as the triangle rule and the parallelogram rule.

1. What is a scalar?

A scalar is a quantity that has magnitude only. Magnitude means size or amount.

Examples of scalar quantities include:

• Mass: \(5\text{ kg}\)
• Temperature: \(22^\circ\text{C}\)
• Time: \(3\text{ s}\)
• Distance: \(12\text{ m}\)
• Speed: \(60\text{ km/h}\)

Notice that each of these tells us how much, but not which way.

For example, if a car travels at \(60\text{ km/h}\), that tells us its speed, but not whether it is moving north, south, east, or west. So speed is a scalar.

2. What is a geometric vector?

A geometric vector is a quantity that has both magnitude and direction.

Examples of vector quantities include:

• Displacement
• Velocity
• Acceleration
• Force

For example, saying “move \(5\text{ m}\) east” gives both a size, \(5\text{ m}\), and a direction, east. That makes it a vector.

Geometrically, a vector is often drawn as a directed line segment, or arrow.

• The length of the arrow represents the magnitude.
• The arrowhead shows the direction.

Vectors are often named using bold letters such as \(\mathbf{a}\), \(\mathbf{v}\), or by two points such as \(\overrightarrow{AB}\).

3. Magnitude and direction

If a vector \(\mathbf{v}\) has magnitude \(7\), we write its magnitude as \(|\mathbf{v}| = 7\).

Two vectors are equal if they have:

• the same magnitude, and
• the same direction.

They do not need to start at the same point. A vector can be shifted to another location without changing the vector, as long as its length and direction stay the same.

4. Scalar vs. vector: important comparisons

• Distance is a scalar because it measures how much ground is covered.
• Displacement is a vector because it measures the change in position and includes direction.

For example, if someone walks \(3\text{ m}\) east and then \(3\text{ m}\) west:

• Total distance traveled is \(6\text{ m}\).
• Total displacement is \(0\text{ m}\), because they end where they started.

Another important pair is:

• Speed: scalar
• Velocity: vector

Speed tells how fast something moves. Velocity tells how fast and in what direction it moves.

5. Representing vectors geometrically

Suppose a vector means “\(4\) units to the right.” We can draw it as an arrow pointing right with length \(4\).

If another vector means “\(3\) units upward,” we draw an arrow pointing up with length \(3\).

These geometric pictures help us combine vectors visually. This is one of the most useful features of vectors.

6. Adding vectors geometrically

When adding vectors, we combine their effects. If one vector says to move one way and another says to move another way, their sum tells the overall result.

There are two common geometric methods for vector addition:

• the triangle rule
• the parallelogram rule

7. Triangle rule for vector addition

To add \(\mathbf{a} + \mathbf{b}\) using the triangle rule:

1. Draw vector \(\mathbf{a}\).
2. Starting at the end of \(\mathbf{a}\), draw vector \(\mathbf{b}\).
3. The vector from the start of \(\mathbf{a}\) to the end of \(\mathbf{b}\) is the sum \(\mathbf{a} + \mathbf{b}\).

This is often called the head-to-tail method.

If you imagine walking according to vector \(\mathbf{a}\), then continuing according to vector \(\mathbf{b}\), the result is your total displacement.

8. Parallelogram rule for vector addition

To add \(\mathbf{a}\) and \(\mathbf{b}\) using the parallelogram rule:

1. Draw both vectors starting from the same point.
2. Complete a parallelogram using copies of the two vectors.
3. The diagonal from the common starting point gives \(\mathbf{a} + \mathbf{b}\).

This method is especially useful when both vectors start at the same point.

9. Subtracting vectors geometrically

Vector subtraction can be understood as adding the opposite vector.

The opposite of vector \(\mathbf{b}\) is written \(-\mathbf{b}\). It has:

• the same magnitude as \(\mathbf{b}\),
• but the opposite direction.

So:

$$ \mathbf{a} - \mathbf{b} = \mathbf{a} + (-\mathbf{b}) $$

To subtract geometrically:

1. Reverse the direction of \(\mathbf{b}\) to get \(-\mathbf{b}\).
2. Add \(-\mathbf{b}\) to \(\mathbf{a}\) using the triangle rule or parallelogram rule.

Another useful geometric idea is this: if vectors \(\mathbf{a}\) and \(\mathbf{b}\) start at the same point, then \(\mathbf{a} - \mathbf{b}\) is the vector from the tip of \(\mathbf{b}\) to the tip of \(\mathbf{a}\).

10. Worked Example 1: Identifying scalars and vectors

Classify each quantity as a scalar or a vector:

• \(8\text{ kg}\)
• \(12\text{ m north}\)
• \(25^\circ\text{C}\)
• \(40\text{ km/h east}\)

Solution

• \(8\text{ kg}\): scalar, because it has magnitude only.
• \(12\text{ m north}\): vector, because it has magnitude and direction.
• \(25^\circ\text{C}\): scalar, because temperature has no direction.
• \(40\text{ km/h east}\): vector, because it gives speed with direction, so it is velocity.

11. Worked Example 2: Vector addition using the triangle rule

A student walks \(4\text{ m}\) east and then \(3\text{ m}\) north. Find the resultant displacement.

Step 1: Draw the first vector

Draw an arrow \(4\text{ m}\) to the east.

Step 2: Draw the second vector from the tip of the first

From the end of that arrow, draw an arrow \(3\text{ m}\) north.

Step 3: Draw the resultant

The resultant vector goes from the starting point to the final point.

This forms a right triangle with side lengths \(4\) and \(3\). The magnitude of the resultant is:

$$ \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 $$

So the resultant displacement has magnitude \(5\text{ m}\).

The direction is north-east. More precisely, it is the direction from the start point to the end point.

Answer: The resultant displacement is \(5\text{ m}\) in a north-east direction.

12. Worked Example 3: Vector addition using the parallelogram rule

Two forces act on an object from the same point:

• \(\mathbf{F_1}\): \(6\text{ N}\) east
• \(\mathbf{F_2}\): \(6\text{ N}\) north

Find the resultant force geometrically.

Solution

Because both vectors start at the same point, the parallelogram rule works well.

1. Draw \(\mathbf{F_1}\) as an arrow \(6\) units east.
2. Draw \(\mathbf{F_2}\) as an arrow \(6\) units north from the same starting point.
3. Complete the parallelogram.
4. The diagonal from the common start point is the resultant.

The diagonal forms a right triangle with sides \(6\) and \(6\), so its magnitude is:

$$ \sqrt{6^2 + 6^2} = \sqrt{72} = 6\sqrt{2} $$

So the resultant force is:

$$ 6\sqrt{2}\text{ N} $$

The direction is halfway between east and north, which is 45^\circ\ north of east.

Answer: The resultant force is \(6\sqrt{2}\text{ N}\) at \(45^\circ\) north of east.

13. Worked Example 4: Vector subtraction

A boat’s displacement is represented by vector \(\mathbf{a}\), \(10\text{ m}\) east. A second vector \(\mathbf{b}\) is \(4\text{ m}\) east. Find \(\mathbf{a} - \mathbf{b}\).

Solution

We use:

$$ \mathbf{a} - \mathbf{b} = \mathbf{a} + (-\mathbf{b}) $$

Since \(\mathbf{b}\) is \(4\text{ m}\) east, the opposite vector \(-\mathbf{b}\) is \(4\text{ m}\) west.

Now add:

• \(10\text{ m}\) east
• \(4\text{ m}\) west

The result is \(6\text{ m}\) east.

Answer:

$$ \mathbf{a} - \mathbf{b} = 6\text{ m east} $$

14. Common mistakes to avoid

• Confusing distance and displacement: distance is scalar, displacement is vector.
• Confusing speed and velocity: speed is scalar, velocity is vector.
• Adding vector magnitudes directly when directions are different. For example, \(4\text{ m east}\) plus \(3\text{ m north}\) is not \(7\text{ m}\); it must be combined geometrically.
• Forgetting that subtraction means adding the opposite vector.
• Ignoring direction. Two vectors with the same magnitude are not equal if their directions are different.

