Precalculus Quiz: Show Scalar Multiplication Visually

20 questions · exam conditions

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Show Scalar Multiplication VisuallyQuestion 1 of 20

Vector $\vec{a} = (5, -3)$ undergoes scalar multiplication by $k$ , resulting in vector $\vec{b} = (-15, 9)$ . A student claims that since the $x$ -component changed from positive to negative and the $y$ -component changed from negative to positive, the scalar $k$ must be positive. What is wrong with this reasoning?

The student correctly identified that

k

is positive; there is no error in the reasoning provided

The student failed to recognize that

k = -3

, which is negative, and negative scalars reverse component signs

The student confused scalar multiplication with vector addition, which explains the sign changes observed

The student incorrectly calculated the components; the actual vector

\vec{b}

should be

(15, -9)

when

k > 0

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Precalculus Quiz

Precalculus Quiz: Show Scalar Multiplication Visually

Practice Show Scalar Multiplication Visually 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 Show Scalar Multiplication Visually, 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

The student correctly identified that $k$ is positive; there is no error in the reasoning provided
The student failed to recognize that $k = -3$ , which is negative, and negative scalars reverse component signs (correct answer)
The student confused scalar multiplication with vector addition, which explains the sign changes observed
The student incorrectly calculated the components; the actual vector $\vec{b}$ should be $(15, -9)$ when $k > 0$

Explanation: To find

k

, we solve

k(5, -3) = (-15, 9)

, which gives us

5k = -15

and

-3k = 9

. Both equations yield

k = -3

. Since

k

is negative, it reverses the direction of the vector, which means each component changes sign and is scaled by the absolute value

|k| = 3

. So

5 \times (-3) = -15

and

(-3) \times (-3) = 9

. The student's error was assuming that sign changes in components indicate a positive scalar, when in fact they indicate a negative scalar. Choice A is wrong because the reasoning is flawed. Choice C incorrectly suggests vector addition. Choice D incorrectly suggests a calculation error when the given values are consistent.

Question 2

Vector $\vec{m}$ has components $(6, -8)$ . When $\vec{m}$ is multiplied by scalar $k = \frac{3}{4}$ , a student calculates the result as $\left(\frac{18}{4}, -\frac{24}{4}\right)$ . Although this can be simplified to $\left(\frac{9}{2}, -6\right)$ , what error did the student make in the initial calculation?

The student multiplied by $\frac{4}{3}$ instead of $\frac{3}{4}$ , resulting in components that are too large
The student applied the scalar to the magnitude rather than to each component individually
The student correctly performed scalar multiplication; $\left(\frac{18}{4}, -\frac{24}{4}\right) = \left(\frac{9}{2}, -6\right)$ is the right answer (correct answer)
The student forgot to reverse the direction when multiplying by a positive scalar less than 1

Explanation: Let's verify the student's work:

\frac{3}{4} \cdot (6, -8) = \left(\frac{3}{4} \cdot 6, \frac{3}{4} \cdot (-8)\right) = \left(\frac{18}{4}, -\frac{24}{4}\right) = \left(\frac{9}{2}, -6\right)

. The student performed the scalar multiplication correctly. The calculation

\frac{3}{4} \cdot 6 = \frac{18}{4}

and

\frac{3}{4} \cdot (-8) = -\frac{24}{4}

are both correct. There was no error in the student's work. Choice A suggests using the reciprocal, which would give different values. Choice B misunderstands scalar multiplication. Choice D incorrectly suggests that positive scalars less than 1 reverse direction, which they don't.

Question 3

If vector $\vec{v} = (a, b)$ where $a > 0$ and $b < 0$ , and scalar multiplication $c\vec{v}$ results in a vector pointing into the second quadrant, which of the following must be true about scalar $c$ ?

