Precalculus Quiz: Magnitude And Direction Of Scaled Vectors

20 questions · exam conditions

0:00

Magnitude And Direction Of Scaled VectorsQuestion 1 of 20

Vector $\mathbf{v}$ has magnitude $12$ and direction $45^\circ$ from the positive $x$ -axis. What are the magnitude and direction of $\frac{1}{2}\mathbf{v}$ ?

Magnitude

6

; direction

45^\circ

Magnitude

24

; direction

45^\circ

Magnitude

6

; direction

225^\circ

Magnitude

12

; direction

45^\circ

← Back to quizzes

Precalculus Quiz

Precalculus Quiz: Magnitude And Direction Of Scaled Vectors

Practice Magnitude And Direction Of Scaled Vectors in Precalculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Magnitude And Direction Of Scaled Vectors, giving you a quick way to practice the rules, question types, and explanations that matter most for Precalculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Vector $\mathbf{v}$ has magnitude $12$ and direction $45^\circ$ from the positive $x$ -axis. What are the magnitude and direction of $\frac{1}{2}\mathbf{v}$ ?

Magnitude $6$ ; direction $45^\circ$ (correct answer)
Magnitude $24$ ; direction $45^\circ$
Magnitude $6$ ; direction $225^\circ$
Magnitude $12$ ; direction $45^\circ$

Explanation: This question tests understanding of how scalar multiplication affects the magnitude and direction of a vector. When a vector v is multiplied by a scalar c, the magnitude of the result is |cv| = |c|·|v| (the absolute value of c times the magnitude of v), and the direction either stays the same (if c > 0) or reverses 180° (if c < 0). The magnitude is |(1/2)v| = |1/2|·12 = (1/2)·12 = 6, and since c is positive, the direction remains the same as the original vector v's direction of 45° from the positive x-axis. Choice A is correct because it properly applies |cv| = |c|·|v| to get magnitude 6 and correctly identifies the direction remains the same based on the positive sign of c. Choice C incorrectly claims the direction reverses to 225° when c = 1/2, but since c is positive, the direction actually stays the same. Key to scalar multiplication: magnitude always scales by |c| (the absolute value), so |cv| = |c|·|v|, while direction depends on the sign of c—positive preserves direction, negative reverses it. Special scalars to remember: c = 1 (no change), c = -1 (flip direction only), c = 2 (double length, same direction), c = -2 (double length, opposite direction), c = 1/2 (half length, same direction).

Question 2

A force vector $\mathbf{v}$ has magnitude $60\ \text{N}$ directed due north. What are the magnitude and compass direction of $-\tfrac{1}{2}\mathbf{v}$ ?

Magnitude $30\ \text{N}$ ; due north
Magnitude $120\ \text{N}$ ; due south
Magnitude $30\ \text{N}$ ; due south (correct answer)
Magnitude $60\ \text{N}$ ; due south

Explanation: This question tests understanding of how scalar multiplication affects the magnitude and direction of a vector. When a vector v is multiplied by a scalar c, the magnitude of the result is |cv| = |c|·|v| (the absolute value of c times the magnitude of v), and the direction either stays the same (if c > 0) or reverses 180° (if c < 0). The magnitude is |cv| = |-1/2|·60 = (1/2)·60 = 30, and since c is negative, the direction is opposite to the original vector v's direction of due north, which is due south. Choice C is correct because it properly applies |cv| = |c|·|v| and correctly identifies the direction based on the sign of c. Choice A incorrectly claims the direction stays the same when c = -1/2, but since c is negative, the direction actually reverses 180°. Special scalars to remember: c = 1 (no change), c = -1 (flip direction only), c = 2 (double length, same direction), c = -2 (double length, opposite direction), c = 1/2 (half length, same direction). For direction: think of the sign of c as a switch—positive means 'keep the same direction,' negative means 'flip 180° to the opposite direction,' and the magnitude of c only affects how much to scale, not which way to point.

Question 3

A velocity vector $\mathbf{v}$ is $50\ \text{km/h}$ east. What are the magnitude and direction of $-2\mathbf{v}$ ?