15. Key ideas to remember

• A scalar has magnitude only.
• A vector has magnitude and direction.
• Vectors are represented by arrows.
• Equal vectors have the same magnitude and direction.
• Vector addition can be done using the triangle rule or parallelogram rule.
• Vector subtraction means adding the opposite vector.

Brief Summary

Scalars describe size only, while geometric vectors describe both size and direction. This difference matters because vectors must be combined using geometric rules, not just ordinary addition. The triangle rule and parallelogram rule help us add vectors, and subtraction is done by adding the opposite vector. Once you keep track of both magnitude and direction, vector problems become much easier to understand.

Algebraic Vectors and Components

In many maths and physics problems, a quantity has both a size and a direction. These quantities are called vectors. Examples include displacement, velocity, force, and acceleration.

In this lesson, you will learn how to write vectors algebraically using components, how to use 6, 7 notation, and how to find a vector6s magnitude and direction angle.

This is an important skill because it lets us work with vectors using algebra instead of always drawing diagrams.

1. What is a vector?

A vector has:

• magnitude: how large it is
• direction: which way it points

A scalar quantity only has size. For example, mass and temperature are scalars, but force and velocity are vectors.

When we draw a vector on a coordinate plane, it can be described by how far it moves:

• horizontally
• vertically

These are called the components of the vector.

2. Writing vectors in component form

If a vector moves:

• 6 units to the right, and
• 3 units up,

then its component form is:

$$\begin{pmatrix} 6 \\ 3 \end{pmatrix}$$

or, using algebraic vector notation,

$$6\mathbf{i} + 3\mathbf{j}$$

Here:

• \(\mathbf{i}\) means 1 unit in the horizontal direction
• \(\mathbf{j}\) means 1 unit in the vertical direction

So:

• \(6\mathbf{i}\) means 6 units horizontally
• \(3\mathbf{j}\) means 3 units vertically

If a vector points left or down, the component will be negative.

For example:

• \(-4\mathbf{i}\) means 4 units left
• \(-2\mathbf{j}\) means 2 units down

3. Position vectors

A position vector describes the location of a point relative to the origin.

If point \(A\) has coordinates \((x, y)\), then the position vector of \(A\) is:

$$\vec{OA} = x\mathbf{i} + y\mathbf{j}$$

For example, if \(A = (4, -2)\), then:

$$\vec{OA} = 4\mathbf{i} - 2\mathbf{j}$$

This means start at the origin, move 4 units right and 2 units down.

4. Vectors between two points

To find the vector from point \(A(x_1, y_1)\) to point \(B(x_2, y_2)\), subtract coordinates:

$$\vec{AB} = \begin{pmatrix} x_2 - x_1 \\ y_2 - y_1 \end{pmatrix}$$

In \(\mathbf{i}, \mathbf{j}\) notation, this is:

$$\vec{AB} = (x_2 - x_1)\mathbf{i} + (y_2 - y_1)\mathbf{j}$$

This is very useful when you know the coordinates of two points and want the vector joining them.

Worked Example 1: Writing a vector from components

A vector has horizontal component 5 and vertical component \(-3\). Write it in:

1. column form
2. \(\mathbf{i}, \mathbf{j}\) form

Solution

The vector has:

• 5 units right
• 3 units down

So the column form is:

$$\begin{pmatrix} 5 \\ -3 \end{pmatrix}$$

And the algebraic form is:

$$5\mathbf{i} - 3\mathbf{j}$$

5. Adding and subtracting vectors

Vectors in component form are added and subtracted by working with matching components.

If

$$\mathbf{a} = 2\mathbf{i} + 5\mathbf{j}, \quad \mathbf{b} = 4\mathbf{i} - \mathbf{j}$$

then:

$$\mathbf{a} + \mathbf{b} = (2+4)\mathbf{i} + (5-1)\mathbf{j} = 6\mathbf{i} + 4\mathbf{j}$$

and

$$\mathbf{a} - \mathbf{b} = (2-4)\mathbf{i} + (5-(-1))\mathbf{j} = -2\mathbf{i} + 6\mathbf{j}$$

This works because vectors combine component by component.

6. Magnitude of a vector

The magnitude of a vector is its length or size. For a vector

$$\mathbf{v} = a\mathbf{i} + b\mathbf{j}$$

the magnitude is written as \(|\mathbf{v}|\) and found using Pythagoras' theorem:

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

This is because the horizontal and vertical components form a right triangle.

For example, if

$$\mathbf{v} = 3\mathbf{i} + 4\mathbf{j}$$

then:

$$|\mathbf{v}| = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5$$

Worked Example 2: Finding a vector between two points and its magnitude

Let \(A(2, 1)\) and \(B(7, 5)\).

Find:

1. \(\vec{AB}\)
2. the magnitude of \(\vec{AB}\)

Solution

First subtract coordinates:

$$\vec{AB} = \begin{pmatrix} 7-2 \\ 5-1 \end{pmatrix} = \begin{pmatrix} 5 \\ 4 \end{pmatrix}$$

So in algebraic form:

$$\vec{AB} = 5\mathbf{i} + 4\mathbf{j}$$

Now find the magnitude:

$$|\vec{AB}| = \sqrt{5^2 + 4^2} = \sqrt{25+16} = \sqrt{41}$$

So the answers are:

• \(\vec{AB} = 5\mathbf{i} + 4\mathbf{j}\)
• \(|\vec{AB}| = \sqrt{41}\)

7. Direction angle of a vector

The direction angle tells us which way a vector points. Usually, this angle is measured from the positive \(x\)-axis.

For a vector

$$\mathbf{v} = a\mathbf{i} + b\mathbf{j}$$

we can use trigonometry:

$$\tan \theta = \frac{b}{a}$$

So:

$$\theta = \tan^{-1}\left(\frac{b}{a}\right)$$

This gives the basic angle, but you must also think about the quadrant of the vector.

Quadrants reminder:

• Quadrant I: right and up
• Quadrant II: left and up
• Quadrant III: left and down
• Quadrant IV: right and down

If both components are positive, the vector is in Quadrant I and the calculator result usually works directly.

If not, you may need to adjust the angle so it matches the correct direction.

Worked Example 3: Magnitude and direction angle

Find the magnitude and direction angle of:

$$\mathbf{v} = 6\mathbf{i} + 8\mathbf{j}$$

Solution

Step 1: Find the magnitude

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

Step 2: Find the direction angle

$$\tan \theta = \frac{8}{6} = \frac{4}{3}$$ $$\theta = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.1^\circ$$

Since both components are positive, the vector is in Quadrant I, so this angle is correct.

Answer:

• magnitude = \(10\)
• direction angle \(\approx 53.1^\circ\)

8. Finding components from magnitude and direction

Sometimes you know the magnitude and angle, and you need the components.

If a vector has magnitude \(r\) and direction angle \(\theta\), then:

$$x = r\cos \theta$$ $$y = r\sin \theta$$

So the vector is:

$$\mathbf{v} = (r\cos\theta)\mathbf{i} + (r\sin\theta)\mathbf{j}$$

This comes from right triangle trigonometry.

Worked Example 4: Finding components from magnitude and angle

A vector has magnitude \(12\) and direction angle \(30^\circ\). Write the vector in \(\mathbf{i}, \mathbf{j}\) form.

Solution

Use:

$$x = r\cos\theta, \quad y = r\sin\theta$$

So:

$$x = 12\cos 30^\circ = 12\left(\frac{\sqrt{3}}{2}\right) = 6\sqrt{3}$$ $$y = 12\sin 30^\circ = 12\left(\frac{1}{2}\right) = 6$$

Therefore:

$$\mathbf{v} = 6\sqrt{3}\mathbf{i} + 6\mathbf{j}$$

9. Important signs and interpretation

Always pay attention to whether components are positive or negative.