$c > 0$ , because positive scalars preserve the original quadrant location of vectors
$c = 0$ , because only the zero vector can change quadrants during scalar multiplication
$c > 1$ , because the vector must be lengthened to reach the second quadrant
$c < 0$ , because the resulting vector has opposite signs for both components compared to $\vec{v}$ (correct answer)

Explanation: When you encounter scalar multiplication problems, focus on how the scalar affects both the direction and magnitude of the original vector. Since

\vec{v} = (a, b)

has

a > 0

and

b < 0

, this vector points into the fourth quadrant (positive x-component, negative y-component). For

c\vec{v}

to point into the second quadrant, the resulting vector must have a negative x-component and positive y-component. When we multiply

\vec{v}

by scalar

c

, we get

c\vec{v} = (ca, cb)

. Since

a > 0

, to make

ca < 0

, we need

c < 0

. Similarly, since

b < 0

, to make

cb > 0

, we also need

c < 0

. A negative scalar flips both components' signs, rotating the vector 180° and placing it in the opposite quadrant. Choice A incorrectly assumes positive scalars preserve quadrant location. While positive scalars preserve direction, they don't change which quadrant a vector occupies, so a positive

c

would keep

\vec{v}

in the fourth quadrant. Choice B wrongly suggests only the zero vector changes quadrants—actually,

c = 0

produces the zero vector at the origin, which isn't in any quadrant. Choice C focuses on magnitude rather than direction. The length of the vector is irrelevant to which quadrant it occupies; only the signs of the components matter. Remember: negative scalars always flip a vector to the opposite quadrant by reversing both components' signs. This is a fundamental property that appears frequently in vector problems.

Question 4

On a coordinate plane, vector $\mathbf{v}$ is the arrow from the origin to $(3,4)$ . Which statement correctly describes the visual effect of scalar multiplication by $2$ (that is, $2\mathbf{v}$ ) compared to $\mathbf{v}$ ?

$2\mathbf{v}$ points in the opposite direction and has the same length as $\mathbf{v}$ .
$2\mathbf{v}$ has the same direction as $\mathbf{v}$ and is twice as long. (correct answer)
$2\mathbf{v}$ has the same direction as $\mathbf{v}$ and is half as long.
$2\mathbf{v}$ changes the angle of $\mathbf{v}$ but keeps the same length.

Explanation: This question tests understanding of how scalar multiplication affects a vector visually and algebraically. When a vector v is multiplied by a scalar k, the result kv has magnitude |kv| = |k|·|v| (the length is scaled by the absolute value of k) and direction that either stays the same (if k > 0) or reverses 180° (if k < 0). Multiplying by k = 2 which is greater than 1 causes the vector to stretch by a factor of 2, so the arrow becomes longer compared to the original. Choice B is correct because it accurately describes both length and direction changes. Choice A claims the direction reverses when k is positive, but direction only reverses when k < 0—positive scalars preserve direction. Key to scalar multiplication: multiply every component by the scalar (kv = ⟨ka, kb⟩), and remember that the magnitude scales by |k| (the absolute value) while direction stays the same if k > 0 or reverses if k < 0. Visually, think of scalar multiplication as stretching (if |k| > 1) or compressing (if |k| < 1) the arrow representing the vector, and flipping it 180° if k is negative—the arrow always remains on the same line through the origin.

Question 5

For vector $\vec v = \langle 3,4\rangle$ drawn as an arrow from the origin to $(3,4)$ on a coordinate plane, what are the components of $2\vec v$ ?

$\langle 5,6\rangle$
$\langle 6,8\rangle$ (correct answer)
$\langle 3,8\rangle$
$\langle 6,4\rangle$

