Distance between two vectors is defined as ||u − v||, the norm of the displacement vector. Here u − v = (1−4, 3−7, −2−(−2)) = (−3, −4, 0), so ||u − v|| = √(9 + 16 + 0) = 5. Subtracting individual norms (options A and D) confuses the distance between vectors with the difference of their lengths — a common error that produces a different, generally meaningless quantity.
Question 2 Multiple Choice
If v = (−3, 4), what is ||−2v||?
A−10, because scaling by −2 makes the norm negative
B0, because −2 and v have opposite signs that cancel
C5, because negating v doesn't change its norm, so multiplying by −2 and negating cancels out
D10, because ||cv|| = |c| ||v||, so ||−2v|| = |−2| × ||v|| = 2 × 5 = 10
The scaling property of norms is ||cv|| = |c| ||v||, using the absolute value of the scalar. Here ||v|| = √(9 + 16) = 5, so ||−2v|| = |−2| × 5 = 10. Norms are always non-negative — a norm can never be negative, so option A is impossible. Option C makes the error of thinking −2 and the negation cancel; they don't, because |−2| = 2, not 0.
Question 3 True / False
The triangle inequality for norms states that ||u + v|| = ||u|| + ||v|| for any two vectors.
TTrue
FFalse
Answer: False
The triangle inequality is ||u + v|| ≤ ||u|| + ||v||, with equality holding only when u and v point in exactly the same direction. Geometrically: the direct path (||u + v||) is never longer than the two-leg detour (going ||u|| then ||v||), but it is shorter whenever the two vectors are not collinear and pointing the same way. The inequality is ≤, not =.
Question 4 True / False
Dividing any nonzero vector by its norm produces a unit vector that points in the same direction as the original vector.
TTrue
FFalse
Answer: True
This is the normalization procedure: v̂ = v/||v||. Its norm is ||v/||v||| = (1/||v||)||v|| = 1, confirming it is a unit vector. Dividing by a positive scalar (||v|| > 0 for nonzero v) preserves direction. Normalization lets you separate a vector's direction from its magnitude, which is fundamental to defining orthonormal bases and unit direction vectors.
Question 5 Short Answer
Why does the scaling property use |c| (absolute value of the scalar) rather than just c, and what geometric fact does this reflect?
Think about your answer, then reveal below.
Model answer: The absolute value is necessary because length is always non-negative. If c = −3, scaling v by −3 reverses its direction but triples its length — the result has length 3||v||, not −3||v||. Geometrically, negating a vector flips its direction but leaves its length unchanged, and multiplying by a negative scalar scales the length by the magnitude of that scalar. The absolute value |c| captures the 'how much stretching' while ignoring the sign that encodes direction reversal.
This is a subtle but important point about how norms interact with scalar multiplication. Norms measure geometric length, which is orientation-independent — the length of a displacement doesn't change if you flip its direction. The absolute value in ||cv|| = |c|||v|| is what ensures norms stay non-negative and measure pure magnitude, consistent with all three norm properties.