Home

Tutoring

Subjects

Live Classes

Study Coach

Essay Review

On-Demand Courses

Colleges

Games


Sign up

Log in

Opening subject page...

Loading your content

Practice

  • All Subjects
  • Algebra Flashcards
  • SAT Math Practice Tests
  • Math Question of the Day
  • Live Classes
  • On-Demand Courses

Varsity Tutors

  • Find a Tutor
  • Test Prep
  • Online Classes
  • K-12 Learning
  • College Search
  • VarsityTutors.com

© 2026 Varsity Tutors. All rights reserved.

← 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.

Question 1 / 20

0 of 20 answered

Vector v⃗\vec{v}v has magnitude 888 and makes an angle of 120°120°120° with the positive xxx-axis. If w⃗=−3v⃗\vec{w} = -3\vec{v}w=−3v, what is the magnitude of w⃗\vec{w}w and the angle it makes with the positive xxx-axis?

Select an answer to continue

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 v⃗\vec{v}v has magnitude 888 and makes an angle of 120°120°120° with the positive xxx-axis. If w⃗=−3v⃗\vec{w} = -3\vec{v}w=−3v, what is the magnitude of w⃗\vec{w}w and the angle it makes with the positive xxx-axis?

  1. Magnitude 242424, angle 300°300°300° (correct answer)
  2. Magnitude 242424, angle 240°240°240°
  3. Magnitude 111111, angle 300°300°300°
  4. Magnitude 555, angle 240°240°240°

Explanation: For a scalar multiple cv⃗c\vec{v}cv, the magnitude is ∣c∣⋅∣∣v⃗∣∣|c| \cdot ||\vec{v}||∣c∣⋅∣∣v∣∣. Here, ∣∣w⃗∣∣=∣−3∣⋅8=24||\vec{w}|| = |-3| \cdot 8 = 24∣∣w∣∣=∣−3∣⋅8=24. Since c=−3<0c = -3 < 0c=−3<0, the direction is opposite to v⃗\vec{v}v. The angle of v⃗\vec{v}v is 120°120°120°, so the angle of w⃗\vec{w}w is 120°+180°=300°120° + 180° = 300°120°+180°=300°. Choice B uses the wrong direction calculation (120°+120°=240°120° + 120° = 240°120°+120°=240°). Choice C adds the scalar to the magnitude (8+3=118 + 3 = 118+3=11). Choice D subtracts the scalar from the magnitude (8−3=58 - 3 = 58−3=5).

Question 2

Given vectors u⃗=⟨4,−3⟩\vec{u} = \langle 4, -3 \rangleu=⟨4,−3⟩ and r⃗=ku⃗\vec{r} = k\vec{u}r=ku where k<0k < 0k<0, if the magnitude of r⃗\vec{r}r is 151515, what is the value of kkk and in which quadrant does r⃗\vec{r}r point?

  1. k=−3k = -3k=−3, Quadrant II (correct answer)
  2. k=3k = 3k=3, Quadrant IV
  3. k=−3k = -3k=−3, Quadrant IV
  4. k=−13k = -\frac{1}{3}k=−31​, Quadrant II

Explanation: First, ∣∣u⃗∣∣=42+(−3)2=5||\vec{u}|| = \sqrt{4^2 + (-3)^2} = 5∣∣u∣∣=42+(−3)2​=5. Since ∣∣r⃗∣∣=∣k∣⋅∣∣u⃗∣∣||\vec{r}|| = |k| \cdot ||\vec{u}||∣∣r∣∣=∣k∣⋅∣∣u∣∣, we have 15=∣k∣⋅515 = |k| \cdot 515=∣k∣⋅5, so ∣k∣=3|k| = 3∣k∣=3. Given k<0k < 0k<0, we have k=−3k = -3k=−3. Since k<0k < 0k<0, r⃗\vec{r}r points opposite to u⃗\vec{u}u. Vector u⃗\vec{u}u is in Quadrant IV (positive xxx, negative yyy), so r⃗=−3⟨4,−3⟩=⟨−12,9⟩\vec{r} = -3\langle 4, -3 \rangle = \langle -12, 9 \rangler=−3⟨4,−3⟩=⟨−12,9⟩ is in Quadrant II. Choice B ignores the constraint k<0k < 0k<0. Choice C has the wrong quadrant. Choice D incorrectly calculates k=−15∣∣u⃗∣∣2k = -\frac{15}{||\vec{u}||^2}k=−∣∣u∣∣215​.