Explanation: This question tests understanding of how scalar multiplication affects a vector visually and algebraically. Scalar multiplication affects vectors component-wise: if v = ⟨a, b⟩, then kv = ⟨ka, kb⟩, meaning each component is multiplied by the scalar k, which visually corresponds to stretching or compressing the arrow by factor |k| and possibly reversing its direction if k is negative. Given v = ⟨3, 4⟩ and scalar k = 2, we compute kv = ⟨2·3, 2·4⟩ = ⟨6, 8⟩, which visually means the endpoint moves from (3, 4) to (6, 8). Choice B is correct because it properly multiplies each component by k. Choice A adds the scalar to each component instead of multiplying, computing ⟨3 + 2, 4 + 2⟩ = ⟨5, 6⟩ when scalar multiplication requires ⟨2·3, 2·4⟩. Key to scalar multiplication: multiply every component by the scalar (kv = ⟨ka, kb⟩), and remember that the magnitude scales by |k| (the absolute value) while direction stays the same if k > 0 or reverses if k < 0. Visually, think of scalar multiplication as stretching (if |k| > 1) or compressing (if |k| < 1) the arrow representing the vector, and flipping it 180° if k is negative—the arrow always remains on the same line through the origin.

Question 6

A coordinate-plane diagram shows $\vec w$ is collinear with $\vec v$ , points in the same direction as $\vec v$ , and appears to be twice as long. If $\vec w = k\vec v$ , what is the value of $k$ ?

$k=-2$
$k=\tfrac{1}{2}$
$k=2$ (correct answer)
$k=-\tfrac{1}{2}$

Explanation: This question tests understanding of how scalar multiplication affects a vector visually and algebraically. Scalar multiplication preserves collinearity: the vector kv is always parallel (if k > 0) or anti-parallel (if k < 0) to the original vector v, meaning they lie on the same line through the origin. Since w is collinear, same direction, and twice as long, k = 2 is positive and scales the length by 2, matching the description. Choice C is correct because it correctly identifies k based on direction and length scaling. Choice A forgets that negative k would reverse the direction, but the question specifies same direction, requiring k > 0. To find the scalar k given two vectors where w = kv, divide any component of w by the corresponding component of v (e.g., if v = ⟨3, 4⟩ and w = ⟨6, 8⟩, then k = 6/3 = 2 or k = 8/4 = 2), or compare magnitudes using k = |w|/|v|. Visually, think of scalar multiplication as stretching (if |k| > 1) or compressing (if |k| < 1) the arrow representing the vector, and flipping it 180° if k is negative—the arrow always remains on the same line through the origin.

Question 7

For vector $\mathbf{v}=\langle 5,0\rangle$ , what is the effect of multiplying $\mathbf{v}$ by $k=-\tfrac{1}{2}$ on its length and direction (visually)?

Same direction, half the length
Opposite direction, half the length (correct answer)
Opposite direction, twice the length
Same direction, same length

Explanation: This question tests understanding of how scalar multiplication affects a vector visually and algebraically. The visual effect of multiplying a vector by scalar k depends on k's value: k > 1 stretches the vector (longer arrow), 0 < k < 1 compresses it (shorter arrow), k < 0 reverses the direction and scales by |k|, and k = 0 gives the zero vector. Since k = -1/2 is negative, the direction of kv is opposite to v (reversed 180°), while the length is multiplied by |-1/2| = 1/2. For k = -1/2, the vector reverses direction and becomes half as long, which is visible in the scaled arrow. Choice B is correct because it accurately describes both length and direction changes: opposite direction (negative k) and half the length (|k| = 1/2). Choice A claims the direction stays the same when k is positive, but direction only reverses when k < 0—positive scalars preserve direction, and here k = -1/2 is negative. Special scalars to remember: k = 1 leaves the vector unchanged, k = -1 reverses direction only (same length), k = 2 doubles the length, k = 1/2 halves the length, and k = 0 gives the zero vector. Visually, think of scalar multiplication as stretching (if |k| > 1) or compressing (if |k| < 1) the arrow representing the vector, and flipping it 180° if k is negative—the arrow always remains on the same line through the origin.

Question 8

Two vectors $\vec{u}$ and $\vec{w}$ are related by $\vec{w} = k\vec{u}$ where $k$ is a scalar. If $\vec{u} = (2, -5)$ and $|\vec{w}| = 2|\vec{u}|$ , which of the following could be the components of $\vec{w}$ ?

