Show Scalar Multiplication Visually - Precalculus

1. For vector $\mathbf{v}=\langle 3,4\rangle$ and scalar $k=\dfrac{1}{3}$, how much longer/shorter is $k\mathbf{v}$ compared to $\mathbf{v}$?
   Answer: It is $\dfrac{1}{3}$ as long. 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.
2. On the coordinate plane, vectors $\mathbf{v}$ and $\mathbf{w}$ are shown from the origin. $\mathbf{v}$ ends at $(2,1)$ and $\mathbf{w}$ ends at $(6,3)$. If $\mathbf{w}=k\mathbf{v}$, what is the value of $k$?
   Answer: $k=3$ 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. To find the scalar k given two vectors where w = kv, divide any component of w by the corresponding component of v: since v = ⟨2, 1⟩ and w = ⟨6, 3⟩, then k = 6/2 = 3 or k = 3/1 = 3. Choice C is correct because k = 3 properly scales v to w: 3⟨2, 1⟩ = ⟨6, 3⟩. Choice A uses the reciprocal 1/3 instead of 3, which would scale by the wrong factor, giving ⟨2/3, 1/3⟩ instead of ⟨6, 3⟩. 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|. Don't confuse scalar multiplication (which scales every component proportionally, keeping the vector on its line) with vector addition (which uses the parallelogram rule and usually changes both magnitude and direction in a different way).
3. For vector $\mathbf{v}=\langle 3,4\rangle$, how does $\tfrac{1}{2}\mathbf{v}$ compare visually to $\mathbf{v}$ on a coordinate plane?
   Answer: Same direction, half the length 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.
4. For vector $\mathbf{v}=\langle 2,1\rangle$, what is the effect of multiplying $\mathbf{v}$ by $k=-2$ on its length and direction (as it would appear on a coordinate plane)?
   Answer: Opposite direction, twice the length 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. Since k = -2 is negative, the direction of kv is opposite to v (reversed 180°), while the length is multiplied by |-2| = 2. Choice C is correct because it accurately describes both length and direction changes: opposite direction (due to negative k) and twice the length (due to |k| = 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 = -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. Don't confuse scalar multiplication (which scales every component proportionally, keeping the vector on its line) with vector addition (which uses the parallelogram rule and usually changes both magnitude and direction in a different way).
5. 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)?
   Answer: $\tfrac{1}{3}\vec v$ points in the same direction and is one-third as long 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.
6. Given $\mathbf{v} = \langle 3,4\rangle$, what are the components of $-2\mathbf{v}$?​
   Answer: $\langle -6,-8\rangle$ 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 C is correct because it properly multiplies each component by k. Choice A 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. Don't confuse scalar multiplication (which scales every component proportionally, keeping the vector on its line) with vector addition (which uses the parallelogram rule and usually changes both magnitude and direction in a different way).
7. Compare $\vec v$ and $-2\vec v$ visually. If $\vec v$ is drawn as an arrow from the origin, which statement is correct?
   Answer: $-2\vec v$ is twice as long as $\vec v$ and points in the opposite direction 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 = -2 which is negative causes the vector to reverse direction and stretch by a factor of 2, so the arrow becomes longer and in the opposite direction compared to the original. Choice A is correct because it accurately describes both length and direction changes. Choice B 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.
8. For vector $\mathbf{v} = \langle 3,4\rangle$ shown on a coordinate plane, how much longer or shorter is $\tfrac{1}{2}\mathbf{v}$ compared to $\mathbf{v}$?​
   Answer: It is half as long as $\mathbf{v}$ 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.
9. For vector $\mathbf{v}=\langle 5,0\rangle$ and scalar $k=\dfrac{1}{2}$, what is the effect of multiplying $\mathbf{v}$ by $k$ on its length and direction?
   Answer: The vector is half as long and points in the same direction. 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.
10. In the coordinate plane shown, vector $\vec{OA}$ has initial point at the origin $O$ and terminal point at $A(4, 3)$. Point $B$ is positioned such that $\vec{OB} = -\frac{1}{3}\vec{OA}$. What are the coordinates of point $B$?
    Answer: $\left(-\frac{4}{3}, -1\right)$, because scalar multiplication affects both magnitude and direction proportionally Vector $\vec{OA} = (4-0, 3-0) = (4, 3)$. When we compute $\vec{OB} = -\frac{1}{3}\vec{OA}$, we get $\vec{OB} = -\frac{1}{3}(4, 3) = \left(-\frac{4}{3}, -1\right)$. Since $\vec{OB}$ starts at the origin, point $B$ has coordinates $\left(-\frac{4}{3}, -1\right)$. The negative scalar reverses the direction and scales the magnitude by $\frac{1}{3}$. Choice B incorrectly ignores the effect of the negative sign on the components. Choice C incorrectly applies only the direction change without the magnitude scaling. Choice D uses incorrect calculations for the scalar multiplication.
11. 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 $2\vec v$?
    Answer: $\langle 6,8\rangle$ 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.
12. Given the vector $\vec v$ points from the origin to $(2,1)$ on a coordinate plane, in what direction does $-\vec v$ point compared to $\vec v$?
    Answer: Opposite direction to $\vec v$, same length 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.
13. 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 $\vec{w}$?
    Answer: Either $(4, -10)$ or $(-4, 10)$, since both $k = 2$ and $k = -2$ satisfy the magnitude condition 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 $k = \frac{1}{2}$.
14. 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 $\vec v$?
    Answer: It is $\tfrac{1}{3}$ as long as $\vec v$ 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.
15. For vector $\vec v = \langle -4,2\rangle$ shown on a coordinate plane, what are the components of $\tfrac{1}{2}\vec v$?
    Answer: $\langle -2,1\rangle$ 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.
16. For vector $\vec v = \langle 5,0\rangle$ shown on a coordinate plane along the positive $x$-axis, what is the direction of $-\vec v$?
    Answer: Along the negative $x$-axis 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.
17. 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 $k$?
    Answer: $k=2$ 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.
18. 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?​
    Answer: $(6,9)$ 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.
19. Given the vector $\mathbf{v}=\langle 3,4\rangle$ (so $|\mathbf{v}|=5$), what is the magnitude of $-2\mathbf{v}$?
    Answer: $10$ 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|.
20. Given the vector $\mathbf{v}=\langle 2,1\rangle$ and scalar $k=-1$, in what direction does $k\mathbf{v}$ point compared to $\mathbf{v}$?
    Answer: Opposite direction to $\mathbf{v}$, and the length is unchanged. 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.

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