Question 3

A vector p⃗\vec{p}p​ has magnitude 666 and points in the direction of angle 45°45°45°. Vector q⃗=cp⃗\vec{q} = c\vec{p}q​=cp​ has the same direction as p⃗\vec{p}p​ but twice the magnitude. If s⃗=−12q⃗\vec{s} = -\frac{1}{2}\vec{q}s=−21​q​, what are the magnitude and direction angle of s⃗\vec{s}s?

  1. Magnitude 121212, direction 225°225°225°
  2. Magnitude 333, direction 45°45°45°
  3. Magnitude 666, direction 45°45°45°
  4. Magnitude 666, direction 225°225°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 p⃗\vec{p}p​ has magnitude 6 and direction 45°. Since q⃗=cp⃗\vec{q} = c\vec{p}q​=cp​ has the same direction but twice the magnitude, q⃗\vec{q}q​ must have magnitude 12 and direction 45°. This means c=2c = 2c=2. Now for s⃗=−12q⃗\vec{s} = -\frac{1}{2}\vec{q}s=−21​q​: The scalar −12-\frac{1}{2}−21​ has absolute value 12\frac{1}{2}21​ and is negative. The magnitude of s⃗\vec{s}s is 12×12=6\frac{1}{2} \times 12 = 621​×12=6. Since we're multiplying by a negative scalar, the direction reverses. Adding 180° to the original direction: 45°+180°=225°45° + 180° = 225°45°+180°=225°. Looking at the wrong answers: Choice A gives the correct direction (225°) but incorrectly calculates magnitude as 12 - this ignores the 12\frac{1}{2}21​ factor. Choice B has magnitude 3 (which would be 12×6\frac{1}{2} \times 621​×6, incorrectly using p⃗\vec{p}p​'s magnitude instead of q⃗\vec{q}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 u⃗\vec{u}u and v⃗=−2.5u⃗\vec{v} = -2.5\vec{u}v=−2.5u are given. If the angle between u⃗\vec{u}u and the positive xxx-axis is θ\thetaθ, and ∣∣u⃗∣∣=4||\vec{u}|| = 4∣∣u∣∣=4, which statement about v⃗\vec{v}v is correct?

  1. ∣∣v⃗∣∣=10||\vec{v}|| = 10∣∣v∣∣=10 and v⃗\vec{v}v makes angle θ\thetaθ with positive xxx-axis
  2. ∣∣v⃗∣∣=6.5||\vec{v}|| = 6.5∣∣v∣∣=6.5 and v⃗\vec{v}v makes angle θ+180°\theta + 180°θ+180° with positive xxx-axis
  3. ∣∣v⃗∣∣=10||\vec{v}|| = 10∣∣v∣∣=10 and v⃗\vec{v}v makes angle θ+180°\theta + 180°θ+180° with positive xxx-axis (correct answer)
  4. ∣∣v⃗∣∣=1.5||\vec{v}|| = 1.5∣∣v∣∣=1.5 and v⃗\vec{v}v makes angle θ+180°\theta + 180°θ+180° with positive xxx-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 v⃗=−2.5u⃗\vec{v} = -2.5\vec{u}v=−2.5u. First, find the magnitude of v⃗\vec{v}v: ∣∣v⃗∣∣=∣∣−2.5u⃗∣∣=∣−2.5∣⋅∣∣u⃗∣∣=2.5×4=10||\vec{v}|| = ||-2.5\vec{u}|| = |-2.5| \cdot ||\vec{u}|| = 2.5 \times 4 = 10∣∣v∣∣=∣∣−2.5u∣∣=∣−2.5∣⋅∣∣u∣∣=2.5×4=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 v⃗\vec{v}v points in the opposite direction from u⃗\vec{u}u. If u⃗\vec{u}u makes angle θ\thetaθ with the positive x-axis, then v⃗\vec{v}v makes angle θ+180°\theta + 180°θ+180° (or θ+π\theta + \piθ+π radians). Now examine each choice: Choice A incorrectly states that v⃗\vec{v}v makes the same angle θ\thetaθ as u⃗\vec{u}u. This ignores the negative scalar's effect on direction. Choice B has the wrong magnitude calculation: 6.5≠2.5×46.5 \neq 2.5 \times 46.5=2.5×4. This appears to come from incorrectly adding rather than multiplying: 4+2.5=6.54 + 2.5 = 6.54+2.5=6.5. Choice C correctly identifies both the magnitude (10) and the direction (θ+180°\theta + 180°θ+180°). Choice D has completely incorrect magnitude (1.5), possibly from subtracting: 4−2.5=1.54 - 2.5 = 1.54−2.5=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 v=⟨3,4⟩\mathbf{v}=\langle 3,4\ranglev=⟨3,4⟩ and scalar c=−1c=-1c=−1, which statement correctly describes cvc\mathbf{v}cv?​