$100\ \text{km/h}$ east
$50\ \text{km/h}$ west
$100\ \text{km/h}$ west (correct answer)
$25\ \text{km/h}$ west

Explanation: This question tests understanding of how scalar multiplication affects the magnitude and direction of a vector. When a vector v is multiplied by a scalar c, the magnitude of the result is |cv| = |c|·|v| (the absolute value of c times the magnitude of v), and the direction either stays the same (if c > 0) or reverses 180° (if c < 0). The magnitude is |-2v| = |-2|·50 = 2·50 = 100 km/h, and since c is negative, the direction is opposite to the original vector v's direction, so west instead of east. Choice C is correct because it properly applies |cv| = |c|·|v| to get magnitude 100 and correctly identifies the opposite direction based on the negative sign of c. Choice A incorrectly claims the direction stays the same when c = -2, but since c is negative, the direction actually reverses 180°. For direction: think of the sign of c as a switch—positive means 'keep the same direction,' negative means 'flip 180° to the opposite direction,' and the magnitude of c only affects how much to scale, not which way to point. Special scalars to remember: c = 1 (no change), c = -1 (flip direction only), c = 2 (double length, same direction), c = -2 (double length, opposite direction), c = 1/2 (half length, same direction).

Question 4

Vector $\mathbf{v}$ has magnitude $12$ and direction $45^\circ$ from the positive $x$ -axis. What are the magnitude and direction of $\frac{1}{2}\mathbf{v}$ ?

Magnitude $6$ ; direction $45^\circ$ (correct answer)
Magnitude $24$ ; direction $45^\circ$
Magnitude $6$ ; direction $225^\circ$
Magnitude $12$ ; direction $45^\circ$

Question 5

A velocity vector $\mathbf{v}$ has magnitude $50\ \text{km/h}$ due east. What are the magnitude and compass direction of $-\mathbf{v}$ ?

Magnitude $50\ \text{km/h}$ ; due east
Magnitude $-50\ \text{km/h}$ ; due west
Magnitude $50\ \text{km/h}$ ; due west (correct answer)
Magnitude $0\ \text{km/h}$ ; direction undefined

Explanation: This question tests understanding of how scalar multiplication affects the magnitude and direction of a vector. For direction, the sign of c determines the result: positive scalars preserve the direction of the original vector, while negative scalars reverse it by 180°, making cv point in exactly the opposite direction from v. The magnitude is |cv| = |-1|·50 = 1·50 = 50, and since c is negative, the direction is opposite to the original vector v's direction of due east, which is due west. Choice C is correct because it properly applies |cv| = |c|·|v| and correctly identifies the direction based on the sign of c. Choice B forgets to take the absolute value of c, computing |cv| = c·|v| = -50, but magnitude must always be positive (|cv| = |c|·|v|). Remember: the absolute value in |cv| = |c|·|v| ensures magnitudes are always positive, so even if c = -1, we have |cv| = 1·|v|, not -1·|v|. For direction: think of the sign of c as a switch—positive means 'keep the same direction,' negative means 'flip 180° to the opposite direction,' and the magnitude of c only affects how much to scale, not which way to point.

Question 6

A velocity vector $\mathbf{v}$ is $50\ \text{km/h}$ east. What are the magnitude and direction of $-3\mathbf{v}$ ?

Magnitude $150\ \text{km/h}$ ; direction west (correct answer)
Magnitude $150\ \text{km/h}$ ; direction east
Magnitude $-150\ \text{km/h}$ ; direction west
Magnitude $50\ \text{km/h}$ ; direction west

Explanation: This question tests understanding of how scalar multiplication affects the magnitude and direction of a vector. When a vector v is multiplied by a scalar c, the magnitude of the result is |cv| = |c|·|v| (the absolute value of c times the magnitude of v), and the direction either stays the same (if c > 0) or reverses 180° (if c < 0). The magnitude is |cv| = |-3|·50 = 3·50 = 150, and since c is negative, the direction is opposite to the original vector v's direction of east, so west. Choice A is correct because it properly applies |cv| = |c|·|v| and correctly identifies direction based on sign of c. Choice C forgets to take the absolute value of c, computing |cv| = c·|v| = -150, but magnitude must always be positive (|cv| = |c|·|v|). Key to scalar multiplication: magnitude always scales by |c| (the absolute value), so |cv| = |c|·|v|, while direction depends on the sign of c—positive preserves direction, negative reverses it. For direction: think of the sign of c as a switch—positive means 'keep the same direction,' negative means 'flip 180° to the opposite direction,' and the magnitude of c only affects how much to scale, not which way to point.