• positive \(\mathbf{i}\): right
• negative \(\mathbf{i}\): left
• positive \(\mathbf{j}\): up
• negative \(\mathbf{j}\): down

For example:

• \(3\mathbf{i} - 2\mathbf{j}\) means right 3, down 2
• \(-4\mathbf{i} + 5\mathbf{j}\) means left 4, up 5

This helps you check whether your angle and drawing make sense.

10. Common mistakes to avoid

• Mixing up coordinates when finding \(\vec{AB}\)
  Always do end point minus start point.
• Forgetting negative signs
  A downward or leftward component must be negative.
• Using the wrong ratio for angle
  For \(a\mathbf{i} + b\mathbf{j}\), use \(\tan \theta = \frac{b}{a}\).
• Ignoring the quadrant
  The calculator gives a reference angle, but the actual direction must match the vector.
• Confusing magnitude with components
  The magnitude is one number; components are the horizontal and vertical parts.

11. Quick practice questions

1. Write the vector with components \((-2, 7)\) in \(\mathbf{i}, \mathbf{j}\) form.
2. Find \(\vec{PQ}\) if \(P(1, -3)\) and \(Q(5, 2)\).
3. Find the magnitude of \(8\mathbf{i} - 6\mathbf{j}\).
4. Find the direction angle of \(3\mathbf{i} + 3\mathbf{j}\).

Answers

1. \(-2\mathbf{i} + 7\mathbf{j}\)
2. \(4\mathbf{i} + 5\mathbf{j}\)
3. \(\sqrt{8^2 + (-6)^2} = \sqrt{100} = 10\)
4. \(\tan\theta = \frac{3}{3} = 1\), so \(\theta = 45^\circ\)

Summary

Vectors can be written using components in column form or in algebraic form using \(\mathbf{i}\) and \(\mathbf{j}\). The \(\mathbf{i}\) component shows horizontal movement, and the \(\mathbf{j}\) component shows vertical movement.

You can find a vector between two points by subtracting coordinates, find its magnitude using

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

and find its direction angle using

$$\theta = \tan^{-1}\left(\frac{b}{a}\right)$$

when the vector is \(a\mathbf{i} + b\mathbf{j}\), remembering to check the quadrant.

These ideas are the foundation for working with vectors in both maths and physics.

Unit Vectors and Normalization

In vector algebra, a vector can tell us how much and which direction. For example, a velocity vector tells both speed and direction, and a force vector tells both strength and direction.

Sometimes, we only care about the direction of a vector and not its size. That is where unit vectors and normalization become useful.

A unit vector is a vector with length 1. It keeps the direction of the original vector but removes its magnitude. Normalization is the process of turning a non-zero vector into a unit vector.

This idea is very important in maths, physics, and 3D geometry because it lets us describe direction clearly and compare vectors without being affected by their lengths.

1. What is the magnitude of a vector?

Before we can make a unit vector, we need to know how to find the magnitude (or length) of a vector.

If a 2D vector is written as \(\vec{v} = \langle a, b \rangle\), then its magnitude is

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

If a 3D vector is written as \(\vec{v} = \langle a, b, c \rangle\), then its magnitude is

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

This comes from the Pythagorean Theorem.

2. What is a unit vector?

A unit vector is any vector whose magnitude is exactly 1.

For example:

• \(\langle 1, 0 \rangle\) is a unit vector because its length is 1.
• \(\langle 0, 1 \rangle\) is also a unit vector.
• \(\left\langle \frac{3}{5}, \frac{4}{5} \right\rangle\) is a unit vector because its magnitude is 1.

We often use special standard unit vectors in coordinate geometry:

• In 2D: \(\mathbf{i} = \langle 1,0 \rangle\), \(\mathbf{j} = \langle 0,1 \rangle\)
• In 3D: \(\mathbf{i} = \langle 1,0,0 \rangle\), \(\mathbf{j} = \langle 0,1,0 \rangle\), \(\mathbf{k} = \langle 0,0,1 \rangle\)

These represent directions along the coordinate axes.

3. What does it mean to normalize a vector?

To normalize a vector means to divide the vector by its own magnitude.

If \(\vec{v}\) is a non-zero vector, then the unit vector in the direction of \(\vec{v}\) is

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

Here, \(\hat{v}\) means “the unit vector in the direction of \(\vec{v}\).”

This works because dividing by the magnitude scales the vector down or up so that its new length becomes 1, while the direction stays the same.

Important: You can only normalize a non-zero vector. The zero vector \(\langle 0,0 \rangle\) or \(\langle 0,0,0 \rangle\) has magnitude 0, and division by 0 is not possible.

4. Why normalization works

Suppose \(\vec{v}\) has magnitude \(|\vec{v}|\). If we divide every component by \(|\vec{v}|\), we get a new vector:

$$\frac{\vec{v}}{|\vec{v}|}$$

The magnitude of this new vector is

$$\left|\frac{\vec{v}}{|\vec{v}|}\right| = \frac{|\vec{v}|}{|\vec{v}|} = 1$$

So the new vector has length 1, which is exactly what we want.

5. Step-by-step method for normalization

1. Write the vector clearly.
2. Find its magnitude.
3. Divide each component of the vector by the magnitude.
4. Check that the new vector has magnitude 1.

Worked Example 1: Normalize a 2D vector

Find the unit vector in the direction of \(\vec{v} = \langle 3, 4 \rangle\).

Step 1: Find the magnitude.

$$|\vec{v}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$

Step 2: Divide the vector by its magnitude.

$$\hat{v} = \frac{\vec{v}}{|\vec{v}|} = \frac{\langle 3,4 \rangle}{5} = \left\langle \frac{3}{5}, \frac{4}{5} \right\rangle$$

Step 3: Check the magnitude.

$$\left|\hat{v}\right| = \sqrt{\left(\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2} = \sqrt{\frac{9}{25} + \frac{16}{25}} = \sqrt{1} = 1$$

So the unit vector is

$$\boxed{\left\langle \frac{3}{5}, \frac{4}{5} \right\rangle}$$

Worked Example 2: Normalize a vector with a negative component

Find the unit vector in the direction of \(\vec{u} = \langle -2, 6 \rangle\).

Step 1: Find the magnitude.

$$|\vec{u}| = \sqrt{(-2)^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40} = 2\sqrt{10}$$

Step 2: Divide by the magnitude.

$$\hat{u} = \frac{\langle -2,6 \rangle}{2\sqrt{10}} = \left\langle \frac{-2}{2\sqrt{10}}, \frac{6}{2\sqrt{10}} \right\rangle = \left\langle \frac{-1}{\sqrt{10}}, \frac{3}{\sqrt{10}} \right\rangle$$

You may also write this by rationalizing the denominators:

$$\hat{u} = \left\langle \frac{-\sqrt{10}}{10}, \frac{3\sqrt{10}}{10} \right\rangle$$

Both forms are correct.

Worked Example 3: Normalize a 3D vector

Find the unit vector in the direction of \(\vec{w} = \langle 2, -1, 2 \rangle\).

Step 1: Find the magnitude.

$$|\vec{w}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$$

Step 2: Divide by the magnitude.

$$\hat{w} = \frac{\langle 2,-1,2 \rangle}{3} = \left\langle \frac{2}{3}, \frac{-1}{3}, \frac{2}{3} \right\rangle$$

Step 3: Check the magnitude.

$$\left|\hat{w}\right| = \sqrt{\left(\frac{2}{3}\right)^2 + \left(\frac{-1}{3}\right)^2 + \left(\frac{2}{3}\right)^2} = \sqrt{\frac{4}{9} + \frac{1}{9} + \frac{4}{9}} = \sqrt{1} = 1$$

So the unit vector is

$$\boxed{\left\langle \frac{2}{3}, \frac{-1}{3}, \frac{2}{3} \right\rangle}$$

Worked Example 4: Use a unit vector to build a vector with a given magnitude

A force acts in the direction of \(\langle 3,4 \rangle\), but its magnitude is 20. Find the force vector.

Step 1: Find the unit vector in the given direction.

