Component Form

A vector from the origin to point $(a, b)$ is: $\mathbf{v} = \langle a, b \rangle$

• $a$ is the horizontal component ($x$-component)
• $b$ is the vertical component ($y$-component)

Vector Between Two Points

The vector from point $P_1(x_1, y_1)$ to $P_2(x_2, y_2)$ is: $\vec{P_1P_2} = \langle x_2 - x_1, y_2 - y_1 \rangle$

Magnitude of a Vector

The magnitude (or length) of vector $\mathbf{v} = \langle a, b \rangle$ is:

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

This is the distance formula!

Direction of a Vector

The direction angle $\theta$ is measured counterclockwise from the positive $x$-axis:

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

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

⚠️ Be careful with quadrants when finding $\theta$!

Vector Operations

Scalar Multiplication

If $\mathbf{v} = \langle a, b \rangle$ and $k$ is a scalar: $k\mathbf{v} = \langle ka, kb \rangle$

• If $k > 0$: same direction, scaled by $k$
• If $k < 0$: opposite direction, scaled by $|k|$

Vector Addition

If $\mathbf{u} = \langle a_1, b_1 \rangle$ and $\mathbf{v} = \langle a_2, b_2 \rangle$: $\mathbf{u} + \mathbf{v} = \langle a_1 + a_2, b_1 + b_2 \rangle$

Geometric interpretation: Tip-to-tail method or parallelogram rule

Vector Subtraction

$\mathbf{u} - \mathbf{v} = \langle a_1 - a_2, b_1 - b_2 \rangle$

Unit Vectors

A unit vector has magnitude 1.

The standard unit vectors are:

• $\mathbf{i} = \langle 1, 0 \rangle$ (horizontal)
• $\mathbf{j} = \langle 0, 1 \rangle$ (vertical)

Any vector can be written as: $\mathbf{v} = a\mathbf{i} + b\mathbf{j}$

Finding a Unit Vector

To find a unit vector in the direction of $\mathbf{v}$: $\mathbf{u} = \frac{\mathbf{v}}{|\mathbf{v}|} = \frac{1}{|\mathbf{v}|}\mathbf{v}$

Dot Product

The dot product of $\mathbf{u} = \langle a_1, b_1 \rangle$ and $\mathbf{v} = \langle a_2, b_2 \rangle$ is:

$\mathbf{u} \cdot \mathbf{v} = a_1a_2 + b_1b_2$

Properties

• $\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}||\mathbf{v}|\cos(\theta)$ where $\theta$ is the angle between them
• If $\mathbf{u} \cdot \mathbf{v} = 0$, the vectors are perpendicular
• The dot product is a scalar, not a vector!

$|\mathbf{v}| = \sqrt{a^2 + b^2} = \sqrt{3^2 + 4^2}$

$= \sqrt{9 + 16} = \sqrt{25} = 5$

Find the direction angle:

$\tan(\theta) = \frac{b}{a} = \frac{4}{3}$

$\theta = \tan^{-1}\left(\frac{4}{3}\right)$

$\theta \approx 53.13°$

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

Answers:

• Magnitude: $5$
• Direction: $\theta \approx 53.13°$ or $0.927$ radians

Part (b): Scalar multiplication and subtraction:

$2\vec{u} = 2\langle 3, -4 \rangle = \langle 6, -8 \rangle$

$3\vec{v} = 3\langle -2, 5 \rangle = \langle -6, 15 \rangle$

$2\vec{u} - 3\vec{v} = \langle 6, -8 \rangle - \langle -6, 15 \rangle = \langle 6-(-6), -8-15 \rangle = \langle 12, -23 \rangle$

Part (c): Magnitude formula: $|\vec{u}| = \sqrt{u_1^2 + u_2^2}$

$|\vec{u}| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

$= \langle 2 + (-1), -3 + 4 \rangle$

$= \langle 1, 1 \rangle$

Part b) $3\mathbf{u} - 2\mathbf{v}$

Step 1: Scalar multiplication. $3\mathbf{u} = 3\langle 2, -3 \rangle = \langle 6, -9 \rangle$

$2\mathbf{v} = 2\langle -1, 4 \rangle = \langle -2, 8 \rangle$

Step 2: Subtract. $3\mathbf{u} - 2\mathbf{v} = \langle 6, -9 \rangle - \langle -2, 8 \rangle$

$= \langle 6 - (-2), -9 - 8 \rangle$

$= \langle 8, -17 \rangle$

Answers:

• a) $\langle 1, 1 \rangle$
• b) $\langle 8, -17 \rangle$

$\vec{w} = \langle 10\cos 120°, 10\sin 120° \rangle$

$\cos 120° = -\frac{1}{2}$ and $\sin 120° = \frac{\sqrt{3}}{2}$

$\vec{w} = \left\langle 10\left(-\frac{1}{2}\right), 10\left(\frac{\sqrt{3}}{2}\right) \right\rangle = \langle -5, 5\sqrt{3} \rangle$

Part (b): A unit vector has magnitude 1. To find the unit vector in the direction of $\vec{w}$:

$\hat{w} = \frac{\vec{w}}{|\vec{w}|} = \frac{\langle -5, 5\sqrt{3} \rangle}{10} = \left\langle -\frac{1}{2}, \frac{\sqrt{3}}{2} \right\rangle$

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

$= (5)(-3) + (2)(4)$

$= -15 + 8$

$= -7$

Are they perpendicular?

Two vectors are perpendicular if and only if their dot product equals zero.

Since $\mathbf{u} \cdot \mathbf{v} = -7 \neq 0$, the vectors are not perpendicular.

Answers:

• Dot product: $-7$
• The vectors are not perpendicular