$(4, -10)$ only, since the scalar must be positive to achieve the magnitude relationship
$(-4, 10)$ only, since negative scalars are required to double the magnitude
Either $(4, -10)$ or $(-4, 10)$ , since both $k = 2$ and $k = -2$ satisfy the magnitude condition (correct answer)
$(1, -2.5)$ only, since scalar multiplication requires reducing each component by the same factor

Explanation: Given

|\vec{w}| = 2|\vec{u}|

and

\vec{w} = k\vec{u}

, we have

|k\vec{u}| = 2|\vec{u}|

, which means

|k| \cdot |\vec{u}| = 2|\vec{u}|

. Therefore

|k| = 2

, so

k = 2

k = -2

. If

k = 2

, then

\vec{w} = 2(2, -5) = (4, -10)

. If

k = -2

, then

\vec{w} = -2(2, -5) = (-4, 10)

. Both vectors have magnitude

\sqrt{16 + 100} = \sqrt{116} = 2\sqrt{29}

, which is indeed twice the magnitude of

\vec{u}

(which is

\sqrt{4 + 25} = \sqrt{29}

). Choices A and B incorrectly eliminate one valid possibility. Choice D gives a vector with half the magnitude, corresponding to

k = \frac{1}{2}

Question 9

Vector $\vec v$ goes from the origin to $(4,6)$ on a coordinate plane. Where does $1.5\vec v$ terminate?

$(6,9)$ (correct answer)
$(5.5,7.5)$
$(2.5,4.5)$
$(4,7.5)$

Explanation: This question tests understanding of how scalar multiplication affects a vector visually and algebraically. Scalar multiplication affects vectors component-wise: if v = ⟨a, b⟩, then kv = ⟨ka, kb⟩, meaning each component is multiplied by the scalar k, which visually corresponds to stretching or compressing the arrow by factor |k| and possibly reversing its direction if k is negative. Given v = ⟨4, 6⟩ and scalar k = 1.5, we compute kv = ⟨1.5·4, 1.5·6⟩ = ⟨6, 9⟩, which visually means the endpoint moves from (4, 6) to (6, 9). Choice A is correct because it properly multiplies each component by k. Choice B adds the scalar to each component instead of multiplying, computing ⟨4 + 1.5, 6 + 1.5⟩ = ⟨5.5, 7.5⟩ when scalar multiplication requires ⟨1.5·4, 1.5·6⟩. Key to scalar multiplication: multiply every component by the scalar (kv = ⟨ka, kb⟩), and remember that the magnitude scales by |k| (the absolute value) while direction stays the same if k > 0 or reverses if k < 0. Visually, think of scalar multiplication as stretching (if |k| > 1) or compressing (if |k| < 1) the arrow representing the vector, and flipping it 180° if k is negative—the arrow always remains on the same line through the origin.

Question 10

Vector $\vec{p}$ has magnitude 10 and makes an angle of $60°$ with the positive $x$ -axis. If $\vec{q} = -0.3\vec{p}$ , what is the magnitude of $\vec{q}$ and what angle does it make with the positive $x$ -axis?

Magnitude 3, angle $240°$ , because the negative scalar reverses direction while scaling magnitude proportionally (correct answer)
Magnitude 3, angle $60°$ , because scalar multiplication preserves both magnitude ratios and direction
Magnitude 7, angle $240°$ , because scalar multiplication reduces magnitude by $10 - 3 = 7$ units
Magnitude 3, angle $120°$ , because negative scalars add $60°$ to the original angle measurement

Explanation: When you encounter vector scalar multiplication problems, remember that multiplying a vector by a scalar affects both its magnitude and potentially its direction in predictable ways. Let's work through this systematically. Vector

\vec{p}

has magnitude 10 and points at

60°

. When we calculate

\vec{q} = -0.3\vec{p}

, two things happen: First, the magnitude: Scalar multiplication scales the magnitude by the absolute value of the scalar. So

|\vec{q}| = |-0.3| \times |\vec{p}| = 0.3 \times 10 = 3

. Second, the direction: The negative sign reverses the vector's direction. A vector pointing at