Question 7

If a vector $\mathbf{v}$ points at direction $60^\circ$ from the positive $x$ -axis, in what direction does $-\mathbf{v}$ point?

$60^\circ$
$120^\circ$
$240^\circ$ (correct answer)
$300^\circ$

Explanation: This question tests understanding of how scalar multiplication affects the magnitude and direction of a vector. For direction, the sign of c determines the result: positive scalars preserve the direction of the original vector, while negative scalars reverse it by 180°, making cv point in exactly the opposite direction from v. Since the scalar c = -1 is negative, the direction of cv is opposite to v (reversed 180°). Specifically, if v points at 60°, then cv points at 60° + 180° = 240°. Choice C is correct because it correctly identifies direction based on sign of c. Choice A incorrectly claims the direction stays the same when c = -1, but since c is negative, the direction actually reverses 180°. For direction: think of the sign of c as a switch—positive means 'keep the same direction,' negative means 'flip 180° to the opposite direction,' and the magnitude of c only affects how much to scale, not which way to point. Key to scalar multiplication: magnitude always scales by |c| (the absolute value), so |cv| = |c|·|v|, while direction depends on the sign of c—positive preserves direction, negative reverses it.

Question 8

Given $\mathbf{v} = \langle 3,4\rangle$ and scalar $c=2$ , what is the magnitude of $c\mathbf{v}$ ?

$5$
$7$
$10$ (correct answer)
$20$

Explanation: This question tests understanding of how scalar multiplication affects the magnitude and direction of a vector. The formula |cv| = |c|·|v| tells us that the magnitude scales by the absolute value of the scalar: |c| > 1 stretches the vector, 0 < |c| < 1 compresses it, and the absolute value ensures the magnitude is always positive regardless of whether c is positive or negative. Given |v| = √(9 + 16) = √25 = 5 and scalar c = 2, we apply the formula: |cv| = |2|·5 = 2·5 = 10. Choice C is correct because it properly applies |cv| = |c|·|v|. Choice D makes an arithmetic error, calculating 4·5 = 20 instead of 2·5 = 10. When computing from components v = ⟨a, b⟩, remember cv = ⟨ca, cb⟩, and then find magnitude using |cv| = √((ca)² + (cb)²) = |c|√(a² + b²), confirming the formula. Key to scalar multiplication: magnitude always scales by |c| (the absolute value), so |cv| = |c|·|v|, while direction depends on the sign of c—positive preserves direction, negative reverses it.

Question 9

Vector $\mathbf{v}$ has magnitude $10$ and direction $45^\circ$ from the positive $x$ -axis. What is the direction of $-2\mathbf{v}$ (angle from the positive $x$ -axis)?

$45^\circ$
$90^\circ$
$225^\circ$ (correct answer)
$315^\circ$

Explanation: This question tests understanding of how scalar multiplication affects the magnitude and direction of a vector. For direction, the sign of c determines the result: positive scalars preserve the direction of the original vector, while negative scalars reverse it by 180°, making cv point in exactly the opposite direction from v. Since the scalar c = -2 is negative, the direction of cv is opposite to v (reversed 180°). Specifically, if v points 45°, then cv points 45° + 180° = 225°. Choice C is correct because it correctly identifies the direction based on the sign of c. Choice A claims the direction stays the same, ignoring the effect of the negative scalar c = -2. For direction: think of the sign of c as a switch—positive means 'keep the same direction,' negative means 'flip 180° to the opposite direction,' and the magnitude of c only affects how much to scale, not which way to point. Key to scalar multiplication: magnitude always scales by |c| (the absolute value), so |cv| = |c|·|v|, while direction depends on the sign of c—positive preserves direction, negative reverses it.

