Precalculus Quiz: Magnitude And Direction Of Scaled Vectors

20 questions · exam conditions

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Magnitude And Direction Of Scaled VectorsQuestion 1 of 20

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°

Magnitude

24

, angle

240°

Magnitude

11

, angle

300°

Magnitude

5

, angle

240°

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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

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 2

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 3

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 4

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 5

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|.

Question 6

Vector $\mathbf{v}$ has magnitude $10$ . How does $\| -\tfrac{1}{2}\mathbf{v} \|$ compare to $\|\mathbf{v}\|$ ?

$\| -\tfrac{1}{2}\mathbf{v} \| = 20$
$\| -\tfrac{1}{2}\mathbf{v} \| = 10$
$\| -\tfrac{1}{2}\mathbf{v} \| = 5$ (correct answer)
$\| -\tfrac{1}{2}\mathbf{v} \| = -5$

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| = 10 and scalar c = -1/2, we apply the formula: |cv| = |-1/2|·10 = (1/2)·10 = 5. Choice C is correct because it properly applies |cv| = |c|·|v| to get 5. Choice D forgets to take the absolute value of c, computing |cv| = c·|v| = -5, 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 = -1/2, we have |cv| = (1/2)·|v|, not - (1/2)·|v|. 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 7

Vector $\vec{a} = \langle -1, 3 \rangle$ is scaled by factor $m$ to produce vector $\vec{b} = m\vec{a}$ . If $\vec{b}$ has magnitude $10\sqrt{10}$ and points in the same general direction as $\vec{a}$ , what is the sum of the components of $\vec{b}$ ?

$4$
$-20$
$10$
$20$ (correct answer)

Explanation: When you encounter vector scaling problems, remember that scalar multiplication affects both magnitude and direction. A positive scalar preserves direction, while a negative scalar reverses it. First, let's find the magnitude of vector

\vec{a} = \langle -1, 3 \rangle

. Using the magnitude formula:

|\vec{a}| = \sqrt{(-1)^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10}

. Since

\vec{b} = m\vec{a}

, we have

|\vec{b}| = |m| \cdot |\vec{a}|

. Given that

|\vec{b}| = 10\sqrt{10}

10\sqrt{10} = |m| \cdot \sqrt{10}

|m| = 10

The key insight is that

\vec{b}

points in the same direction as

\vec{a}

. Since scalar multiplication by a positive number preserves direction, we need

m = +10

(not

m = -10

, which would reverse direction). Therefore:

\vec{b} = 10 \langle -1, 3 \rangle = \langle -10, 30 \rangle

The sum of components is

-10 + 30 = 20

. Looking at the wrong answers: Choice A (

4

) likely comes from incorrectly calculating the original vector's component sum (

-1 + 3 = 2

) and making computational errors. Choice B (

-20

) results from using

m = -10

, ignoring the "same direction" constraint. Choice C (

10

) might come from confusing the scaling factor with the final answer. Strategy tip: Always check direction constraints carefully. "Same direction" means the scalar must be positive, while "opposite direction" requires a negative scalar. The phrase "general direction" is key to determining the sign of your scaling factor.

Question 8

Vector $\vec{p} = \langle 8, 6 \rangle$ is scaled by a factor $c$ to produce $\vec{q} = c\vec{p}$ . If the magnitude of $\vec{q}$ is $5$ and $\vec{q}$ points into the third quadrant, what is the $y$ -component of $\vec{q}$ ?

$3$
$-3$ (correct answer)
$-4$
$-2.4$

Explanation: When you see vector scaling problems, you're working with the fundamental relationship that scaling a vector by factor

c

multiplies both its components and its magnitude by

|c|

. The key insight is determining whether the scaling factor is positive or negative based on the quadrant information. Start by finding the magnitude of the original vector:

|\vec{p}| = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = 10

. Since

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

has magnitude 5, we know

|c| \cdot 10 = 5

, so

|c| = 0.5

. Now for the crucial step: the original vector

\vec{p} = \langle 8, 6 \rangle

points into the first quadrant (both components positive), but

\vec{q}

points into the third quadrant (both components negative). This means

c

must be negative, so

c = -0.5

. Therefore:

\vec{q} = -0.5 \langle 8, 6 \rangle = \langle -4, -3 \rangle

. The

y

-component is

-3

. Choice A gives

3

, which would be correct if you forgot that the vector points into the third quadrant and used

c = +0.5

. Choice C gives

-4

, which is actually the

x

-component of

\vec{q}

—a common mix-up. Choice D gives

-2.4

, which might result from incorrectly calculating the scaling factor or confusing the relationship between components. Strategy tip: Always check quadrants carefully in vector problems. When a scaled vector changes quadrants from the original, the scaling factor must be negative, which flips the signs of all components.