60°

will point in the opposite direction when multiplied by a negative scalar. The opposite direction is

60° + 180° = 240°

. Choice A correctly identifies both the magnitude (3) and angle (

240°

), explaining that negative scalars reverse direction while scaling magnitude proportionally. Choice B gets the magnitude right but keeps the original angle, missing that negative scalars reverse direction. Choice C has the correct angle but incorrectly calculates magnitude as

10 - 3 = 7

. This reflects a misunderstanding—scalar multiplication doesn't subtract values; it multiplies them. Choice D gets the magnitude right but calculates the angle as

120°

, which would be

60° + 60°

. This shows confusion about how direction reversal works. Remember: When multiplying vectors by scalars, the magnitude gets multiplied by the scalar's absolute value, and negative scalars add

180°

to reverse direction.

Question 11

For vector $\mathbf{v}=\langle 5,0\rangle$ and scalar $k=\tfrac{1}{2}$ , what are the components of $\tfrac{1}{2}\mathbf{v}$ ?

$\langle \tfrac{5}{2},0\rangle$ (correct answer)
$\langle 5,\tfrac{1}{2}\rangle$
$\langle 5,0\rangle$
$\langle \tfrac{11}{2},\tfrac{1}{2}\rangle$

Explanation: This question tests understanding of how scalar multiplication affects a vector visually and algebraically. When a vector v is multiplied by a scalar k, the result kv has magnitude |kv| = |k|·|v| (the length is scaled by the absolute value of k) and direction that either stays the same (if k > 0) or reverses 180° (if k < 0). Given v = ⟨5, 0⟩ and scalar k = 1/2, we compute kv = ⟨k·5, k·0⟩ = ⟨(1/2)·5, (1/2)·0⟩ = ⟨5/2, 0⟩, which visually means the endpoint moves from (5, 0) to (5/2, 0). Choice A is correct because it properly multiplies each component by k, giving ⟨5/2, 0⟩. Choice D adds the scalar to each component instead of multiplying, computing ⟨5 + 1/2, 0 + 1/2⟩ = ⟨11/2, 1/2⟩ when scalar multiplication requires ⟨(1/2)·5, (1/2)·0⟩. Visually, think of scalar multiplication as stretching (if |k| > 1) or compressing (if |k| < 1) the arrow representing the vector, and flipping it 180° if k is negative—the arrow always remains on the same line through the origin. Special scalars to remember: k = 1 leaves the vector unchanged, k = -1 reverses direction only (same length), k = 2 doubles the length, k = 1/2 halves the length, and k = 0 gives the zero vector.

Question 12

On the same coordinate axes, $\mathbf{w}$ appears to be collinear with $\mathbf{v}$ but points in the opposite direction and is three times as long. If $\mathbf{w}=k\mathbf{v}$ , what is the value of $k$ ?

$k=3$
$k=-3$ (correct answer)
$k=-\tfrac{1}{3}$
$k=\tfrac{1}{3}$

Explanation: This question tests understanding of how scalar multiplication affects a vector visually and algebraically. Scalar multiplication preserves collinearity: the vector kv is always parallel (if k > 0) or anti-parallel (if k < 0) to the original vector v, meaning they lie on the same line through the origin. Since w is collinear with v but points in the opposite direction and is three times as long, k must be negative with |k| = 3, so k = -3. Choice B is correct because it accurately describes both length and direction changes. Choice A claims the direction remains the same when k is positive, but direction only reverses when k < 0—positive scalars preserve direction. To find the scalar k given two vectors where w = kv, divide any component of w by the corresponding component of v (e.g., if v = ⟨3, 4⟩ and w = ⟨6, 8⟩, then k = 6/3 = 2 or k = 8/4 = 2), or compare magnitudes using k = |w|/|v|. Visually, think of scalar multiplication as stretching (if |k| > 1) or compressing (if |k| < 1) the arrow representing the vector, and flipping it 180° if k is negative—the arrow always remains on the same line through the origin.