From Example 1, the unit vector in the direction of \(\langle 3,4 \rangle\) is

$$\left\langle \frac{3}{5}, \frac{4}{5} \right\rangle$$

Step 2: Multiply the unit vector by the required magnitude.

$$\vec{F} = 20\left\langle \frac{3}{5}, \frac{4}{5} \right\rangle = \langle 12,16 \rangle$$

So the force vector is

$$\boxed{\langle 12,16 \rangle}$$

This shows why unit vectors are useful: once we know the pure direction, we can scale it to any length we need.

6. Geometric meaning

When you normalize a vector, you do not change its direction. You only change its length to 1.

You can imagine the original vector and its unit vector as arrows pointing exactly the same way, but the unit vector is shorter or longer so that its final length is exactly 1.

This is especially useful in 3D geometry, where we often want a clean direction vector for a line, plane, motion, or force.

7. Common mistakes to avoid

• Forgetting to find the magnitude first. Do not divide by one component; divide by the whole magnitude.
• Using the wrong magnitude formula. In 3D, remember to include all three squared components.
• Changing the direction by mistake. A unit vector must point in the same direction as the original vector.
• Normalizing the zero vector. This is impossible because its magnitude is 0.
• Assuming a vector is already a unit vector. Always check its magnitude if you are unsure.

8. Quick check: Is this a unit vector?

To test whether a vector is a unit vector, find its magnitude. If the magnitude is 1, then it is a unit vector.

For example, is \(\left\langle \frac{1}{2}, \frac{\sqrt{3}}{2} \right\rangle\) a unit vector?

$$\sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1$$

Yes, it is a unit vector.

9. Key ideas to remember

• A vector has both magnitude and direction.
• A unit vector has magnitude 1.
• Normalization means turning a non-zero vector into a unit vector.
• The formula is $$\hat{v} = \frac{\vec{v}}{|\vec{v}|}$$
• Normalization keeps direction the same but changes the length to 1.
• The zero vector cannot be normalized.

Brief Summary

Unit vectors are vectors of length 1 that show direction only. To normalize a non-zero vector, divide it by its magnitude. This process is useful whenever we want pure direction, and it also helps us create vectors of any required magnitude in the same direction.

The Dot Product (Scalar Product)

In vector algebra, we often work with quantities that have both magnitude and direction. These quantities are called vectors. The dot product, also called the scalar product, is a way of combining two vectors to get a single number.

This idea is important because it connects algebra and geometry. Algebraically, we can calculate the dot product using vector components. Geometrically, it tells us about the angle between two vectors.

By the end of this lesson, you should be able to:

• calculate the dot product of two vectors in 2D and 3D,
• use the dot product formula involving the angle between vectors,
• find the angle between two vectors,
• recognize when two vectors are perpendicular.

1. What is the dot product?

Suppose we have two vectors

$$\vec{a} = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix}, \quad \vec{b} = \begin{pmatrix} b_1 \\ b_2 \end{pmatrix}$$

Then their dot product is defined by multiplying matching components and adding the results:

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

In 3D, if

$$\vec{a} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}, \quad \vec{b} = \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}$$

then

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

The answer is a scalar, which means it is a single number, not a vector.

2. Geometric meaning of the dot product

The dot product can also be written using the magnitudes of the vectors and the angle between them:

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

Here:

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

This formula is very important because it shows how the dot product measures how much one vector points in the same direction as another.

Notice what happens for special angles:

• If \(\theta = 0^\circ\), then \(\cos\theta = 1\), so the dot product is positive and as large as possible.
• If \(\theta = 90^\circ\), then \(\cos\theta = 0\), so the dot product is \(0\).
• If \(\theta > 90^\circ\), then \(\cos\theta\) is negative, so the dot product is negative.

3. Magnitude of a vector

To use the angle formula, we need to know how to find the magnitude of a vector.

For a 2D vector

$$\vec{a} = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix}$$

the magnitude is

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

For a 3D vector

$$\vec{a} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}$$

the magnitude is

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

4. Two key formulas to remember

• Component formula: $$\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3$$
• Angle formula: $$\vec{a} \cdot \vec{b} = |\vec{a}|\,|\vec{b}|\cos\theta$$

These two formulas describe the same dot product. This means we can connect algebra and geometry by setting them equal:

$$a_1b_1 + a_2b_2 + a_3b_3 = |\vec{a}|\,|\vec{b}|\cos\theta$$

From this, we can find the angle between two vectors:

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

5. When are two vectors perpendicular?

Two vectors are perpendicular if the angle between them is \(90^\circ\).

Since \(\cos 90^\circ = 0\), the dot product of perpendicular vectors is:

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

So a very useful test is:

If \(\vec{a} \cdot \vec{b} = 0\), then the vectors are perpendicular.

6. Important properties of the dot product

• It is commutative: $$\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$$
• Dot product with itself gives the square of the magnitude: $$\vec{a} \cdot \vec{a} = |\vec{a}|^2$$
• If one vector is the zero vector, the dot product is zero.

For example,

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

This is a useful way to check your work.

7. Worked Example 1: Basic 2D dot product

Find the dot product of

$$\vec{a} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}, \quad \vec{b} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$$

Step 1: Multiply matching components.

$$3(2) = 6, \quad 4(-1) = -4$$

Step 2: Add the results.

$$\vec{a} \cdot \vec{b} = 6 + (-4) = 2$$

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

The answer is a number, so it is a scalar.

8. Worked Example 2: Dot product in 3D

Find the dot product of

$$\vec{p} = \begin{pmatrix} 1 \\ -2 \\ 5 \end{pmatrix}, \quad \vec{q} = \begin{pmatrix} 4 \\ 3 \\ -1 \end{pmatrix}$$

Step 1: Multiply matching components.

$$1(4) = 4, \quad (-2)(3) = -6, \quad 5(-1) = -5$$

Step 2: Add them.

$$\vec{p} \cdot \vec{q} = 4 - 6 - 5 = -7$$

Answer: $$\vec{p} \cdot \vec{q} = -7$$

Since the result is negative, the angle between the vectors is greater than \(90^\circ\).

9. Worked Example 3: Finding the angle between two vectors

Find the angle between

$$\vec{a} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}, \quad \vec{b} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$$

Step 1: Find the dot product.

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

Step 2: Find the magnitudes.

$$|\vec{a}| = \sqrt{1^2 + 2^2} = \sqrt{5}$$

$$|\vec{b}| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$$

Step 3: Use the angle formula.

$$\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|\,|\vec{b}|} = \frac{11}{5\sqrt{5}}$$

Step 4: Find the angle.

$$\theta = \cos^{-1}\left(\frac{11}{5\sqrt{5}}\right)$$

Using a calculator,

$$\theta \approx 10.3^\circ$$

Answer: The angle between the vectors is approximately $$10.3^\circ$$

10. Worked Example 4: Testing for perpendicular vectors

Show whether the vectors

$$\vec{u} = \begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix}, \quad \vec{v} = \begin{pmatrix} 4 \\ -2 \\ 2 \end{pmatrix}$$

are perpendicular.

Step 1: Find the dot product.

$$\vec{u} \cdot \vec{v} = 2(4) + 1(-2) + (-3)(2)$$

$$= 8 - 2 - 6 = 0$$

Step 2: Interpret the result.

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

Answer: Yes, the vectors are perpendicular.

11. Common mistakes to avoid

• Do not add components first. Multiply matching components first, then add.
• Do not confuse dot product with multiplication of vectors. The dot product gives a scalar, not a vector.
• Be careful with negative signs. A small sign error can change the whole answer.
• Use the correct angle formula. $$\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|\,|\vec{b}|}$$
• Make sure your calculator is in the correct mode (degrees or radians), depending on what your course expects. In school problems, degrees are usually used unless stated otherwise.

12. Quick practice ideas

To get comfortable with dot product, practice these types of questions:

• Find the dot product of two vectors in 2D.
• Find the dot product of two vectors in 3D.
• Use the dot product to find the angle between two vectors.
• Check whether two vectors are perpendicular.

