Precalculus Quiz: Show Scalar Multiplication Visually

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Show Scalar Multiplication VisuallyQuestion 1 of 20

Given the vector $\mathbf{v}=\langle 3,4\rangle$ and scalar multiplication by $k=2$ , what are the components of $2\mathbf{v}$ ?

\langle 5,6\rangle

\langle 6,8\rangle

\langle 3,8\rangle

\langle 6,4\rangle

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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.

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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.

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Question 1

Given the vector $\mathbf{v}=\langle 3,4\rangle$ and scalar multiplication by $k=2$ , what are the components of $2\mathbf{v}$ ?

$\langle 5,6\rangle$
$\langle 6,8\rangle$ (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. 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 = ⟨3, 4⟩ and scalar k = 2, we compute kv = ⟨k·3, k·4⟩ = ⟨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, giving ⟨6, 8⟩. 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 2

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 3

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 4

For vector $\vec v=\langle 3,4\rangle$ shown from the origin on a coordinate plane, how much longer or shorter is $\tfrac{1}{3}\vec v$ compared to $\vec v$ ?

It is $3$ times as long as $\vec v$
It is $\tfrac{1}{3}$ as long as $\vec v$ (correct answer)
It has the same length as $\vec v$
It is $\tfrac{3}{2}$ as long as $\vec v$

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. The original vector v = ⟨3, 4⟩ has magnitude |v| = √(9 + 16) = 5. When multiplied by scalar k = 1/3, the new magnitude is |kv| = |1/3|·5 = 5/3, which is 1/3 as long. Choice B is correct because it correctly calculates magnitude as |k| times original magnitude. Choice A uses the reciprocal 1/k instead of k, which would scale by the wrong factor. 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

On a coordinate plane, vector $\mathbf{v}$ is shown from the origin to $(4,6)$ . Where does $1.5\mathbf{v}$ terminate when drawn from the origin?

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

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 D uses the reciprocal 1/k instead of k, which would scale by the wrong factor. 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 vector $\mathbf{v}=\langle 6,2\rangle$ is shown on a coordinate plane. What are the components of $\tfrac{1}{2}\mathbf{v}$ ?

$\langle 12,4\rangle$
$\langle 3,1\rangle$ (correct answer)
$\langle 6,1\rangle$
$\langle 6.5,2.5\rangle$

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 = ⟨6, 2⟩ and scalar k = 1/2, we compute kv = ⟨(1/2)·6, (1/2)·2⟩ = ⟨3, 1⟩, which visually means the endpoint moves from (6, 2) to (3, 1). Choice B is correct because it properly multiplies each component by k. Choice A uses the reciprocal 1/k instead of k, which would scale by the wrong factor. 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 7

On a coordinate plane, a vector $\mathbf{v}$ is drawn from the origin to $(4,6)$ . 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. 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 = ⟨4, 6⟩ and scalar k = 1.5, we compute kv = ⟨k·4, k·6⟩ = ⟨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 = 1.5, giving the terminal point (6, 9). Choice C uses the reciprocal 1/k instead of k, computing ⟨4/1.5, 6/1.5⟩ ≈ ⟨2.67, 4⟩ which would scale by the wrong factor. 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. 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|.

Question 8

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 9

For vector $\mathbf{v}=\langle 3,4\rangle$ , how does $\tfrac{1}{2}\mathbf{v}$ compare visually to $\mathbf{v}$ on a coordinate plane?

Same direction, twice the length
Opposite direction, half the length
Same direction, half the length (correct answer)
Same length, rotated $90^\circ$

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. Multiplying by k = 1/2, which is between 0 and 1, causes the vector to compress by a factor of 1/2, so the arrow becomes shorter compared to the original. Since k = 1/2 is positive, the direction of kv remains the same as v, while the length is multiplied by 1/2. Choice C is correct because it accurately describes both length and direction changes: same direction (positive k) and half the length (k = 1/2). Choice B 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. 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 10

For vector $\mathbf{v}=\langle 3,4\rangle$ and scalar $k=\dfrac{1}{3}$ , how much longer/shorter is $k\mathbf{v}$ compared to $\mathbf{v}$ ?

It is $3$ times as long.
It is $\dfrac{1}{3}$ as long. (correct answer)
It is $\dfrac{2}{3}$ as long.
It has the 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. For k = 1/3, the vector becomes 1/3 as long in the same direction, which is visible in the scaled arrow. Choice B is correct because it accurately describes the length change: kv is 1/3 as long as v. Choice A claims it is 3 times as long, which would be true if k = 3, not k = 1/3. 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 11

Given the vector $\mathbf{v}=\langle 2,1\rangle$ and scalar $k=-1$ , what are the components of $-\mathbf{v}$ ?

$\langle -2,1\rangle$
$\langle 1,2\rangle$
$\langle -2,-1\rangle$ (correct answer)
$\langle 2,0\rangle$

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 = ⟨2, 1⟩ and scalar k = -1, we compute kv = ⟨k·2, k·1⟩ = ⟨(-1)·2, (-1)·1⟩ = ⟨-2, -1⟩, which visually means the endpoint moves from (2, 1) to (-2, -1). Choice C is correct because it properly multiplies each component by k = -1, giving ⟨-2, -1⟩. Choice A reverses only the first component, computing ⟨-2, 1⟩ instead of multiplying both components by -1, which would change the direction of the vector rather than just reversing it. 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 12

For vector $\vec v=\langle -2,5\rangle$ shown on a coordinate plane, what are the components of $-\vec v$ ?

$\langle -3,4\rangle$
$\langle 2,5\rangle$
$\langle 2,-5\rangle$ (correct answer)
$\langle -2,-5\rangle$

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 = ⟨-2, 5⟩ and scalar k = -1, we compute kv = ⟨-1·-2, -1·5⟩ = ⟨2, -5⟩, which visually means the endpoint moves from (-2, 5) to (2, -5). Choice C is correct because it properly multiplies each component by k. Choice B claims kv = v (no change), ignoring the effect of the scalar k = -1 entirely. 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 13

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)
$\langle 3,8\rangle$
$\langle 6,4\rangle$

Question 14

On a coordinate plane, a position vector $\mathbf{v}$ goes from the origin to $(4,6)$ . If $\mathbf{w}=1.5\mathbf{v}$ , where does $\mathbf{w}$ 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. 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 = ⟨4, 6⟩ and scalar k = 1.5, we compute kv = ⟨k·4, k·6⟩ = ⟨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 = 1.5, giving the terminal point (6, 9). Choice B multiplies only partially or incorrectly, computing ⟨5.5, 7.5⟩ instead of ⟨6, 9⟩. 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. 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|.

Question 15

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 16

For vector $\mathbf{v}=\langle 3,4\rangle$ and scalar $k=2$ , what are the components of $k\mathbf{v}$ ?

$\langle 5,6\rangle$
$\langle 6,8\rangle$ (correct answer)
$\langle 3,8\rangle$
$\langle 6,7\rangle$

Question 17

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 (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 18

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 19

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 20

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.