Question 10

Given vector $\mathbf{v}$ has direction $30^\circ$ from the positive $x$ -axis. In what direction does $-\mathbf{v}$ point (as an angle from the positive $x$ -axis)?

$30^\circ$
$60^\circ$
$150^\circ$
$210^\circ$ (correct answer)

Explanation: This question tests understanding of how scalar multiplication affects the magnitude and direction of a vector. For direction, the sign of c determines the result: positive scalars preserve the direction of the original vector, while negative scalars reverse it by 180°, making cv point in exactly the opposite direction from v. Since the scalar c = -1 is negative, the direction of cv is opposite to v (reversed 180°). Specifically, if v points 30° from the positive x-axis, then cv points 30° + 180° = 210° from the positive x-axis. Choice D is correct because it correctly identifies the direction reverses 180° based on the negative sign of c. Choice A claims the direction stays the same when c = -1, but since c is negative, the direction actually reverses 180°. For direction: think of the sign of c as a switch—positive means 'keep the same direction,' negative means 'flip 180° to the opposite direction,' and the magnitude of c only affects how much to scale, not which way to point. Special scalars to remember: c = 1 (no change), c = -1 (flip direction only), c = 2 (double length, same direction), c = -2 (double length, opposite direction), c = 1/2 (half length, same direction).

Question 11

For vector $\mathbf{v}$ , how does $\|\tfrac{1}{3}\mathbf{v}\|$ compare to $\|\mathbf{v}\|$ ?

$\|\tfrac{1}{3}\mathbf{v}\| = 3\|\mathbf{v}\|$
$\|\tfrac{1}{3}\mathbf{v}\| = \tfrac{1}{3}\|\mathbf{v}\|$ (correct answer)
$\|\tfrac{1}{3}\mathbf{v}\| = \|\mathbf{v}\| + \tfrac{1}{3}$
$\|\tfrac{1}{3}\mathbf{v}\| = \|\mathbf{v}\|$

Explanation: This question tests understanding of how scalar multiplication affects the magnitude and direction of a vector. The formula |cv| = |c|·|v| tells us that the magnitude scales by the absolute value of the scalar: |c| > 1 stretches the vector, 0 < |c| < 1 compresses it, and the absolute value ensures the magnitude is always positive regardless of whether c is positive or negative. Given scalar c = 1/3, we apply the formula: |cv| = |1/3|·|v| = (1/3)·|v|. Choice B is correct because it properly applies |cv| = |c|·|v|. Choice A uses the wrong scaling factor, computing |cv| as 3·|v| instead of |c|·|v|. Key to scalar multiplication: magnitude always scales by |c| (the absolute value), so |cv| = |c|·|v|, while direction depends on the sign of c—positive preserves direction, negative reverses it. Special scalars to remember: c = 1 (no change), c = -1 (flip direction only), c = 2 (double length, same direction), c = -2 (double length, opposite direction), c = 1/2 (half length, same direction).

Question 12

Given vector $\mathbf{v}$ has direction $30^\circ$ from the positive $x$ -axis. In what direction does $-\mathbf{v}$ point (as an angle from the positive $x$ -axis)?

$30^\circ$
$60^\circ$
$150^\circ$
$210^\circ$ (correct answer)

Explanation: This question tests understanding of how scalar multiplication affects the magnitude and direction of a vector. For direction, the sign of c determines the result: positive scalars preserve the direction of the original vector, while negative scalars reverse it by 180°, making cv point in exactly the opposite direction from v. Since the scalar c = -1 is negative, the direction of cv is opposite to v (reversed 180°). Specifically, if v points 30° from the positive x-axis, then cv points 30° + 180° = 210° from the positive x-axis. Choice D is correct because it correctly identifies the direction reverses 180° based on the negative sign of c. Choice A claims the direction stays the same when c = -1, but since c is negative, the direction actually reverses 180°. For direction: think of the sign of c as a switch—positive means 'keep the same direction,' negative means 'flip 180° to the opposite direction,' and the magnitude of c only affects how much to scale, not which way to point. Special scalars to remember: c = 1 (no change), c = -1 (flip direction only), c = 2 (double length, same direction), c = -2 (double length, opposite direction), c = 1/2 (half length, same direction).