13. Final summary

The dot product is a way of combining two vectors to produce a single number. Using components, we multiply matching entries and add them. Using geometry, the dot product is also equal to the product of the magnitudes times the cosine of the angle between the vectors.

This makes the dot product very useful for finding angles and checking perpendicularity. If the dot product is zero, the vectors are perpendicular. If the dot product is positive, the angle is acute; if it is negative, the angle is obtuse.

Key formulas:

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

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

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

$$\vec{a} \cdot \vec{b} = 0 \Rightarrow \text{vectors are perpendicular}$$

Orthogonal and Parallel Vectors

Vectors are quantities that have both magnitude and direction. In coordinate form, a vector can be written as \\(\vec{v} = \langle a, b \rangle\\) in 2D or \\(\vec{v} = \langle a, b, c \rangle\\) in 3D.

Two important relationships between vectors are parallel and orthogonal. Understanding these ideas helps in geometry, physics, and later work with lines, planes, and forces.

In this lesson, you will learn how to:

• recognize when two vectors are parallel,
• recognize when two vectors are orthogonal,
• use scalar multiples to prove parallelism,
• use the dot product to prove orthogonality.

1. What does it mean for vectors to be parallel?

Two vectors are parallel if they point in the same direction or exactly opposite directions. This happens when one vector is a scalar multiple of the other.

If \\(\vec{u}\\) and \\(\vec{v}\\) are vectors, then they are parallel if there exists a number \\(k\\) such that

$$\vec{u} = k\vec{v}.$$

The number \\(k\\) is called a scalar. If \\(k > 0\\), the vectors point in the same direction. If \\(k < 0\\), they point in opposite directions.

For example, if

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

then

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

so the vectors are parallel.

If instead

$$\vec{c} = \langle -2, -3 \rangle,$$

then

$$\vec{c} = -1\vec{a},$$

so \\(\vec{a}\\) and \\(\vec{c}\\) are also parallel, but they point in opposite directions.

How to test for parallel vectors

In 2D, compare the ratios of corresponding components:

$$\vec{u} = \langle a, b \rangle, \quad \vec{v} = \langle c, d \rangle.$$

If

$$\frac{a}{c} = \frac{b}{d}$$

for the same scalar value, then the vectors are parallel.

In 3D, the idea is the same. For

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

the vectors are parallel if all corresponding components have the same scale factor:

$$\frac{a}{d} = \frac{b}{e} = \frac{c}{f}.$$

Be careful: this method only works when the denominators are not zero. Often, the safest method is to check directly whether one vector can be written as \\(k\\) times the other.

2. What does it mean for vectors to be orthogonal?

Two vectors are orthogonal if they meet at a right angle. In other words, the angle between them is \\(90^\circ\\).

The main test for orthogonality uses the dot product.

For vectors in 2D:

$$\langle a, b \rangle \cdot \langle c, d \rangle = ac + bd$$

For vectors in 3D:

$$\langle a, b, c \rangle \cdot \langle d, e, f \rangle = ad + be + cf$$

Two vectors are orthogonal if and only if their dot product is zero:

$$\vec{u} \cdot \vec{v} = 0.$$

For example, let

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

Then

$$\vec{p} \cdot \vec{q} = (2)(3) + (3)(-2) = 6 - 6 = 0.$$

So \\(\vec{p}\\) and \\(\vec{q}\\) are orthogonal.

Why does a zero dot product mean a right angle?

The dot product is also connected to the angle \\(\theta\\) between two vectors:

$$\vec{u} \cdot \vec{v} = |\vec{u}|\,|\vec{v}|\cos\theta.$$

If the vectors are not zero vectors and \\(\vec{u} \cdot \vec{v} = 0\\), then

$$|\vec{u}|\,|\vec{v}|\cos\theta = 0.$$

Since the magnitudes are not zero, it must be that

$$\cos\theta = 0,$$

which means

$$\theta = 90^\circ.$$

So the vectors are orthogonal.

3. Important difference: parallel vs orthogonal

• Parallel vectors: one is a scalar multiple of the other.
• Orthogonal vectors: their dot product is zero.

These are different ideas. A pair of vectors usually cannot be both parallel and orthogonal, unless one of the vectors is the zero vector.

4. The zero vector

The zero vector is

$$\vec{0} = \langle 0,0 \rangle \quad \text{or} \quad \langle 0,0,0 \rangle.$$

Its magnitude is zero, so it has no direction. Because of this, questions involving the zero vector need care.

• The zero vector has dot product zero with every vector.
• But in school mathematics, we usually focus on non-zero vectors when talking about direction, parallelism, and angles.

So when you test whether vectors are parallel or orthogonal, it is usually assumed the vectors are non-zero unless stated otherwise.

5. Worked Examples

Example 1: Testing for parallel vectors in 2D

Determine whether \\(\vec{u} = \langle 3, 6 \rangle\\) and \\(\vec{v} = \langle 1, 2 \rangle\\) are parallel.

Step 1: Check if one vector is a scalar multiple of the other.

Multiply \\(\vec{v}\\) by 3:

$$3\vec{v} = 3\langle 1, 2 \rangle = \langle 3, 6 \rangle.$$

This matches \\(\vec{u}\\).

Conclusion:

$$\vec{u} = 3\vec{v},$$

so the vectors are parallel.

Example 2: Testing for orthogonal vectors in 2D

Determine whether \\(\vec{a} = \langle 4, 1 \rangle\\) and \\(\vec{b} = \langle 2, -8 \rangle\\) are orthogonal.

Step 1: Find the dot product.

$$\vec{a} \cdot \vec{b} = (4)(2) + (1)(-8) = 8 - 8 = 0.$$

Conclusion: Since the dot product is zero, the vectors are orthogonal.

Example 3: Decide whether vectors are parallel, orthogonal, or neither

Let

$$\vec{p} = \langle 2, -3, 1 \rangle, \quad \vec{q} = \langle 4, -6, 2 \rangle.$$

Step 1: Test for parallelism.

Compare components:

$$\frac{4}{2} = 2, \quad \frac{-6}{-3} = 2, \quad \frac{2}{1} = 2.$$

All components have the same scale factor.

So

$$\vec{q} = 2\vec{p}.$$

Therefore, the vectors are parallel.

Step 2: Test for orthogonality if needed.

$$\vec{p} \cdot \vec{q} = (2)(4) + (-3)(-6) + (1)(2) = 8 + 18 + 2 = 28.$$

The dot product is not zero, so they are not orthogonal.

Conclusion: The vectors are parallel, not orthogonal.

Example 4: Find a value that makes vectors orthogonal

Find the value of \\(k\\) so that

$$\vec{u} = \langle k, 5 \rangle \quad \text{and} \quad \vec{v} = \langle 2, -4 \rangle$$

are orthogonal.

Step 1: Use the condition for orthogonality.

$$\vec{u} \cdot \vec{v} = 0.$$

Step 2: Compute the dot product.

$$\langle k, 5 \rangle \cdot \langle 2, -4 \rangle = 2k + 5(-4).$$ $$2k - 20 = 0.$$

Step 3: Solve for \\(k\\).

$$2k = 20$$ $$k = 10$$

Conclusion: The vectors are orthogonal when

$$k = 10.$$

6. Common mistakes to avoid

• Mixing up the tests: Parallel vectors use scalar multiples; orthogonal vectors use dot product.
• Only checking one pair of components: In 3D, all three components must match the same scale factor for parallelism.
• Arithmetic errors in the dot product: Watch negative signs carefully.
• Assuming perpendicular-looking vectors in a sketch are orthogonal: Always prove it with the dot product.

7. Quick strategy guide

If a question asks whether two vectors are parallel:

1. Try to write one vector as \\(k\\) times the other.
2. If that works, they are parallel.
3. If it does not, they are not parallel.

If a question asks whether two vectors are orthogonal:

1. Calculate the dot product.
2. If the result is 0, they are orthogonal.
3. If the result is not 0, they are not orthogonal.