Question 13

Given $\mathbf{v}=\langle 5,0\rangle$ shown as an arrow to the right along the $x$ -axis, what are the components of $-2\mathbf{v}$ ?

$\langle -10,0\rangle$ (correct answer)
$\langle 3,0\rangle$
$\langle -7,-2\rangle$
$\langle 10,0\rangle$

Explanation: This question tests understanding of how scalar multiplication affects a vector visually and algebraically. Scalar multiplication affects vectors component-wise: if v = ⟨a, b⟩, then kv = ⟨ka, kb⟩, meaning each component is multiplied by the scalar k, which visually corresponds to stretching or compressing the arrow by factor |k| and possibly reversing its direction if k is negative. Given v = ⟨5, 0⟩ and scalar k = -2, we compute kv = ⟨-2·5, -2·0⟩ = ⟨-10, 0⟩, which visually means the endpoint moves from (5, 0) to (-10, 0). Choice A is correct because it properly multiplies each component by k. Choice D forgets to take the absolute value of k when calculating magnitude, using |kv| = k|v| instead of |k|·|v|, which gives a negative magnitude when k is negative. Key to scalar multiplication: multiply every component by the scalar (kv = ⟨ka, kb⟩), and remember that the magnitude scales by |k| (the absolute value) while direction stays the same if k > 0 or reverses if k < 0. Special scalars to remember: k = 1 leaves the vector unchanged, k = -1 reverses direction only (same length), k = 2 doubles the length, k = 1/2 halves the length, and k = 0 gives the zero vector.

Question 14

A velocity vector $\mathbf{v}$ is shown as an arrow pointing left with length 5 units on a diagram. What is the visual effect of multiplying by $-1$ (that is, $-\mathbf{v}$ ) on the vector's length and direction?

Same direction, same length.
Opposite direction, same length. (correct answer)
Opposite direction, double the length.
Same direction, half the length.

Explanation: This question tests understanding of how scalar multiplication affects a vector visually and algebraically. When a vector v is multiplied by a scalar k, the result kv has magnitude |kv| = |k|·|v| (the length is scaled by the absolute value of k) and direction that either stays the same (if k > 0) or reverses 180° (if k < 0). Multiplying by k = -1 which is negative causes the vector to reverse direction by a factor of 1, so the arrow becomes opposite direction and same length compared to the original. Choice B is correct because it accurately describes both length and direction changes. Choice C claims the length doubles when k = -1, but since |k| = 1, the vector is actually the same length, not doubled. Key to scalar multiplication: multiply every component by the scalar (kv = ⟨ka, kb⟩), and remember that the magnitude scales by |k| (the absolute value) while direction stays the same if k > 0 or reverses if k < 0. Special scalars to remember: k = 1 leaves the vector unchanged, k = -1 reverses direction only (same length), k = 2 doubles the length, k = 1/2 halves the length, and k = 0 gives the zero vector.

Question 15

Vector $\mathbf{v}$ goes from the origin to $(4,6)$ on a coordinate plane. Where does $1.5\mathbf{v}$ terminate?

$(6,9)$ (correct answer)
$(5.5,7.5)$
$(2.5,4.5)$
$(4,7.5)$

Explanation: This question tests understanding of how scalar multiplication affects a vector visually and algebraically. Scalar multiplication affects vectors component-wise: if v = ⟨a, b⟩, then kv = ⟨ka, kb⟩, meaning each component is multiplied by the scalar k, which visually corresponds to stretching or compressing the arrow by factor |k| and possibly reversing its direction if k is negative. Given v = ⟨4, 6⟩ and scalar k = 1.5, we compute kv = ⟨1.5·4, 1.5·6⟩ = ⟨6, 9⟩, which visually means the endpoint moves from (4, 6) to (6, 9). Choice A is correct because it properly multiplies each component by k. Choice B multiplies only one component by k, not both, which would change the direction of the vector rather than just scaling it. Key to scalar multiplication: multiply every component by the scalar (kv = ⟨ka, kb⟩), and remember that the magnitude scales by |k| (the absolute value) while direction stays the same if k > 0 or reverses if k < 0. Visually, think of scalar multiplication as stretching (if |k| > 1) or compressing (if |k| < 1) the arrow representing the vector, and flipping it 180° if k is negative—the arrow always remains on the same line through the origin.