Question 13

A velocity vector $\mathbf{v}$ is $50 \text{km/h}$ east. What are the magnitude and direction of $-2\mathbf{v}$ ?

$100 \text{km/h}$ east
$50 \text{km/h}$ west
$100 \text{km/h}$ west (correct answer)
$25 \text{km/h}$ west

|cv| = |c| \cdot |v|

(the absolute value of c times the magnitude of v), and the direction either stays the same (if c > 0) or reverses 180° (if c < 0). The magnitude is

|-2\mathbf{v}| = |-2| \cdot 50 = 2 \cdot 50 = 100

km/h, and since c is negative, the direction is opposite to the original vector v's direction, so west instead of east. Choice C is correct because it properly applies

|cv| = |c| \cdot |v|

to get magnitude 100 and correctly identifies the opposite direction based on the negative sign of c. Choice A incorrectly claims the direction stays the same when c = -2, but since c is negative, the direction actually reverses 180°. For direction: think of the sign of c as a switch—positive means 'keep the same direction,' negative means 'flip 180° to the opposite direction,' and the magnitude of c only affects how much to scale, not which way to point. Special scalars to remember: c = 1 (no change), c = -1 (flip direction only), c = 2 (double length, same direction), c = -2 (double length, opposite direction), c = 1/2 (half length, same direction).

Question 14

A force vector $\mathbf{v}$ has magnitude $18\ \text{N}$ and points north. What are the magnitude and direction of $\tfrac{1}{3}\mathbf{v}$ ?

$6\ \text{N}$ north (correct answer)
$6\ \text{N}$ south
$54\ \text{N}$ north
$18\ \text{N}$ north

Explanation: This question tests understanding of how scalar multiplication affects the magnitude and direction of a vector. When a vector v is multiplied by a scalar c, the magnitude of the result is |cv| = |c|·|v| (the absolute value of c times the magnitude of v), and the direction either stays the same (if c > 0) or reverses 180° (if c < 0). The magnitude is |(1/3)v| = |1/3|·18 = (1/3)·18 = 6 N, and since c is positive, the direction remains the same as the original vector v's direction of north. Choice A is correct because it properly applies |cv| = |c|·|v| to get magnitude 6 and correctly identifies the direction remains the same based on the positive sign of c. Choice B incorrectly claims the direction reverses to south when c = 1/3, but since c is positive, the direction actually stays the same. Key to scalar multiplication: magnitude always scales by |c| (the absolute value), so |cv| = |c|·|v|, while direction depends on the sign of c—positive preserves direction, negative reverses it. Special scalars to remember: c = 1 (no change), c = -1 (flip direction only), c = 2 (double length, same direction), c = -2 (double length, opposite direction), c = 1/2 (half length, same direction).

Question 15

If $\mathbf{v}$ has magnitude $10$ and direction $30^\circ$ from the positive $x$ -axis, what are the magnitude and direction of $2\mathbf{v}$ ?

Magnitude $5$ ; direction $30^\circ$
Magnitude $20$ ; direction $210^\circ$
Magnitude $20$ ; direction $30^\circ$ (correct answer)
Magnitude $12$ ; direction $30^\circ$

Explanation: This question tests understanding of how scalar multiplication affects the magnitude and direction of a vector. When a vector v is multiplied by a scalar c, the magnitude of the result is |cv| = |c|·|v| (the absolute value of c times the magnitude of v), and the direction either stays the same (if c > 0) or reverses 180° (if c < 0). The magnitude is |cv| = |2|·10 = 2·10 = 20, and since c is positive, the direction remains the same as the original vector v's direction of 30°. Choice C is correct because it properly applies |cv| = |c|·|v| and correctly identifies direction based on sign of c. Choice B incorrectly claims the direction reverses to 210° when c = 2, but since c is positive, the direction actually stays the same. Special scalars to remember: c = 1 (no change), c = -1 (flip direction only), c = 2 (double length, same direction), c = -2 (double length, opposite direction), c = 1/2 (half length, same direction). For direction: think of the sign of c as a switch—positive means 'keep the same direction,' negative means 'flip 180° to the opposite direction,' and the magnitude of c only affects how much to scale, not which way to point.