If a question asks whether vectors are parallel, orthogonal, or neither:

1. Check for scalar multiples first.
2. Then check the dot product.
3. State the correct relationship clearly.

8. Summary

To prove two vectors are parallel, show that one is a scalar multiple of the other:

$$\vec{u} = k\vec{v}.$$

To prove two vectors are orthogonal, show that their dot product is zero:

$$\vec{u} \cdot \vec{v} = 0.$$

These two tests are simple and powerful. Once you know which test to use, you can quickly decide the relationship between vectors in both 2D and 3D.

Resolving Vectors and Physical Applications

In many real-life situations, a quantity has both size and direction. Such a quantity is called a vector. Examples include force, velocity, displacement, and acceleration.

Often, a single vector acts at an angle. To understand its effect clearly, we break it into parts along chosen directions, usually horizontal and vertical. This process is called resolving a vector into components.

Resolving vectors is very important in physics. It helps us analyze forces on an object, motion on an incline, the pull of ropes, and velocity in different directions.

1. What does it mean to resolve a vector?

To resolve a vector means to split it into two or more perpendicular parts whose combined effect is the same as the original vector.

In 2D, we usually resolve a vector into:

• a horizontal component
• a vertical component

If a vector has magnitude \(V\) and makes an angle \(\theta\) with the positive horizontal axis, then:

$$V_x = V\cos\theta$$ $$V_y = V\sin\theta$$

Here:

• \(V_x\) is the horizontal component
• \(V_y\) is the vertical component

These formulas come from right-triangle trigonometry. The cosine gives the side adjacent to the angle, and the sine gives the side opposite the angle.

2. Visual idea of components

Imagine pushing a box with a force directed upward and to the right. The force does two things at once:

• it pushes the box forward
• it also tends to lift it upward

These two effects are described by the horizontal and vertical components of the force.

So, instead of working with one slanted vector, we work with two simpler perpendicular vectors. This makes calculations much easier.

3. Signs of components

The formulas \(V_x = V\cos\theta\) and \(V_y = V\sin\theta\) give magnitudes based on the angle, but the direction matters too.

Use signs carefully:

• right is usually positive \((+)\)
• left is usually negative \((- )\)
• up is usually positive \((+)\)
• down is usually negative \((- )\)

For example, if a vector points down and to the right, then its horizontal component is positive and its vertical component is negative.

4. Resultant vectors from components

If the horizontal and vertical components of a vector are known, we can rebuild the original vector. This original vector is called the resultant.

If the components are \(x\) and \(y\), then the magnitude of the resultant is:

$$R = \sqrt{x^2 + y^2}$$

The direction angle \(\theta\), measured from the horizontal, is found using:

$$\tan\theta = \frac{y}{x}$$

So:

$$\theta = \tan^{-1}\left(\frac{y}{x}\right)$$

Always think about the quadrant to make sure the direction is correct.

5. Why resolving vectors matters in physics

Many physics problems involve forces or velocities acting at angles. Objects usually move or balance along horizontal and vertical directions, so resolving vectors lets us study each direction separately.

This is useful in situations such as:

• finding the effect of a force pulling at an angle
• analyzing tension in ropes
• studying motion on a slope
• combining velocities in different directions

6. Important idea: perpendicular directions are independent

Horizontal and vertical components do not interfere with each other. This means we can analyze them separately.

For example, if a force has a horizontal part of \(20\text{ N}\) and a vertical part of \(15\text{ N}\), we treat these as separate effects in their own directions.

This is one of the main reasons vector resolution is so powerful.

7. Worked Example 1: Resolving a force into components

A force of \(50\text{ N}\) acts at an angle of \(30^\circ\) above the horizontal. Find its horizontal and vertical components.

Step 1: Write the formulas

$$F_x = F\cos\theta$$ $$F_y = F\sin\theta$$

Step 2: Substitute the values

$$F_x = 50\cos 30^\circ$$ $$F_y = 50\sin 30^\circ$$

Step 3: Use trig values

$$\cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866$$ $$\sin 30^\circ = 0.5$$

So:

$$F_x = 50(0.866) = 43.3\text{ N}$$ $$F_y = 50(0.5) = 25\text{ N}$$

Answer:

• Horizontal component = \(43.3\text{ N}\) to the right
• Vertical component = \(25\text{ N}\) upward

This means the single \(50\text{ N}\) force has the same effect as two perpendicular forces of \(43.3\text{ N}\) and \(25\text{ N}\).

8. Worked Example 2: Finding a resultant from components

A velocity has components \(12\text{ m/s}\) east and \(5\text{ m/s}\) north. Find the magnitude and direction of the velocity.

Step 1: Find the magnitude

$$R = \sqrt{12^2 + 5^2}$$ $$R = \sqrt{144 + 25}$$ $$R = \sqrt{169} = 13\text{ m/s}$$

Step 2: Find the direction

$$\tan\theta = \frac{5}{12}$$ $$\theta = \tan^{-1}\left(\frac{5}{12}\right)$$ $$\theta \approx 22.6^\circ$$

Answer:

The velocity is \(13\text{ m/s}\) at \(22.6^\circ\) north of east.

9. Worked Example 3: Force on an inclined plane

A block rests on a smooth slope inclined at \(20^\circ\) to the horizontal. Its weight is \(100\text{ N}\). Resolve the weight into:

• a component parallel to the slope
• a component perpendicular to the slope

When dealing with an incline, it is often best to choose axes:

• along the slope
• perpendicular to the slope

For a slope angle \(\theta\):

• component of weight parallel to slope = \(W\sin\theta\)
• component of weight perpendicular to slope = \(W\cos\theta\)

Here, \(W = 100\text{ N}\) and \(\theta = 20^\circ\).

Parallel component:

$$W_{\parallel} = 100\sin 20^\circ$$ $$W_{\parallel} = 100(0.342) = 34.2\text{ N}$$

Perpendicular component:

$$W_{\perp} = 100\cos 20^\circ$$ $$W_{\perp} = 100(0.940) = 94.0\text{ N}$$

Answer:

• Component down the slope = \(34.2\text{ N}\)
• Component into the slope = \(94.0\text{ N}\)

The parallel component tends to make the block slide down the slope. The perpendicular component presses the block against the surface.

10. Worked Example 4: Tension and equilibrium

A lamp is hanging at rest from a rope. The tension in the rope is \(80\text{ N}\), and the rope makes an angle of \(40^\circ\) with the ceiling. Find the vertical and horizontal components of the tension.

Since the ceiling is horizontal, the rope makes a \(40^\circ\) angle with the horizontal.

Horizontal component:

$$T_x = 80\cos 40^\circ$$ $$T_x = 80(0.766) = 61.3\text{ N}$$

Vertical component:

$$T_y = 80\sin 40^\circ$$ $$T_y = 80(0.643) = 51.4\text{ N}$$

Answer:

• Horizontal component = \(61.3\text{ N}\)
• Vertical component = \(51.4\text{ N}\)

In equilibrium problems, vertical components often balance weight, and horizontal components often balance each other.

11. Resolving vectors in 3D

In three dimensions, a vector can have components in the \(x\)-, \(y\)-, and \(z\)-directions.

A vector may be written as:

$$\vec{V} = \langle V_x, V_y, V_z \rangle$$

Its magnitude is:

$$|\vec{V}| = \sqrt{V_x^2 + V_y^2 + V_z^2}$$

In physical applications, this is useful when motion or force is not limited to a flat surface.

For example, if a drone moves with velocity components \(3\text{ m/s}\) east, \(4\text{ m/s}\) north, and \(12\text{ m/s}\) upward, then the speed is:

$$|\vec{V}| = \sqrt{3^2 + 4^2 + 12^2}$$ $$|\vec{V}| = \sqrt{9 + 16 + 144}$$ $$|\vec{V}| = \sqrt{169} = 13\text{ m/s}$$

This is the 3D version of Pythagoras' theorem.