Question 16

For vector $\vec v = \langle 5,0\rangle$ shown on a coordinate plane along the positive $x$ -axis, what is the direction of $-\vec v$ ?

Along the positive $x$ -axis
Along the positive $y$ -axis
Along the negative $x$ -axis (correct answer)
Along the negative $y$ -axis

Explanation: This question tests understanding of how scalar multiplication affects a vector visually and algebraically. When a vector v is multiplied by a scalar k, the result kv has magnitude |kv| = |k|·|v| (the length is scaled by the absolute value of k) and direction that either stays the same (if k > 0) or reverses 180° (if k < 0). Since k = -1 is negative, the direction of kv is opposite to v (reversed 180°), while the length is multiplied by |-1| = 1, remaining the same. Choice C is correct because it accurately describes both length and direction changes. Choice A claims the direction remains the same when k is negative, but direction only reverses when k < 0—positive scalars preserve direction. Special scalars to remember: k = 1 leaves the vector unchanged, k = -1 reverses direction only (same length), k = 2 doubles the length, k = 1/2 halves the length, and k = 0 gives the zero vector. Visually, think of scalar multiplication as stretching (if |k| > 1) or compressing (if |k| < 1) the arrow representing the vector, and flipping it 180° if k is negative—the arrow always remains on the same line through the origin.

Question 17

Based on the visual representation of scalar multiplication, which statement correctly describes $\tfrac{1}{3}\vec v$ compared to $\vec v$ (with $\vec v$ drawn as an arrow on a coordinate plane)?

$\tfrac{1}{3}\vec v$ points in the opposite direction and is three times as long
$\tfrac{1}{3}\vec v$ points in the same direction and is one-third as long (correct answer)
$\tfrac{1}{3}\vec v$ points in the same direction and is three times as long
$\tfrac{1}{3}\vec v$ changes direction to a different angle (not collinear with $\vec v$ )

Explanation: This question tests understanding of how scalar multiplication affects a vector visually and algebraically. The visual effect of multiplying a vector by scalar k depends on k's value: k > 1 stretches the vector (longer arrow), 0 < k < 1 compresses it (shorter arrow), k < 0 reverses the direction and scales by |k|, and k = 0 gives the zero vector. For k = 1/3 which is between 0 and 1, the vector compresses by a factor of 1/3, so the arrow becomes shorter in the same direction compared to the original. Choice B is correct because it accurately describes both length and direction changes. Choice A claims the direction reverses when k is positive, but direction only reverses when k < 0—positive scalars preserve direction. Scalar multiplication preserves collinearity: the vector kv is always parallel (if k > 0) or anti-parallel (if k < 0) to the original vector v, meaning they lie on the same line through the origin. Visually, think of scalar multiplication as stretching (if |k| > 1) or compressing (if |k| < 1) the arrow representing the vector, and flipping it 180° if k is negative—the arrow always remains on the same line through the origin.

Question 18

For vector $\vec v = \langle -4,2\rangle$ shown on a coordinate plane, what are the components of $\tfrac{1}{2}\vec v$ ?