Question 9

Consider vectors $\vec{u} = \langle 3, -4 \rangle$ and $\vec{v} = k\vec{u}$ where $k \neq 0$ . If the dot product $\vec{u} \cdot \vec{v} = -100$ , what is the magnitude of $\vec{v}$ and in which direction does it point relative to $\vec{u}$ ?

Magnitude $20$ , same direction as $\vec{u}$
Magnitude $20$ , opposite direction to $\vec{u}$ (correct answer)
Magnitude $4$ , opposite direction to $\vec{u}$
Magnitude $25$ , opposite direction to $\vec{u}$

Explanation: When you see vectors where one is a scalar multiple of another, you're dealing with parallel vectors that either point in the same direction or opposite directions. The key is using the dot product formula and understanding what the sign tells you about direction. Since

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

, we have

\vec{v} = k\langle 3, -4 \rangle = \langle 3k, -4k \rangle

. The dot product becomes:

\vec{u} \cdot \vec{v} = \langle 3, -4 \rangle \cdot \langle 3k, -4k \rangle = 3(3k) + (-4)(-4k) = 9k + 16k = 25k

Setting this equal to the given value:

25k = -100

, so

k = -4

. Since

k < 0

, vector

\vec{v}

points in the opposite direction to

\vec{u}

. We have

\vec{v} = -4\langle 3, -4 \rangle = \langle -12, 16 \rangle

, giving us magnitude

|\vec{v}| = \sqrt{(-12)^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20

. Choice A gives the correct magnitude but wrong direction—it ignores that negative

k

means opposite direction. Choice C has the wrong magnitude (likely confusing the absolute value of

k

with the vector magnitude) but correct direction. Choice D has the wrong magnitude—this might come from mistakenly using

|\vec{u}| = 5

and multiplying by

|k| = 4

incorrectly, but gets the direction right. The correct answer is B: magnitude 20, opposite direction. Remember: when one vector is a scalar multiple of another, the sign of the scalar determines direction (negative means opposite), while the dot product can help you find that scalar efficiently.

Question 10

A vector $\vec{r}$ has magnitude $7$ and direction angle $150°$ . Vector $\vec{s} = \frac{2}{7}\vec{r}$ is then scaled by factor $-1.5$ to produce vector $\vec{t}$ . What is the magnitude of $\vec{t}$ and its direction angle?

Magnitude $3$ , direction $330°$ (correct answer)
Magnitude $3$ , direction $150°$
Magnitude $21$ , direction $330°$
Magnitude $1.5$ , direction $330°$

Explanation: First,

||\vec{s}|| = \frac{2}{7} \cdot 7 = 2

. Since the scalar

\frac{2}{7} > 0

\vec{s}

has the same direction as

\vec{r}

, so direction angle is

150°

. Then

\vec{t} = -1.5\vec{s}

, so

||\vec{t}|| = |-1.5| \cdot 2 = 3

. Since

-1.5 < 0

\vec{t}

points opposite to

\vec{s}

. The direction angle of

\vec{t}

150° + 180° = 330°

. Choice B forgets the direction reversal from the negative scalar. Choice C incorrectly multiplies magnitudes (

7 \times 3 = 21

). Choice D uses only the magnitude of the final scalar factor.

Question 11

Vector $\vec{w} = \langle 5, -12 \rangle$ is transformed to $\vec{z} = k\vec{w}$ where $k$ is a real number. If $\vec{z}$ has magnitude $39$ and the $x$ -component of $\vec{z}$ is negative, what are the components of $\vec{z}$ ?

$\langle -39, 0 \rangle$
$\langle 15, -36 \rangle$
$\langle -15, 36 \rangle$ (correct answer)
$\langle -13, 36 \rangle$

Explanation: When you see scalar multiplication of vectors, remember that multiplying a vector by scalar

k

scales both components by

k

and changes the magnitude by a factor of

|k|

. First, find the magnitude of the original vector

\vec{w} = \langle 5, -12 \rangle

|\vec{w}| = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13

Since

\vec{z} = k\vec{w}

, we have

|\vec{z}| = |k| \cdot |\vec{w}|

. Given that

|\vec{z}| = 39

39 = |k| \cdot 13

|k| = 3

This means

k = 3

k = -3

. Since the problem states that the

x

-component of

\vec{z}

is negative, and the

x

-component of

\vec{w}

is positive (5), we need

k < 0

. Therefore,

k = -3

. So

\vec{z} = -3\langle 5, -12 \rangle = \langle -15, 36 \rangle

. Choice A

\langle -39, 0 \rangle

has the wrong direction entirely—it lies on the

x

-axis rather than being a scalar multiple of

\vec{w}

. Choice B

\langle 15, -36 \rangle

corresponds to

k = 3

, which would make the

x

-component positive, contradicting the given condition. Choice D

\langle -13, 36 \rangle

uses the original magnitude (13) as the

x

-component rather than applying the scalar correctly. The correct answer is C. Study tip: Always check that your scalar multiple maintains the direction relationship between components. If the original vector points "right and down," a negative scalar should make it point "left and up."

