Vectors

1. A vector goes from $(x_1, y_1)$ to $(x_2, y_2)$→its components are $\left\langle x_2 - x_1, y_2 - y_1 \right\rangle$, wherever it is drawn
2. Square the components, add, take the square root→the magnitude $\left\| \mathbf{v} \right\|$, the vector's length
3. Place the second vector at the tip of the first→the sum runs from the first tail to the second tip; in components, add matching entries
4. Multiply matching components and add→the dot product; it is $0$ exactly when the vectors are perpendicular

• A vector has a magnitude and a direction. $\mathbf{v} = \left\langle a, b \right\rangle = a\mathbf{i} + b\mathbf{j}$ moves $a$ across and $b$ up; its magnitude is $\left\| \mathbf{v} \right\| = \sqrt{a^2 + b^2}$.
• A vector of magnitude $m$ at angle $\theta$ from the positive $x$-axis is $\left\langle m\cos\theta, m\sin\theta \right\rangle$.
• Add and subtract component by component. A scalar multiple $k\mathbf{v}$ stretches by $|k|$ and reverses if $k < 0$. A unit vector $\frac{\mathbf{v}}{\left\| \mathbf{v} \right\|}$ has length $1$.
• Dot product: $\left\langle a, b \right\rangle \cdot \left\langle c, d \right\rangle$$= ac$$+ bd$$= \left\| \mathbf{u} \right\| \left\| \mathbf{v} \right\| \cos\theta$, $\theta$ the angle between them.

Exam-style questions. Anything you miss goes into your review queue.

Graph

Two vectors on a grid

Vector $\mathbf{u}$ is drawn from $(0, 0)$ to $(4, 1)$; vector $\mathbf{v}$ is drawn from $(2, 1)$ to $(1, 4)$.

1. What are the components of $\mathbf{v}$?

1. $\left\langle -1, 3 \right\rangle$
2. $\left\langle 1, -3 \right\rangle$
3. $\left\langle 1, 4 \right\rangle$
4. $\left\langle 3, -1 \right\rangle$

Show the answer

Terminal minus initial: $\left\langle 1 - 2, 4 - 1 \right\rangle = \left\langle -1, 3 \right\rangle$, one unit left and three up.

• Correct: $\left\langle -1, 3 \right\rangle$: $1 - 2 = -1$ across, $4 - 1 = 3$ up.
• $\left\langle 1, -3 \right\rangle$: This subtracts the other way, from tip to tail; it points the opposite direction.
• $\left\langle 1, 4 \right\rangle$: $(1, 4)$ is where the arrow ends; subtract where it starts.
• $\left\langle 3, -1 \right\rangle$: The components are swapped.

2. What is $\mathbf{u} + \mathbf{v}$?

1. $\left\langle 3, 4 \right\rangle$
2. $\left\langle 5, -2 \right\rangle$
3. $\left\langle -4, 3 \right\rangle$
4. $\left\langle 4, 4 \right\rangle$

Show the answer

$\left\langle 4 + (-1), 1 + 3 \right\rangle$$= \left\langle 3, 4 \right\rangle$, adding matching components.

• Correct: $\left\langle 3, 4 \right\rangle$: $4 - 1 = 3$ and $1 + 3 = 4$.
• $\left\langle 5, -2 \right\rangle$: This is $\mathbf{u} - \mathbf{v}$.
• $\left\langle -4, 3 \right\rangle$: This multiplies matching components; add them.
• $\left\langle 4, 4 \right\rangle$: $4 + (-1) = 3$, not $4$.

3. What is $\left\| \mathbf{u} + \mathbf{v} \right\|$?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$\left\| \left\langle 3, 4 \right\rangle \right\| = \sqrt{9 + 16} = 5$.

• Answer: 5

4. What is $\mathbf{u} \cdot \mathbf{v}$?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$4(-1) + 1(3)$$= -4 + 3$$= -1$. It is negative, so the angle between $\mathbf{u}$ and $\mathbf{v}$ is more than $\frac{\pi}{2}$.

• Answer: -1

5. What is the unit vector in the direction of $\left\langle 5, -12 \right\rangle$?

1. $\left\langle \frac{5}{13}, -\frac{12}{13} \right\rangle$
2. $\left\langle \frac{5}{17}, -\frac{12}{17} \right\rangle$
3. $\left\langle 1, -1 \right\rangle$
4. $\left\langle \frac{5}{7}, -\frac{12}{7} \right\rangle$

Show the answer

The length is $\sqrt{25 + 144} = 13$. Divide each component by $13$.

• Correct: $\left\langle \frac{5}{13}, -\frac{12}{13} \right\rangle$: Each component over the length $13$.
• $\left\langle \frac{5}{17}, -\frac{12}{17} \right\rangle$: $17 = 5 + 12$ is not the length; use $\sqrt{5^2 + 12^2}$.
• $\left\langle 1, -1 \right\rangle$: This changes the direction; the ratio of components must stay $5 : -12$.
• $\left\langle \frac{5}{7}, -\frac{12}{7} \right\rangle$: $7 = 12 - 5$ is not the length.

6. Which vector is perpendicular to $\left\langle 2, 3 \right\rangle$?

1. $\left\langle -3, 2 \right\rangle$
2. $\left\langle 3, 2 \right\rangle$
3. $\left\langle -2, -3 \right\rangle$
4. $\left\langle 2, -3 \right\rangle$

Show the answer

$\left\langle 2, 3 \right\rangle \cdot \left\langle -3, 2 \right\rangle$$= -6 + 6$$= 0$, so the vectors are perpendicular.

• Correct: $\left\langle -3, 2 \right\rangle$: Dot product $0$.
• $\left\langle 3, 2 \right\rangle$: $6 + 6 = 12$, not $0$.
• $\left\langle -2, -3 \right\rangle$: This points the opposite way: dot product $-13$.
• $\left\langle 2, -3 \right\rangle$: $4 - 9 = -5$, not $0$.

• Next: Vector-Valued Functions (4.9)