Precalculus Quiz: Show Scalar Multiplication Visually

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.

Question 1

1. The student correctly identified that $k$ is positive; there is no error in the reasoning provided
2. The student failed to recognize that $k = -3$, which is negative, and negative scalars reverse component signs (correct answer)
3. The student confused scalar multiplication with vector addition, which explains the sign changes observed
4. 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

1. The student multiplied by $\frac{4}{3}$ instead of $\frac{3}{4}$, resulting in components that are too large
2. The student applied the scalar to the magnitude rather than to each component individually
3. 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)
4. 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

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

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 4

1. The vector is twice as long and points in the opposite direction.
2. The vector is half as long and points in the same direction. (correct answer)
3. The vector keeps the same length but reverses direction.
4. The vector becomes the zero vector.

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 half as long compared to the original. Choice B is correct because it accurately describes both length and direction changes: the vector is half as long and points in the same direction. Choice A describes the vector as stretched when k = 1/2, but since 0 < k < 1, the vector is actually compressed (shortened), not stretched. 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 5

1. Same direction as $\vec v$, same length
2. Same direction as $\vec v$, half the length
3. Opposite direction to $\vec v$, same length (correct answer)
4. Opposite direction to $\vec v$, double 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). Since k = -1 is negative, the direction of kv is opposite to v (reversed 180°), while the length is multiplied by |k| = 1, so it remains the same. Choice C 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. 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 6

1. $c > 0$, because positive scalars preserve the original quadrant location of vectors
2. $c = 0$, because only the zero vector can change quadrants during scalar multiplication
3. $c > 1$, because the vector must be lengthened to reach the second quadrant
4. $c < 0$, 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 7

1. $2\mathbf{v}$ points in the opposite direction and has the same length as $\mathbf{v}$.
2. $2\mathbf{v}$ has the same direction as $\mathbf{v}$ and is twice as long. (correct answer)
3. $2\mathbf{v}$ has the same direction as $\mathbf{v}$ and is half as long.
4. $2\mathbf{v}$ 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 8

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

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 9

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

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 10

1. Same direction, half the length
2. Opposite direction, half the length (correct answer)
3. Opposite direction, twice the length
4. 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 11

1. $-15$
2. $8$
3. $15$ (correct answer)
4. $\tfrac{5}{3}$

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 C 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.

Question 12

1. $(4, -10)$ only, since the scalar must be positive to achieve the magnitude relationship
2. $(-4, 10)$ only, since negative scalars are required to double the magnitude
3. Either $(4, -10)$ or $(-4, 10)$, since both $k = 2$ and $k = -2$ satisfy the magnitude condition (correct answer)
4. $(1, -2.5)$ 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 $k = 2$ or $k = -2$. 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 13

1. $(6,9)$ (correct answer)
2. $(5.5,7.5)$
3. $(2.5,4.5)$
4. $(4,7.5)$

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 14

1. $-10$
2. $7$
3. $3$
4. $10$ (correct answer)

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| = √(3² + 4²) = √(9 + 16) = √25 = 5. When multiplied by scalar k = -2, the new magnitude is |kv| = |-2|·5 = 2·5 = 10. Choice D is correct because it correctly calculates magnitude as |k| times original magnitude, giving 10. Choice A forgets to take the absolute value of k when calculating magnitude, using |kv| = k|v| = -2·5 = -10 instead of |k|·|v| = 2·5 = 10, 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. 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

1. $\langle 0,0\rangle$ (correct answer)
2. $\langle 3,4\rangle$
3. $\langle 0,4\rangle$
4. $\langle 3,0\rangle$

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. Given v = ⟨3, 4⟩ and scalar k = 0, we compute kv = ⟨k·3, k·4⟩ = ⟨0·3, 0·4⟩ = ⟨0, 0⟩, which visually means the endpoint moves from (3, 4) to (0, 0). For k = 0, the vector becomes the zero vector, which is visible as a point at the origin with no length or direction. Choice A is correct because it properly multiplies each component by k = 0, giving ⟨0, 0⟩. Choice B claims kv = v (no change), ignoring the effect of the scalar k = 0 entirely. 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. 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.

Question 16

1. It is twice as long as $\mathbf{v}$
2. It is half as long as $\mathbf{v}$ (correct answer)
3. It has the same length as $\mathbf{v}$
4. It is 3 units longer than $\mathbf{v}$

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 = 1/2, the new magnitude is |kv| = (1/2)·5 = 2.5, so it is half as long. Choice B is correct because it correctly calculates magnitude as |k| times original magnitude. Choice D treats scalar multiplication as addition to the magnitude, claiming |kv| = |v| + k, which doesn't match how scalar multiplication works. 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 17

1. Same direction as $\mathbf{v}$, and the length doubles.
2. Same direction as $\mathbf{v}$, and the length is unchanged.
3. Opposite direction to $\mathbf{v}$, and the length is unchanged. (correct answer)
4. Opposite direction to $\mathbf{v}$, and the length is cut in half.

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 k = -1 is negative, the direction of kv is opposite to v (reversed 180°), while the length is multiplied by |-1| = 1. Choice C is correct because it accurately describes both length and direction changes: opposite direction to v, and the length is unchanged. Choice A claims the direction stays the same when k is negative, but direction only stays the same when k > 0—negative scalars reverse 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 18

1. $-2\mathbf{v}$ is compressed to half the length and points in the same direction.
2. $-2\mathbf{v}$ is stretched to twice the length and points in the opposite direction. (correct answer)
3. $-2\mathbf{v}$ keeps the same length but points in the opposite direction.
4. $-2\mathbf{v}$ is stretched to twice the length and points in the same direction.

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 negative and has absolute value greater than 1, causes the vector to reverse direction and stretch by a factor of |-2| = 2, so the arrow becomes opposite direction and twice as long compared to the original. Choice B is correct because it accurately describes both length and direction changes: stretched to twice the length and points in the opposite direction. Choice A claims the vector is compressed to half the length, but since |-2| = 2 > 1, the vector is actually stretched (lengthened), not compressed. 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 19

1. $-10$
2. $10$ (correct answer)
3. $5$
4. $7$

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 $|kv| = |k| \cdot |v|$ (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 $v = \langle 3, 4 \rangle$ has magnitude $|v| = \sqrt{3^2 + 4^2} = 5$. When multiplied by scalar $k = -2$, the new magnitude is $|kv| = |-2| \cdot 5 = 10$. 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 $|k| \cdot |v|$, which gives a negative magnitude when $k$ is negative. Key to scalar multiplication: multiply every component by the scalar ($kv = \langle ka, kb \rangle$), 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 20

1. $(6,9)$ (correct answer)
2. $(5.5,7.5)$
3. $(4,9)$
4. $(2.5,4.5)$

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.