Question 16

Vector $\vec{v}$ has magnitude $8$ and makes an angle of $120°$ with the positive $x$ -axis. If $\vec{w} = -3\vec{v}$ , what is the magnitude of $\vec{w}$ and the angle it makes with the positive $x$ -axis?

Magnitude $24$ , angle $300°$ (correct answer)
Magnitude $24$ , angle $240°$
Magnitude $11$ , angle $300°$
Magnitude $5$ , angle $240°$

Explanation: For a scalar multiple

c\vec{v}

, the magnitude is

|c| \cdot ||\vec{v}||

. Here,

||\vec{w}|| = |-3| \cdot 8 = 24

. Since

c = -3 < 0

, the direction is opposite to

\vec{v}

. The angle of

\vec{v}

120°

, so the angle of

\vec{w}

120° + 180° = 300°

. Choice B uses the wrong direction calculation (

120° + 120° = 240°

). Choice C adds the scalar to the magnitude (

8 + 3 = 11

). Choice D subtracts the scalar from the magnitude (

8 - 3 = 5

Question 17

Given vectors $\vec{u} = \langle 4, -3 \rangle$ and $\vec{r} = k\vec{u}$ where $k < 0$ , if the magnitude of $\vec{r}$ is $15$ , what is the value of $k$ and in which quadrant does $\vec{r}$ point?

$k = -3$ , Quadrant II (correct answer)
$k = 3$ , Quadrant IV
$k = -3$ , Quadrant IV
$k = -\frac{1}{3}$ , Quadrant II

Explanation: First,

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

. Since

||\vec{r}|| = |k| \cdot ||\vec{u}||

, we have

15 = |k| \cdot 5

, so

|k| = 3

. Given

k < 0

, we have

k = -3

. Since

k < 0

\vec{r}

points opposite to

\vec{u}

. Vector

\vec{u}

is in Quadrant IV (positive

x

, negative

y

), so

\vec{r} = -3\langle 4, -3 \rangle = \langle -12, 9 \rangle

is in Quadrant II. Choice B ignores the constraint

k < 0

. Choice C has the wrong quadrant. Choice D incorrectly calculates

k = -\frac{15}{||\vec{u}||^2}

Question 18

A vector $\vec{p}$ has magnitude $6$ and points in the direction of angle $45°$ . Vector $\vec{q} = c\vec{p}$ has the same direction as $\vec{p}$ but twice the magnitude. If $\vec{s} = -\frac{1}{2}\vec{q}$ , what are the magnitude and direction angle of $\vec{s}$ ?

Magnitude $12$ , direction $225°$
Magnitude $3$ , direction $45°$
Magnitude $6$ , direction $45°$
Magnitude $6$ , direction $225°$ (correct answer)

Explanation: When working with vector operations, remember that scalar multiplication affects both magnitude and direction predictably: positive scalars preserve direction while negative scalars reverse it. Let's trace through each step systematically. Vector

\vec{p}

has magnitude 6 and direction 45°. Since

\vec{q} = c\vec{p}

has the same direction but twice the magnitude,

\vec{q}

must have magnitude 12 and direction 45°. This means

c = 2

. Now for

\vec{s} = -\frac{1}{2}\vec{q}

: The scalar

-\frac{1}{2}

has absolute value

\frac{1}{2}

and is negative. The magnitude of

\vec{s}

\frac{1}{2} \times 12 = 6

. Since we're multiplying by a negative scalar, the direction reverses. Adding 180° to the original direction:

45° + 180° = 225°

. Looking at the wrong answers: Choice A gives the correct direction (225°) but incorrectly calculates magnitude as 12 - this ignores the

\frac{1}{2}

factor. Choice B has magnitude 3 (which would be

\frac{1}{2} \times 6

, incorrectly using

\vec{p}

's magnitude instead of

\vec{q}

's) and direction 45°, missing the sign reversal entirely. Choice C gives magnitude 6 but direction 45°, correctly finding the magnitude but forgetting that negative scalars reverse direction. Study tip: When multiplying vectors by scalars, handle magnitude and direction separately. The magnitude gets multiplied by the absolute value of the scalar, while negative scalars always add 180° to the direction angle.

