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.

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.

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|.

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).

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.

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.

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.

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.

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.

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 = 1/3 which is between 0 and 1 causes the vector to compress by a factor of 1/3, so the arrow becomes shorter 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. 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.