  1. Same magnitude as v\mathbf{v}v and same direction as v\mathbf{v}v
  2. Double the magnitude of v\mathbf{v}v and opposite direction to v\mathbf{v}v
  3. Same magnitude as v\mathbf{v}v and opposite direction to v\mathbf{v}v (correct answer)
  4. Half the magnitude of v\mathbf{v}v and same direction as v\mathbf{v}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 6

A force vector v\mathbf{v}v has magnitude 60 N60\ \text{N}60 N directed due north. What are the magnitude and compass direction of −12v-\tfrac{1}{2}\mathbf{v}−21​v?

  1. Magnitude 30 N30\ \text{N}30 N; due north
  2. Magnitude 120 N120\ \text{N}120 N; due south
  3. Magnitude 30 N30\ \text{N}30 N; due south (correct answer)
  4. Magnitude 60 N60\ \text{N}60 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 7

Vector v\mathbf{v}v has magnitude 121212 and direction 45∘45^\circ45∘ from the positive xxx-axis. What are the magnitude and direction of 12v\frac{1}{2}\mathbf{v}21​v?​

  1. Magnitude 666; direction 45∘45^\circ45∘ (correct answer)
  2. Magnitude 242424; direction 45∘45^\circ45∘
  3. Magnitude 666; direction 225∘225^\circ225∘
  4. Magnitude 121212; direction 45∘45^\circ45∘

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 8

Vector v\mathbf{v}v has magnitude 121212 and direction 45∘45^\circ45∘ from the positive xxx-axis. What are the magnitude and direction of 12v\frac{1}{2}\mathbf{v}21​v?

  1. Magnitude 666; direction 45∘45^\circ45∘ (correct answer)
  2. Magnitude 242424; direction 45∘45^\circ45∘
  3. Magnitude 666; direction 225∘225^\circ225∘
  4. Magnitude 121212; direction 45∘45^\circ45∘

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 9

Given the vector v=⟨3,4⟩\mathbf{v}=\langle 3,4\ranglev=⟨3,4⟩ and scalar c=−2c=-2c=−2, what are the magnitude and direction of cvc\mathbf{v}cv relative to v\mathbf{v}v?

  1. Magnitude 101010; same direction as v\mathbf{v}v
  2. Magnitude 101010; opposite direction to v\mathbf{v}v (correct answer)
  3. Magnitude −10-10−10; opposite direction to v\mathbf{v}v
  4. Magnitude 555; opposite direction to v\mathbf{v}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 10

Vector v\mathbf{v}v has magnitude 101010. How does ∥−12v∥\| -\tfrac{1}{2}\mathbf{v} \|∥−21​v∥ compare to ∥v∥\|\mathbf{v}\|∥v∥?