Question 19

Two vectors $\vec{u}$ and $\vec{v} = -2.5\vec{u}$ are given. If the angle between $\vec{u}$ and the positive $x$ -axis is $\theta$ , and $||\vec{u}|| = 4$ , which statement about $\vec{v}$ is correct?

$||\vec{v}|| = 10$ and $\vec{v}$ makes angle $\theta$ with positive $x$ -axis
$||\vec{v}|| = 6.5$ and $\vec{v}$ makes angle $\theta + 180°$ with positive $x$ -axis
$||\vec{v}|| = 10$ and $\vec{v}$ makes angle $\theta + 180°$ with positive $x$ -axis (correct answer)
$||\vec{v}|| = 1.5$ and $\vec{v}$ makes angle $\theta + 180°$ with positive $x$ -axis

Explanation: When you encounter vector scaling problems, focus on two key effects: how scalar multiplication affects magnitude and direction. Let's analyze what happens when

\vec{v} = -2.5\vec{u}

. First, find the magnitude of

\vec{v}

||\vec{v}|| = ||-2.5\vec{u}|| = |-2.5| \cdot ||\vec{u}|| = 2.5 \times 4 = 10

The absolute value of the scalar gives us the magnitude scaling factor. Next, consider the direction. Since we're multiplying by a negative scalar (-2.5), vector

\vec{v}

points in the opposite direction from

\vec{u}

. If

\vec{u}

makes angle

\theta

with the positive x-axis, then

\vec{v}

makes angle

\theta + 180°

(or

\theta + \pi

radians). Now examine each choice: Choice A incorrectly states that

\vec{v}

makes the same angle

\theta

\vec{u}

. This ignores the negative scalar's effect on direction. Choice B has the wrong magnitude calculation:

6.5 \neq 2.5 \times 4

. This appears to come from incorrectly adding rather than multiplying:

4 + 2.5 = 6.5

. Choice C correctly identifies both the magnitude (10) and the direction (

\theta + 180°

). Choice D has completely incorrect magnitude (1.5), possibly from subtracting:

4 - 2.5 = 1.5

. Study tip: Remember that scalar multiplication affects magnitude by the absolute value of the scalar, while negative scalars flip the vector's direction by 180°. Always use multiplication for magnitude scaling, never addition or subtraction.

Question 20

Given the vector $\mathbf{v}=\langle 3,4\rangle$ and scalar $c=-2$ , what are the magnitude and direction of $c\mathbf{v}$ relative to $\mathbf{v}$ ?

Magnitude $10$ ; same direction as $\mathbf{v}$
Magnitude $10$ ; opposite direction to $\mathbf{v}$ (correct answer)
Magnitude $-10$ ; opposite direction to $\mathbf{v}$
Magnitude $5$ ; opposite direction to $\mathbf{v}$

Explanation: This question tests understanding of how scalar multiplication affects the magnitude and direction of a vector. When a vector v is multiplied by a scalar c, the magnitude of the result is |cv| = |c|·|v| (the absolute value of c times the magnitude of v), and the direction either stays the same (if c > 0) or reverses 180° (if c < 0). The magnitude is |cv| = |-2|·|⟨3,4⟩| = 2·5 = 10, and since c is negative, the direction is opposite to the original vector v's direction of northeast in the first quadrant. Choice B is correct because it properly applies |cv| = |c|·|v| and correctly identifies direction based on sign of c. Choice C forgets to take the absolute value of c, computing |cv| = c·|v| = -10, but magnitude must always be positive (|cv| = |c|·|v|). Key to scalar multiplication: magnitude always scales by |c| (the absolute value), so |cv| = |c|·|v|, while direction depends on the sign of c—positive preserves direction, negative reverses it. Remember: the absolute value in |cv| = |c|·|v| ensures magnitudes are always positive, so even if c = -3, we have |cv| = 3|v|, not -3|v|.