Question 12

Which statement correctly describes the vector $c\mathbf{v}$ when $c=-1$ and $\mathbf{v}$ is any nonzero vector?

$c\mathbf{v}$ has the same magnitude as $\mathbf{v}$ and the same direction as $\mathbf{v}$
$c\mathbf{v}$ has twice the magnitude of $\mathbf{v}$ and the opposite direction
$c\mathbf{v}$ has magnitude $0$ and direction undefined
$c\mathbf{v}$ has the same magnitude as $\mathbf{v}$ and the opposite direction (correct answer)

Explanation: This question tests understanding of how scalar multiplication affects the magnitude and direction of a vector. For special scalar c = -1, the magnitude stays the same because |c| = 1, giving |cv| = 1·|v| = |v|, and the direction reverses because c is negative. Since the scalar c = -1 is negative, the direction of cv is opposite to v (reversed 180°), while the magnitude remains |v|. Choice D is correct because it correctly states both magnitude and direction. Choice A incorrectly claims the direction stays the same when c = -1, 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). 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 13

Given $\mathbf{v}=\langle 3,4\rangle$ and scalar $c=\tfrac{1}{2}$ , what are the magnitude and direction of $c\mathbf{v}$ relative to $\mathbf{v}$ ?

Magnitude $\tfrac{5}{2}$ ; same direction as $\mathbf{v}$ (correct answer)
Magnitude $5$ ; same direction as $\mathbf{v}$
Magnitude $\tfrac{5}{2}$ ; opposite direction to $\mathbf{v}$
Magnitude $\tfrac{1}{2}$ ; same direction as $\mathbf{v}$

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. The magnitude is |cv| = |1/2|·|⟨3,4⟩| = (1/2)·5 = 5/2, and since c is positive, the direction remains the same as the original vector v's direction. Choice A is correct because it properly applies |cv| = |c|·|v| and correctly identifies direction based on sign of c. Choice B uses the wrong scaling factor, computing |cv| as 1·|v| = 5 instead of (1/2)·|v|. 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). 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 14

For vector $\mathbf{v} = \langle 5,12\rangle$ and scalar $c=\tfrac{1}{2}$ , what is the magnitude of $c\mathbf{v}$ ?

$\tfrac{13}{2}$ (correct answer)
$26$
$\tfrac{13}{4}$
$13$

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| = √(25 + 144) = √169 = 13 and scalar c = 1/2, we apply the formula: |cv| = |1/2|·13 = (1/2)·13 = 13/2. Choice A is correct because it properly applies |cv| = |c|·|v|. Choice B uses the wrong scaling factor, computing |cv| as (1/2)·|v| but then doubling or miscalculating to 26 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. 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.

Question 15

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 16

Vector $\mathbf{v}$ has magnitude $10$ . How does $\| -\tfrac{1}{2}\mathbf{v} \|$ compare to $\|\mathbf{v}\|$ ?

$\| -\tfrac{1}{2}\mathbf{v} \| = 20$
$\| -\tfrac{1}{2}\mathbf{v} \| = 10$
$\| -\tfrac{1}{2}\mathbf{v} \| = 5$ (correct answer)
$\| -\tfrac{1}{2}\mathbf{v} \| = -5$

Question 17

Given $\mathbf{v}=\langle 3,4\rangle$ and scalar $c=-1$ , which statement correctly describes $c\mathbf{v}$ ?

Same magnitude as $\mathbf{v}$ and same direction as $\mathbf{v}$
Double the magnitude of $\mathbf{v}$ and opposite direction to $\mathbf{v}$
Same magnitude as $\mathbf{v}$ and opposite direction to $\mathbf{v}$ (correct answer)
Half the magnitude of $\mathbf{v}$ and same direction as $\mathbf{v}$

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). For scalar c = -1, the magnitude stays the same because |c| = 1, giving |cv| = 1·|v| = |v|, and the direction reverses because c is negative. Choice C is correct because it correctly states both magnitude and direction. Choice A incorrectly claims the direction stays the same when c = -1, 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). 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 18

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 19

Given $\mathbf{v}=\langle 3,4\rangle$ and scalar $c=-1$ , which statement correctly describes $c\mathbf{v}$ ?

Same magnitude as $\mathbf{v}$ and same direction as $\mathbf{v}$
Double the magnitude of $\mathbf{v}$ and opposite direction to $\mathbf{v}$
Same magnitude as $\mathbf{v}$ and opposite direction to $\mathbf{v}$ (correct answer)
Half the magnitude of $\mathbf{v}$ and same direction as $\mathbf{v}$

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). For scalar c = -1, the magnitude stays the same because |c| = 1, giving |cv| = 1·|v| = |v|, and the direction reverses because c is negative. Choice C is correct because it correctly states both magnitude and direction. Choice A incorrectly claims the direction stays the same when c = -1, 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). 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 20

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.