  1. ∥−12v∥=20\| -\tfrac{1}{2}\mathbf{v} \| = 20∥−21​v∥=20
  2. ∥−12v∥=10\| -\tfrac{1}{2}\mathbf{v} \| = 10∥−21​v∥=10
  3. ∥−12v∥=5\| -\tfrac{1}{2}\mathbf{v} \| = 5∥−21​v∥=5 (correct answer)
  4. ∥−12v∥=−5\| -\tfrac{1}{2}\mathbf{v} \| = -5∥−21​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 11

A velocity vector v\mathbf{v}v is 50 km/h50\ \text{km/h}50 km/h east. What are the magnitude and direction of −3v-3\mathbf{v}−3v?

  1. Magnitude 150 km/h150\ \text{km/h}150 km/h; direction west (correct answer)
  2. Magnitude 150 km/h150\ \text{km/h}150 km/h; direction east
  3. Magnitude −150 km/h-150\ \text{km/h}−150 km/h; direction west
  4. Magnitude 50 km/h50\ \text{km/h}50 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 12

Vector a⃗=⟨−1,3⟩\vec{a} = \langle -1, 3 \ranglea=⟨−1,3⟩ is scaled by factor mmm to produce vector b⃗=ma⃗\vec{b} = m\vec{a}b=ma. If b⃗\vec{b}b has magnitude 101010\sqrt{10}1010​ and points in the same general direction as a⃗\vec{a}a, what is the sum of the components of b⃗\vec{b}b?

  1. 444
  2. −20-20−20
  3. 101010
  4. 202020 (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 a⃗=⟨−1,3⟩\vec{a} = \langle -1, 3 \ranglea=⟨−1,3⟩. Using the magnitude formula: ∣a⃗∣=(−1)2+32=1+9=10|\vec{a}| = \sqrt{(-1)^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10}∣a∣=(−1)2+32​=1+9​=10​. Since b⃗=ma⃗\vec{b} = m\vec{a}b=ma, we have ∣b⃗∣=∣m∣⋅∣a⃗∣|\vec{b}| = |m| \cdot |\vec{a}|∣b∣=∣m∣⋅∣a∣. Given that ∣b⃗∣=1010|\vec{b}| = 10\sqrt{10}∣b∣=1010​: 1010=∣m∣⋅1010\sqrt{10} = |m| \cdot \sqrt{10}1010​=∣m∣⋅10​ ∣m∣=10|m| = 10∣m∣=10 The key insight is that b⃗\vec{b}b points in the same direction as a⃗\vec{a}a. Since scalar multiplication by a positive number preserves direction, we need m=+10m = +10m=+10 (not m=−10m = -10m=−10, which would reverse direction). Therefore: b⃗=10⟨−1,3⟩=⟨−10,30⟩\vec{b} = 10 \langle -1, 3 \rangle = \langle -10, 30 \rangleb=10⟨−1,3⟩=⟨−10,30⟩ The sum of components is −10+30=20-10 + 30 = 20−10+30=20. Looking at the wrong answers: Choice A (444) likely comes from incorrectly calculating the original vector's component sum (−1+3=2-1 + 3 = 2−1+3=2) and making computational errors. Choice B (−20-20−20) results from using m=−10m = -10m=−10, ignoring the "same direction" constraint. Choice C (101010) 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 13

If a vector v\mathbf{v}v points at direction 60∘60^\circ60∘ from the positive xxx-axis, in what direction does −v-\mathbf{v}−v point?

  1. 60∘60^\circ60∘
  2. 120∘120^\circ120∘
  3. 240∘240^\circ240∘ (correct answer)
  4. 300∘300^\circ300∘

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 14

Vector p⃗=⟨8,6⟩\vec{p} = \langle 8, 6 \ranglep​=⟨8,6⟩ is scaled by a factor ccc to produce q⃗=cp⃗\vec{q} = c\vec{p}q​=cp​. If the magnitude of q⃗\vec{q}q​ is 555 and q⃗\vec{q}q​ points into the third quadrant, what is the yyy-component of q⃗\vec{q}q​?