12. Common mistakes to avoid

• Mixing up sine and cosine: use cosine for the side adjacent to the given angle, sine for the opposite side.
• Ignoring signs: left/down may need negative values.
• Using the wrong angle: make sure you know whether the angle is measured from the horizontal, vertical, or slope.
• Forgetting units: include units like N, m/s, or m.
• Not checking reasonableness: each component should usually be smaller than the original vector.

13. Problem-solving steps for resolving vectors

1. Draw a clear diagram.
2. Mark the angle carefully.
3. Choose axes, usually horizontal/vertical or parallel/perpendicular to a slope.
4. Use sine and cosine to find components.
5. Add signs if needed based on direction.
6. For resultants, use Pythagoras for magnitude and inverse tangent for direction.
7. Check that your answer makes physical sense.

14. Key formulas

• Horizontal component: \(V_x = V\cos\theta\)
• Vertical component: \(V_y = V\sin\theta\)
• Resultant magnitude in 2D: \(R = \sqrt{x^2 + y^2}\)
• Direction in 2D: \(\theta = \tan^{-1}(y/x)\)
• Magnitude in 3D: \(|\vec{V}| = \sqrt{V_x^2 + V_y^2 + V_z^2}\)

15. Brief summary

Resolving vectors means splitting a vector into perpendicular components, usually horizontal and vertical. If a vector of magnitude \(V\) makes an angle \(\theta\) with the horizontal, its components are \(V\cos\theta\) and \(V\sin\theta\).

These components help us solve physics problems involving force, tension, motion, and slopes. Once components are known, we can also combine them to find a resultant using Pythagoras' theorem and trigonometry.

With careful diagrams, correct trig ratios, and attention to signs, resolving vectors becomes a very useful tool in both mathematics and physics.

3D Cartesian Coordinates help us describe the position of points in space. In 2D, a point is written as \((x,y)\) and shows where the point is on a flat plane. In 3D, we need one more number, so a point is written as \((x,y,z)\).

This extra coordinate lets us describe positions in space, not just on a sheet of paper. 3D coordinates are used in geometry, physics, engineering, computer graphics, and many real-world models.

In this lesson, you will learn how the three axes work, how to plot and describe points in 3D, and how to find the distance and midpoint between two points in space.

1. The 3D Coordinate System

A 3D Cartesian coordinate system has three number lines, called axes:

• x-axis: usually drawn left-right
• y-axis: usually drawn sideways on the page
• z-axis: usually drawn up-down

These three axes meet at one point called the origin. The origin has coordinates:

$$ (0,0,0) $$

Each point in space is described by an ordered triple:

$$ (x,y,z) $$

This means:

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

The order matters. The point \((2,1,3)\) is different from \((1,2,3)\).

2. Understanding the Coordinates

For a point \((x,y,z)\):

• \(x\) tells how far the point is in the x-direction,
• \(y\) tells how far the point is in the y-direction,
• \(z\) tells how far the point is above or below the xy-plane.

The xy-plane is the flat plane where \(z=0\). Similarly:

• the yz-plane is where \(x=0\),
• the xz-plane is where \(y=0\).

These planes are the 3D versions of the coordinate axes in 2D. They help us picture where a point is located.

For example:

• If a point is \((4,-2,0)\), it lies on the xy-plane because \(z=0\).
• If a point is \((0,3,5)\), it lies on the yz-plane because \(x=0\).
• If a point is \((-1,0,2)\), it lies on the xz-plane because \(y=0\).

3. How to Plot a Point in 3D

To plot a point \((x,y,z)\):

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

Because 3D drawings are shown on flat paper, sketches may look slanted. The picture is only a visual aid, but the coordinates still follow the same exact rules.

4. Distance from the Origin in 3D

In 2D, the distance from the origin to \((x,y)\) is found using the Pythagorean theorem:

$$ d=\sqrt{x^2+y^2} $$

In 3D, we extend this idea one more step. For a point \((x,y,z)\), the distance from the origin is:

$$ d=\sqrt{x^2+y^2+z^2} $$

This works because we first find the distance in the xy-plane, then use the Pythagorean theorem again with the z-value.

5. Distance Between Two Points in 3D

Suppose we have two points:

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

The distance between them is:

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

This is the 3D version of the distance formula you already know from 2D. The only difference is the extra \(z\)-term.

6. Midpoint of a Line Segment in 3D

The midpoint is the point exactly halfway between two endpoints.

If the endpoints are:

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

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 means we average each coordinate separately.

7. Worked Examples

Example 1: Identifying position in space

Consider the point \(P(3,-2,4)\).

• The x-coordinate is \(3\), so the point is 3 units in the positive x-direction.
• The y-coordinate is \(-2\), so it is 2 units in the negative y-direction.
• The z-coordinate is \(4\), so it is 4 units above the xy-plane.

So \(P(3,-2,4)\) is a point in space that is right/left according to \(x\), sideways according to \(y\), and upward according to \(z\).

Example 2: Distance from the origin

Find the distance from the origin to the point \(Q(2,3,6)\).

Use the formula:

$$ d=\sqrt{x^2+y^2+z^2} $$

Substitute the values:

$$ d=\sqrt{2^2+3^2+6^2} $$

$$ d=\sqrt{4+9+36} $$

$$ d=\sqrt{49} $$

$$ d=7 $$

Answer: The distance from the origin to \(Q\) is \(7\) units.

Example 3: Distance between two points

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

Use the 3D distance formula:

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

Substitute the coordinates:

$$ AB=\sqrt{(5-1)^2+(-1-2)^2+(7-3)^2} $$

$$ AB=\sqrt{4^2+(-3)^2+4^2} $$

$$ AB=\sqrt{16+9+16} $$

$$ AB=\sqrt{41} $$

Answer: The distance between \(A\) and \(B\) is \(\sqrt{41}\) units.

Example 4: Midpoint in 3D

Find the midpoint of the segment joining \(C(2,4,-1)\) and \(D(6,0,5)\).

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:

$$ M\left(\frac{2+6}{2},\frac{4+0}{2},\frac{-1+5}{2}\right) $$

$$ M\left(\frac{8}{2},\frac{4}{2},\frac{4}{2}\right) $$

$$ M(4,2,2) $$

Answer: The midpoint is \((4,2,2)\).

8. Common Mistakes to Avoid

• Mixing up the order: Always write coordinates as \((x,y,z)\), not in any other order.
• Forgetting the third term: In 3D distance, include \((z_2-z_1)^2\).
• Sign errors: Be careful when subtracting negative numbers, such as \(-1-2=-3\).
• Not averaging each coordinate: In the midpoint formula, divide each sum by 2.

9. Quick Check for Understanding

• What plane does the point \((3,5,0)\) lie on?
• Find the distance from the origin to \((-2,1,2)\).
• Find the midpoint of \((1,1,1)\) and \((3,5,7)\).

Answers:

• \((3,5,0)\) lies on the xy-plane.
• Distance: $$\sqrt{(-2)^2+1^2+2^2}=\sqrt{4+1+4}=3$$
• Midpoint: $$\left(\frac{1+3}{2},\frac{1+5}{2},\frac{1+7}{2}\right)=(2,3,4)$$

10. Summary

In 3D Cartesian coordinates, a point is written as \((x,y,z)\). The three coordinates tell us the point’s position relative to the x-, y-, and z-axes.

You can find distance in 3D by extending the Pythagorean theorem:

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

You can find the midpoint by averaging each coordinate:

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

These ideas are important for 3D geometry and also connect directly to vectors, motion, and spatial problems in mathematics and physics.

Vectors in 3D Space help us describe movement, force, position, and direction when things are not limited to a flat plane. In 2D, we use the horizontal and vertical directions. In 3D, we add a third direction, usually called the depth or height direction.

A 3D vector tells us how far and in what direction something goes in space. For example, a drone might move 4 units east, 3 units north, and 2 units up. That movement is a vector in 3D space.