$\langle -2,1\rangle$ (correct answer)
$\langle -4,1\rangle$
$\langle -3,\tfrac{5}{2}\rangle$
$\langle -\tfrac{7}{2},\tfrac{3}{2}\rangle$

Explanation: This question tests understanding of how scalar multiplication affects a vector visually and algebraically. Scalar multiplication affects vectors component-wise: if v = ⟨a, b⟩, then kv = ⟨ka, kb⟩, meaning each component is multiplied by the scalar k, which visually corresponds to stretching or compressing the arrow by factor |k| and possibly reversing its direction if k is negative. Given v = ⟨-4, 2⟩ and scalar k = 1/2, we compute kv = ⟨(1/2)·(-4), (1/2)·2⟩ = ⟨-2, 1⟩, which visually means the endpoint moves from (-4, 2) to (-2, 1). Choice A is correct because it properly multiplies each component by k. Choice B multiplies only one component by k, not both, which would change the direction of the vector rather than just scaling it. Key to scalar multiplication: multiply every component by the scalar (kv = ⟨ka, kb⟩), and remember that the magnitude scales by |k| (the absolute value) while direction stays the same if k > 0 or reverses if k < 0. Special scalars to remember: k = 1 leaves the vector unchanged, k = -1 reverses direction only (same length), k = 2 doubles the length, k = 1/2 halves the length, and k = 0 gives the zero vector.

Question 19

On a coordinate plane, vector $\mathbf{v}$ is drawn from the origin to the point $(3,4)$ . Which point is the terminal point of $2\mathbf{v}$ when it is drawn from the origin on the same axes?

$(5,6)$
$(6,8)$ (correct answer)
$(3,8)$
$(1.5,2)$

Explanation: This question tests understanding of how scalar multiplication affects a vector visually and algebraically. Scalar multiplication affects vectors component-wise: if v = ⟨a, b⟩, then kv = ⟨ka, kb⟩, meaning each component is multiplied by the scalar k, which visually corresponds to stretching or compressing the arrow by factor |k| and possibly reversing its direction if k is negative. Given v = ⟨3, 4⟩ and scalar k = 2, we compute kv = ⟨2·3, 2·4⟩ = ⟨6, 8⟩, which visually means the endpoint moves from (3, 4) to (6, 8). Choice B is correct because it properly multiplies each component by k. Choice A adds the scalar to each component instead of multiplying, computing ⟨a + k, b + k⟩ when scalar multiplication requires ⟨ka, kb⟩. Key to scalar multiplication: multiply every component by the scalar (kv = ⟨ka, kb⟩), and remember that the magnitude scales by |k| (the absolute value) while direction stays the same if k > 0 or reverses if k < 0. Visually, think of scalar multiplication as stretching (if |k| > 1) or compressing (if |k| < 1) the arrow representing the vector, and flipping it 180° if k is negative—the arrow always remains on the same line through the origin.

Question 20

Vector $\mathbf{v}$ has magnitude $5$ and points in some direction on the coordinate plane. What is the magnitude of $-3\mathbf{v}$ ?

$-15$
$15$ (correct answer)
$8$
$\tfrac{5}{3}$

Explanation: This question tests understanding of how scalar multiplication affects a vector visually and algebraically. When a vector v is multiplied by a scalar k, the result kv has magnitude |kv| = |k|·|v| (the length is scaled by the absolute value of k) and direction that either stays the same (if k > 0) or reverses 180° (if k < 0). The original vector v has magnitude |v| = 5. When multiplied by scalar k = -3, the new magnitude is |kv| = |-3|·5 = 3·5 = 15. Choice B is correct because it correctly calculates magnitude as |k| times original magnitude. Choice A forgets to take the absolute value of k when calculating magnitude, using |kv| = k|v| instead of |k|·|v|, which gives a negative magnitude when k is negative. Key to scalar multiplication: multiply every component by the scalar (kv = ⟨ka, kb⟩), and remember that the magnitude scales by |k| (the absolute value) while direction stays the same if k > 0 or reverses if k < 0. Visually, think of scalar multiplication as stretching (if |k| > 1) or compressing (if |k| < 1) the arrow representing the vector, and flipping it 180° if k is negative—the arrow always remains on the same line through the origin.