  1. 333
  2. −3-3−3 (correct answer)
  3. −4-4−4
  4. −2.4-2.4−2.4

Explanation: When you see vector scaling problems, you're working with the fundamental relationship that scaling a vector by factor ccc multiplies both its components and its magnitude by ∣c∣|c|∣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: ∣p⃗∣=82+62=64+36=10|\vec{p}| = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = 10∣p​∣=82+62​=64+36​=10. Since q⃗=cp⃗\vec{q} = c\vec{p}q​=cp​ has magnitude 5, we know ∣c∣⋅10=5|c| \cdot 10 = 5∣c∣⋅10=5, so ∣c∣=0.5|c| = 0.5∣c∣=0.5. Now for the crucial step: the original vector p⃗=⟨8,6⟩\vec{p} = \langle 8, 6 \ranglep​=⟨8,6⟩ points into the first quadrant (both components positive), but q⃗\vec{q}q​ points into the third quadrant (both components negative). This means ccc must be negative, so c=−0.5c = -0.5c=−0.5. Therefore: q⃗=−0.5⟨8,6⟩=⟨−4,−3⟩\vec{q} = -0.5 \langle 8, 6 \rangle = \langle -4, -3 \rangleq​=−0.5⟨8,6⟩=⟨−4,−3⟩. The yyy-component is −3-3−3. Choice A gives 333, which would be correct if you forgot that the vector points into the third quadrant and used c=+0.5c = +0.5c=+0.5. Choice C gives −4-4−4, which is actually the xxx-component of q⃗\vec{q}q​—a common mix-up. Choice D gives −2.4-2.4−2.4, 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 15

Given v=⟨3,4⟩\mathbf{v} = \langle 3,4\ranglev=⟨3,4⟩ and scalar c=2c=2c=2, what is the magnitude of cvc\mathbf{v}cv?

  1. 555
  2. 777
  3. 101010 (correct answer)
  4. 202020

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 16

Vector v\mathbf{v}v has magnitude 101010 and direction 45∘45^\circ45∘ from the positive xxx-axis. What is the direction of −2v-2\mathbf{v}−2v (angle from the positive xxx-axis)?

  1. 45∘45^\circ45∘
  2. 90∘90^\circ90∘
  3. 225∘225^\circ225∘ (correct answer)
  4. 315∘315^\circ315∘

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 17

Given v=⟨8,15⟩\mathbf{v} = \langle 8,15\ranglev=⟨8,15⟩ and scalar c=−12c=-\tfrac{1}{2}c=−21​, which statement correctly describes cvc\mathbf{v}cv?

  1. Magnitude is multiplied by 12\tfrac{1}{2}21​ and direction is unchanged
  2. Magnitude is multiplied by 12\tfrac{1}{2}21​ and direction is reversed 180∘180^\circ180∘ (correct answer)
  3. Magnitude is multiplied by 222 and direction is reversed 180∘180^\circ180∘
  4. Magnitude is unchanged and direction is undefined

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 = -1/2, the magnitude halves because |c| = 1/2, giving |cv| = (1/2)·|v|, and the direction reverses because c is negative. Choice B is correct because it correctly states both magnitude and direction. 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 18

Consider vectors u⃗=⟨3,−4⟩\vec{u} = \langle 3, -4 \rangleu=⟨3,−4⟩ and v⃗=ku⃗\vec{v} = k\vec{u}v=ku where k≠0k \neq 0k=0. If the dot product u⃗⋅v⃗=−100\vec{u} \cdot \vec{v} = -100u⋅v=−100, what is the magnitude of v⃗\vec{v}v and in which direction does it point relative to u⃗\vec{u}u?

