Vector Components & Magnitude - Calculus 3

1. A vector $\vec{r}$ makes equal angles with the positive $x$, $y$, and $z$ axes. If the $z$-component of $\vec{r}$ is $4$, what is the magnitude of $\vec{r}$?
   Answer: $4\sqrt{3}$ If $\vec{r}$ makes equal angles with all three coordinate axes, then $\vec{r} = (a,a,a)$ for some value $a$. Since the $z$-component is 4, we have $a = 4$, so $\vec{r} = (4,4,4)$. The magnitude is $|\vec{r}| = \sqrt{4^2 + 4^2 + 4^2} = \sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3}$. Choice B assumes doubling error, choice C assumes $|\vec{r}| = 3 \times 4 = 12$, and choice D uses $\sqrt{6}$ instead of $\sqrt{3}$.
2. Vector $\vec{v}$ has magnitude 12 and makes an angle of $60°$ with the positive $x$-axis and $45°$ with the positive $y$-axis. What is the $z$-component of $\vec{v}$?
   Answer: $\pm 6$ If $\vec{v} = (x,y,z)$ with $|\vec{v}| = 12$, then $x = 12\cos(60°) = 12 \cdot \frac{1}{2} = 6$ and $y = 12\cos(45°) = 12 \cdot \frac{\sqrt{2}}{2} = 6\sqrt{2}$. Using $x^2 + y^2 + z^2 = 144$: $36 + 72 + z^2 = 144$, so $z^2 = 36$ and $z = \pm 6$. Choice A assumes $y = 6$ instead of $6\sqrt{2}$, choice B results from computational error, and choice D assumes $x = 4\sqrt{3}$.
3. Vector $\vec{d} = (2, -3, 1)$ is expressed as $\vec{d} = \alpha\vec{i} + \beta\vec{j} + \gamma\vec{k}$ where $\vec{i}$, $\vec{j}$, and $\vec{k}$ are the standard unit vectors. If a new vector $\vec{e}$ is formed by rotating $\vec{d}$ by $90°$ about the $z$-axis, what is $|\vec{e}|$?
   Answer: $\sqrt{14}$ A $90°$ rotation about the $z$-axis transforms $(x,y,z) \mapsto (-y,x,z)$. So $\vec{e} = (-(-3), 2, 1) = (3, 2, 1)$. The magnitude is $|\vec{e}| = \sqrt{3^2 + 2^2 + 1^2} = \sqrt{9 + 4 + 1} = \sqrt{14}$. Note that rotation preserves magnitude, so $|\vec{e}| = |\vec{d}| = \sqrt{4 + 9 + 1} = \sqrt{14}$. Choice A uses $\sqrt{13}$ (computational error), choice C uses $\sqrt{15}$, and choice D represents $2\sqrt{7}$ which equals $\sqrt{28}$.
4. If $\vec{p} = (3, 4, 0)$ and $\vec{q} = (0, 3, 4)$, and vector $\vec{n}$ is a unit vector in the direction of $\vec{p} + \vec{q}$, what is the $y$-component of $\vec{n}$?
   Answer: $\frac{7}{\sqrt{74}}$ First calculate $\vec{p} + \vec{q} = (3,4,0) + (0,3,4) = (3,7,4)$. The magnitude is $|\vec{p} + \vec{q}| = \sqrt{3^2 + 7^2 + 4^2} = \sqrt{9 + 49 + 16} = \sqrt{74}$. The unit vector is $\vec{n} = \frac{(3,7,4)}{\sqrt{74}} = (\frac{3}{\sqrt{74}}, \frac{7}{\sqrt{74}}, \frac{4}{\sqrt{74}})$. Therefore, the $y$-component is $\frac{7}{\sqrt{74}}$. Choice B uses $\sqrt{65}$ (error in magnitude calculation), choice C assumes equal components, and choice D uses wrong magnitude $\sqrt{29}$.
5. If vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ satisfy $\vec{a} + \vec{b} + \vec{c} = \vec{0}$, and $|\vec{a}| = 3$, $|\vec{b}| = 4$, $|\vec{c}| = 5$, what is $\vec{a} \cdot \vec{b}$?
   Answer: $0$ This problem tests your understanding of vector relationships and dot products when vectors form a closed triangle. When you see vectors that sum to zero with given magnitudes, think about the geometric interpretation: these three vectors form a triangle when placed head-to-tail. Since $\vec{a} + \vec{b} + \vec{c} = \vec{0}$, we can rearrange this to $\vec{a} + \vec{b} = -\vec{c}$. Taking the dot product of both sides with itself: $(\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = (-\vec{c}) \cdot (-\vec{c})$. Expanding the left side using the distributive property: $\vec{a} \cdot \vec{a} + 2\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} = \vec{c} \cdot \vec{c}$. Since $\vec{v} \cdot \vec{v} = |\vec{v}|^2$, this becomes: $|\vec{a}|^2 + 2\vec{a} \cdot \vec{b} + |\vec{b}|^2 = |\vec{c}|^2$. Substituting the given magnitudes: $3^2 + 2\vec{a} \cdot \vec{b} + 4^2 = 5^2$, which gives us $9 + 2\vec{a} \cdot \vec{b} + 16 = 25$. Solving: $25 + 2\vec{a} \cdot \vec{b} = 25$, so $\vec{a} \cdot \vec{b} = 0$. Choice A (-1) and choice B (-½) result from calculation errors or incorrect setup of the constraint equation. Choice C (½) might come from sign errors when manipulating the vector equation or forgetting the factor of 2 in the expansion. The correct answer is D: $0$. Strategy tip: When vectors sum to zero, use the algebraic manipulation $\vec{a} + \vec{b} = -\vec{c}$ and take dot products to create solvable equations. This technique works for any "closed polygon" vector problem.
6. A particle moves along a curve such that its position vector is $\vec{r}(t) = (t^2 - 1, 2t, t^2 + 1)$. At what value of $t$ is the magnitude of the position vector minimized?
   Answer: $t = 0$ When finding the minimum magnitude of a position vector, you're looking for the point where the particle is closest to the origin. This requires setting up an optimization problem using calculus techniques. The magnitude of the position vector is $|\vec{r}(t)| = \sqrt{(t^2-1)^2 + (2t)^2 + (t^2+1)^2}$. To minimize this, it's easier to minimize the squared magnitude: $|\vec{r}(t)|^2 = (t^2-1)^2 + 4t^2 + (t^2+1)^2$. Expanding this expression: $|\vec{r}(t)|^2 = t^4 - 2t^2 + 1 + 4t^2 + t^4 + 2t^2 + 1 = 2t^4 + 4t^2 + 2$ Taking the derivative: $\frac{d}{dt}|\vec{r}(t)|^2 = 8t^3 + 8t = 8t(t^2 + 1)$ Setting this equal to zero: $8t(t^2 + 1) = 0$. Since $t^2 + 1 > 0$ for all real $t$, we must have $t = 0$. To confirm this is a minimum, check the second derivative: $\frac{d^2}{dt^2}|\vec{r}(t)|^2 = 24t^2 + 8$. At $t = 0$, this equals 8, which is positive, confirming a minimum. Choice A ($t = \frac{1}{2}$) and choice B ($t = 1$) are positive values where the derivative is positive, meaning the function is increasing. Choice C ($t = -1$) gives a negative derivative value, so the function is decreasing there, not at a minimum. Remember: when optimizing vector magnitudes, work with the squared magnitude to avoid messy square root derivatives, then use standard calculus optimization techniques.

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