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Vector Operations - Calculus 3

  1. Points AA, BB, and CC have position vectors a⃗=⟨3,−1,2⟩\vec{a} = \langle 3, -1, 2 \rangle, b⃗=⟨1,4,−3⟩\vec{b} = \langle 1, 4, -3 \rangle, and c⃗=⟨5,2,1⟩\vec{c} = \langle 5, 2, 1 \rangle respectively. If point DD is such that AD→=2AB→+12AC→\overrightarrow{AD} = 2\overrightarrow{AB} + \frac{1}{2}\overrightarrow{AC}, what is the zz-component of the position vector of point DD?
    Answer: −8.5-8.5 First, find the direction vectors: AB→=b⃗−a⃗=⟨1,4,−3⟩−⟨3,−1,2⟩=⟨−2,5,−5⟩\overrightarrow{AB} = \vec{b} - \vec{a} = \langle 1, 4, -3 \rangle - \langle 3, -1, 2 \rangle = \langle -2, 5, -5 \rangle and AC→=c⃗−a⃗=⟨5,2,1⟩−⟨3,−1,2⟩=⟨2,3,−1⟩\overrightarrow{AC} = \vec{c} - \vec{a} = \langle 5, 2, 1 \rangle - \langle 3, -1, 2 \rangle = \langle 2, 3, -1 \rangle. Then AD→=2⟨−2,5,−5⟩+12⟨2,3,−1⟩=⟨−4,10,−10⟩+⟨1,1.5,−0.5⟩=⟨−3,11.5,−10.5⟩\overrightarrow{AD} = 2\langle -2, 5, -5 \rangle + \frac{1}{2}\langle 2, 3, -1 \rangle = \langle -4, 10, -10 \rangle + \langle 1, 1.5, -0.5 \rangle = \langle -3, 11.5, -10.5 \rangle. Therefore, d⃗=a⃗+AD→=⟨3,−1,2⟩+⟨−3,11.5,−10.5⟩=⟨0,10.5,−8.5⟩\vec{d} = \vec{a} + \overrightarrow{AD} = \langle 3, -1, 2 \rangle + \langle -3, 11.5, -10.5 \rangle = \langle 0, 10.5, -8.5 \rangle. The zz-component is −8.5-8.5. Choice A results from computational errors in the scalar multiplication. Choice B comes from incorrectly computing 2+(−10)+(−1)=−92 + (-10) + (-1) = -9 then subtracting 2. Choice D results from forgetting the 12\frac{1}{2} factor in the second term.
  2. Three vectors p⃗\vec{p}, q⃗\vec{q}, and r⃗\vec{r} satisfy the equation 3p⃗+2q⃗−r⃗=0⃗3\vec{p} + 2\vec{q} - \vec{r} = \vec{0}. If p⃗=⟨1,−2,3⟩\vec{p} = \langle 1, -2, 3 \rangle and q⃗=⟨2,1,−1⟩\vec{q} = \langle 2, 1, -1 \rangle, what is the component of r⃗\vec{r} in the direction of the unit vector i^\hat{i}?
    Answer: 77 From 3p⃗+2q⃗−r⃗=0⃗3\vec{p} + 2\vec{q} - \vec{r} = \vec{0}, we get r⃗=3p⃗+2q⃗\vec{r} = 3\vec{p} + 2\vec{q}. Computing: r⃗=3⟨1,−2,3⟩+2⟨2,1,−1⟩=⟨3,−6,9⟩+⟨4,2,−2⟩=⟨7,−4,7⟩\vec{r} = 3\langle 1, -2, 3 \rangle + 2\langle 2, 1, -1 \rangle = \langle 3, -6, 9 \rangle + \langle 4, 2, -2 \rangle = \langle 7, -4, 7 \rangle. The component in the direction of i^=⟨1,0,0⟩\hat{i} = \langle 1, 0, 0 \rangle is the xx-component, which is 77. Choice B results from a sign error in rearranging the equation. Choice C comes from confusing the coefficient of p⃗\vec{p} with the final answer. Choice D results from both sign and coefficient errors.
  3. Let u⃗=3a⃗−2b⃗\vec{u} = 3\vec{a} - 2\vec{b} and v⃗=a⃗+4b⃗\vec{v} = \vec{a} + 4\vec{b}, where a⃗\vec{a} and b⃗\vec{b} are non-zero vectors. If 2u⃗−v⃗=ka⃗+mb⃗2\vec{u} - \vec{v} = k\vec{a} + m\vec{b} for scalars kk and mm, what is the value of k+mk + m?
    Answer: −3-3 First, substitute the given expressions: 2u⃗−v⃗=2(3a⃗−2b⃗)−(a⃗+4b⃗)=6a⃗−4b⃗−a⃗−4b⃗=5a⃗−8b⃗2\vec{u} - \vec{v} = 2(3\vec{a} - 2\vec{b}) - (\vec{a} + 4\vec{b}) = 6\vec{a} - 4\vec{b} - \vec{a} - 4\vec{b} = 5\vec{a} - 8\vec{b}. Therefore k=5k = 5 and m=−8m = -8, so k+m=5+(−8)=−3k + m = 5 + (-8) = -3. Choice A results from incorrectly computing 6−1−4−4=−36-1-4-4 = -3 but then taking the absolute value. Choice B comes from finding k=5k = 5 but forgetting the negative sign on mm. Choice D results from computing k+mk + m as −5+(−3)=−8-5 + (-3) = -8 due to sign errors in the algebra.
  4. Given vectors a⃗=⟨2,−1,4⟩\vec{a} = \langle 2, -1, 4 \rangle and b⃗=⟨−3,2,1⟩\vec{b} = \langle -3, 2, 1 \rangle, find the scalar tt such that the vector w⃗=(1−t)a⃗+tb⃗\vec{w} = (1-t)\vec{a} + t\vec{b} has its yy-component equal to zero.
    Answer: t=13t = \frac{1}{3} The yy-component of w⃗\vec{w} is (1−t)(−1)+t(2)=−1+t+2t=−1+3t(1-t)(-1) + t(2) = -1 + t + 2t = -1 + 3t. Setting this equal to zero: −1+3t=0-1 + 3t = 0, so 3t=13t = 1 and t=13t = \frac{1}{3}. Choice B comes from incorrectly setting up the equation as 1−3t=01 - 3t = 0. Choice C results from a sign error when expanding (1−t)(−1)(1-t)(-1). Choice D combines both the setup error and sign error.

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