  1. Magnitude 202020, same direction as u⃗\vec{u}u
  2. Magnitude 202020, opposite direction to u⃗\vec{u}u (correct answer)
  3. Magnitude 444, opposite direction to u⃗\vec{u}u
  4. Magnitude 252525, opposite direction to u⃗\vec{u}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 v⃗=ku⃗\vec{v} = k\vec{u}v=ku, we have v⃗=k⟨3,−4⟩=⟨3k,−4k⟩\vec{v} = k\langle 3, -4 \rangle = \langle 3k, -4k \ranglev=k⟨3,−4⟩=⟨3k,−4k⟩. The dot product becomes: u⃗⋅v⃗=⟨3,−4⟩⋅⟨3k,−4k⟩=3(3k)+(−4)(−4k)=9k+16k=25k\vec{u} \cdot \vec{v} = \langle 3, -4 \rangle \cdot \langle 3k, -4k \rangle = 3(3k) + (-4)(-4k) = 9k + 16k = 25ku⋅v=⟨3,−4⟩⋅⟨3k,−4k⟩=3(3k)+(−4)(−4k)=9k+16k=25k Setting this equal to the given value: 25k=−10025k = -10025k=−100, so k=−4k = -4k=−4. Since k<0k < 0k<0, vector v⃗\vec{v}v points in the opposite direction to u⃗\vec{u}u. We have v⃗=−4⟨3,−4⟩=⟨−12,16⟩\vec{v} = -4\langle 3, -4 \rangle = \langle -12, 16 \ranglev=−4⟨3,−4⟩=⟨−12,16⟩, giving us magnitude ∣v⃗∣=(−12)2+162=144+256=400=20|\vec{v}| = \sqrt{(-12)^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20∣v∣=(−12)2+162​=144+256​=400​=20. Choice A gives the correct magnitude but wrong direction—it ignores that negative kkk means opposite direction. Choice C has the wrong magnitude (likely confusing the absolute value of kkk with the vector magnitude) but correct direction. Choice D has the wrong magnitude—this might come from mistakenly using ∣u⃗∣=5|\vec{u}| = 5∣u∣=5 and multiplying by ∣k∣=4|k| = 4∣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 19

A vector r⃗\vec{r}r has magnitude 777 and direction angle 150°150°150°. Vector s⃗=27r⃗\vec{s} = \frac{2}{7}\vec{r}s=72​r is then scaled by factor −1.5-1.5−1.5 to produce vector t⃗\vec{t}t. What is the magnitude of t⃗\vec{t}t and its direction angle?

  1. Magnitude 333, direction 330°330°330° (correct answer)
  2. Magnitude 333, direction 150°150°150°
  3. Magnitude 212121, direction 330°330°330°
  4. Magnitude 1.51.51.5, direction 330°330°330°

Explanation: First, ∣∣s⃗∣∣=27⋅7=2||\vec{s}|| = \frac{2}{7} \cdot 7 = 2∣∣s∣∣=72​⋅7=2. Since the scalar 27>0\frac{2}{7} > 072​>0, s⃗\vec{s}s has the same direction as r⃗\vec{r}r, so direction angle is 150°150°150°. Then t⃗=−1.5s⃗\vec{t} = -1.5\vec{s}t=−1.5s, so ∣∣t⃗∣∣=∣−1.5∣⋅2=3||\vec{t}|| = |-1.5| \cdot 2 = 3∣∣t∣∣=∣−1.5∣⋅2=3. Since −1.5<0-1.5 < 0−1.5<0, t⃗\vec{t}t points opposite to s⃗\vec{s}s. The direction angle of t⃗\vec{t}t is 150°+180°=330°150° + 180° = 330°150°+180°=330°. Choice B forgets the direction reversal from the negative scalar. Choice C incorrectly multiplies magnitudes (7×3=217 \times 3 = 217×3=21). Choice D uses only the magnitude of the final scalar factor.

Question 20

Given vector v\mathbf{v}v has direction 30∘30^\circ30∘ from the positive xxx-axis. In what direction does −v-\mathbf{v}−v point (as an angle from the positive xxx-axis)?​

  1. 30∘30^\circ30∘
  2. 60∘60^\circ60∘
  3. 150∘150^\circ150∘
  4. 210∘210^\circ210∘ (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).