In this lesson, you will learn how to write 3D vectors using unit vectors \(\mathbf{i}, \mathbf{j}, \mathbf{k}\), find their magnitude, calculate direction cosines, and use the dot product in three dimensions.

1. Writing vectors in 3D

In 3D space, we usually use three perpendicular axes:

• x-axis: left-right direction
• y-axis: forward-back direction
• z-axis: up-down direction

The three 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

A vector with components \(a\), \(b\), and \(c\) is written as

$$\vec{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}$$

It can also be written in coordinate form as

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

These two forms mean the same thing.

For example,

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

means the vector has components:

• x-component: \(2\)
• y-component: \(-3\)
• z-component: \(5\)

So in coordinate form,

$$\vec{v} = \langle 2, -3, 5 \rangle$$

2. Position vectors in 3D

If a point in space has coordinates \((x, y, z)\), then the vector from the origin to that point is called its position vector.

For a point \(P(4, -2, 7)\), the position vector is

$$\overrightarrow{OP} = 4\mathbf{i} - 2\mathbf{j} + 7\mathbf{k}$$

This vector starts at the origin \((0,0,0)\) and ends at \((4,-2,7)\).

3. Vector between two points

To find the vector from point \(A(x_1,y_1,z_1)\) to point \(B(x_2,y_2,z_2)\), subtract the coordinates of \(A\) from the coordinates of \(B\):

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

In unit vector form,

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

This is very useful in geometry and physics because it tells both the distance and direction from one point to another.

4. Magnitude of a 3D vector

The magnitude of a vector is its length or size. For a vector

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

the magnitude is

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

This formula comes from the 3D version of the Pythagorean theorem.

For example, if \(\vec{v} = \langle 3,4,12 \rangle\), then

$$|\vec{v}| = \sqrt{3^2+4^2+12^2} = \sqrt{9+16+144} = \sqrt{169} = 13$$

So the vector has length 13.

5. Unit vectors in the direction of a vector

A unit vector is a vector with magnitude 1. To find the unit vector in the direction of a vector \(\vec{v}\), divide the vector by its magnitude:

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

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

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

This keeps the direction the same but changes the length to 1.

6. Direction cosines

In 3D, a vector can make angles with all three axes. Suppose a vector \(\vec{v}\) makes angles \(\alpha\), \(\beta\), and \(\gamma\) with the positive x-, y-, and z-axes.

The direction cosines are:

$$\cos\alpha, \quad \cos\beta, \quad \cos\gamma$$

For a vector \(\vec{v} = \langle a,b,c \rangle\), the direction cosines are found by dividing each component by the magnitude:

$$\cos\alpha = \frac{a}{|\vec{v}|}, \quad \cos\beta = \frac{b}{|\vec{v}|}, \quad \cos\gamma = \frac{c}{|\vec{v}|}$$

Notice that these are exactly the components of the unit vector in the direction of \(\vec{v}\).

An important fact is:

$$\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$$

This is a useful check for your work.

7. Dot product in 3D

The dot product combines two vectors and gives a number, not another vector.

If

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

then their dot product is

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

In unit vector form, if

$$\vec{a} = a_1\mathbf{i}+a_2\mathbf{j}+a_3\mathbf{k}$$

and

$$\vec{b} = b_1\mathbf{i}+b_2\mathbf{j}+b_3\mathbf{k}$$

then

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

The dot product is also related to the angle \(\theta\) between the vectors:

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

This formula is useful for finding the angle between two vectors.

8. When vectors are perpendicular

Two vectors are perpendicular if they meet at a right angle. In 3D, just like in 2D, this happens when their dot product is zero:

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

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

Worked Example 1: Writing a vector and finding its magnitude

Write the vector \(\langle -2, 6, 3 \rangle\) in unit vector form, and find its magnitude.

Step 1: Write in unit vector form

$$\vec{v} = -2\mathbf{i} + 6\mathbf{j} + 3\mathbf{k}$$

Step 2: Use the magnitude formula

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

$$= \sqrt{4+36+9}$$

$$= \sqrt{49} = 7$$

Answer: The vector is \(-2\mathbf{i}+6\mathbf{j}+3\mathbf{k}\), and its magnitude is \(7\).

Worked Example 2: Finding a vector between two points

Find the vector from \(A(1,-2,4)\) to \(B(5,3,-1)\).

Step 1: Subtract coordinates of A from coordinates of B

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

$$= \langle 4, 5, -5 \rangle$$

Step 2: Write in unit vector form if needed

$$\overrightarrow{AB} = 4\mathbf{i} + 5\mathbf{j} - 5\mathbf{k}$$

Answer: $$\overrightarrow{AB} = \langle 4,5,-5 \rangle$$

Worked Example 3: Finding direction cosines

Find the direction cosines of the vector

$$\vec{v} = 2\mathbf{i} - \mathbf{j} + 2\mathbf{k}$$

Step 1: Identify the components

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

Step 2: Find the magnitude

$$|\vec{v}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4+1+4} = \sqrt{9} = 3$$

Step 3: Divide each component by the magnitude

$$\cos\alpha = \frac{2}{3}, \quad \cos\beta = -\frac{1}{3}, \quad \cos\gamma = \frac{2}{3}$$

Check:

$$\left(\frac{2}{3}\right)^2 + \left(-\frac{1}{3}\right)^2 + \left(\frac{2}{3}\right)^2 = \frac{4}{9}+\frac{1}{9}+\frac{4}{9} = 1$$

Answer: The direction cosines are \(\frac{2}{3}\), \(-\frac{1}{3}\), and \(\frac{2}{3}\).

Worked Example 4: Dot product and angle between two vectors

Find the dot product of

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

Then determine whether the vectors are perpendicular.

Step 1: Compute the dot product

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

$$= 2+2-4 = 0$$

Step 2: Interpret the result

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

Optional angle check: If \(\vec{a} \cdot \vec{b} = |\vec{a}|\,|\vec{b}|\cos\theta = 0\), then \(\cos\theta=0\), so \(\theta = 90^\circ\).

Answer: The dot product is \(0\), so the vectors are perpendicular.

9. Common mistakes to avoid

• Forgetting the third component: In 3D, every vector has x-, y-, and z-components.
• Mixing up point coordinates and vector components: A point is a location like \((2,3,4)\), while a vector describes movement like \(\langle 2,3,4 \rangle\).
• Subtracting points in the wrong order: \(\overrightarrow{AB} = B-A\), not \(A-B\).
• Making sign errors: Be careful with negative coordinates and components.
• Using the wrong magnitude formula: In 3D, use \(\sqrt{a^2+b^2+c^2}\), not just two terms.

10. Key formulas

• Vector form: $$\vec{v} = a\mathbf{i}+b\mathbf{j}+c\mathbf{k} = \langle a,b,c \rangle$$
• Vector from \(A(x_1,y_1,z_1)\) to \(B(x_2,y_2,z_2)\): $$\overrightarrow{AB} = \langle x_2-x_1,\ y_2-y_1,\ z_2-z_1 \rangle$$
• Magnitude: $$|\vec{v}| = \sqrt{a^2+b^2+c^2}$$
• Unit vector in direction of \(\vec{v}\): $$\hat{v} = \frac{\vec{v}}{|\vec{v}|}$$
• Direction cosines: $$\cos\alpha = \frac{a}{|\vec{v}|}, \quad \cos\beta = \frac{b}{|\vec{v}|}, \quad \cos\gamma = \frac{c}{|\vec{v}|}$$
• Dot product: $$\vec{a} \cdot \vec{b} = a_1b_1+a_2b_2+a_3b_3$$
• Angle formula: $$\vec{a} \cdot \vec{b} = |\vec{a}|\,|\vec{b}|\cos\theta$$

Brief Summary

Vectors in 3D space use three components to describe direction and size. We write them using \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\), find their lengths using the 3D magnitude formula, and describe their directions using direction cosines.

The dot product helps compare two vectors and can show whether they are perpendicular or help find the angle between them. These skills are important in geometry, mechanics, and any situation involving motion or force in space.