Vector Components & Magnitude - Multivariable Calculus

1. A vector $\vec{v}$ begins at the origin and ends at a point $P(a,b,c)$ on the surface of a sphere of radius $R$ centered at the origin. Let $\vec{u} = \langle a, b, 0 \rangle$ be the projection of $\vec{v}$ onto the $xy$-plane. If the magnitude of $\vec{u}$ is half the magnitude of $\vec{v}$, what is the absolute value of the $z$-component, $|c|$, in terms of $R$?
   Answer: $|c| = \frac{\sqrt{3}}{2} R$ The magnitude of vector $\vec{v}$ is $|\vec{v}| = \sqrt{a^2+b^2+c^2}$. Since its endpoint lies on a sphere of radius $R$ centered at the origin, $|\vec{v}| = R$. The magnitude of vector $\vec{u}$ is $|\vec{u}| = \sqrt{a^2+b^2}$. We are given that $|\vec{u}| = \frac{1}{2}|\vec{v}|$, so $\sqrt{a^2+b^2} = \frac{1}{2}R$. Squaring this equation gives $a^2+b^2 = \frac{1}{4}R^2$. We also know that $R^2 = |\vec{v}|^2 = a^2+b^2+c^2$. Substituting the expression for $a^2+b^2$, we get $R^2 = \frac{1}{4}R^2 + c^2$. Solving for $c^2$ yields $c^2 = R^2 - \frac{1}{4}R^2 = \frac{3}{4}R^2$. Taking the square root of both sides gives $|c| = \sqrt{\frac{3}{4}R^2} = \frac{\sqrt{3}}{2}R$.
2. A vector $\vec{w}$ has magnitude $4$ and makes angles of $60°$ with the positive $x$-axis and $45°$ with the positive $y$-axis. What is the $z$-component of $\vec{w}$?
   Answer: $\pm 2$ If $\vec{w} = (w_x, w_y, w_z)$ with $|\vec{w}| = 4$, then $w_x = 4\cos(60°) = 4 \cdot \frac{1}{2} = 2$ and $w_y = 4\cos(45°) = 4 \cdot \frac{\sqrt{2}}{2} = 2\sqrt{2}$. Using $w_x^2 + w_y^2 + w_z^2 = 16$: $4 + 8 + w_z^2 = 16$, so $w_z^2 = 4$ and $w_z = \pm 2$. Choice A incorrectly uses $\pm 2\sqrt{2}$. Choice C gives $\pm \sqrt{2}$. Choice D gives $\pm 4$, which would make the magnitude too large.
3. Let $\vec{u}$ and $\vec{v}$ be vectors with magnitudes $|\vec{u}|=3$ and $|\vec{v}|=4$. Which of the following is NOT a possible magnitude for the vector $\vec{w} = \vec{u} - \vec{v}$?
   Answer: 8 The triangle inequality for vectors states that $|\vec{a} + \vec{b}| \le |\vec{a}| + |\vec{b}|$ and $|\vec{a} - \vec{b}| \ge ||\vec{a}| - |\vec{b}||$. Applying this to $\vec{w} = \vec{u} - \vec{v}$, we have $||\vec{u}| - |\vec{v}|| \le |\vec{u} - \vec{v}| \le |\vec{u}| + |\vec{v}|$. Substituting the given magnitudes, we get $|3 - 4| \le |\vec{w}| \le 3 + 4$, which simplifies to $1 \le |\vec{w}| \le 7$. The magnitude of $\vec{w}$ must be in the closed interval $[1, 7]$. Of the given choices, 8 is the only value outside this interval and is therefore not a possible magnitude.
4. For which real value of $k$ is the vector $\vec{v} = \langle k, 2k, -2 \rangle$ a unit vector?
   Answer: For no real value of $k$. For $\vec{v}$ to be a unit vector, its magnitude must be 1. The square of its magnitude is $|\vec{v}|^2 = k^2 + (2k)^2 + (-2)^2 = k^2 + 4k^2 + 4 = 5k^2 + 4$. Setting the magnitude to 1 is equivalent to setting the squared magnitude to $1^2 = 1$. So, we must solve the equation $5k^2 + 4 = 1$. Subtracting 4 from both sides gives $5k^2 = -3$, which leads to $k^2 = -3/5$. Since the square of a real number cannot be negative, there is no real value of $k$ that satisfies this condition.
5. The vector $\vec{w} = \langle 2, -4, 4 \rangle$ has a magnitude of 6. What are the components of a vector $\vec{z}$ that has magnitude 2 and is in the direction opposite to $\vec{w}$?
   Answer: $\left\langle -\frac{2}{3}, \frac{4}{3}, -\frac{4}{3} \right\rangle$ First, find the unit vector in the direction of $\vec{w}$ by dividing $\vec{w}$ by its magnitude, which is given as 6. $\hat{w} = \frac{1}{6}\langle 2, -4, 4 \rangle = \left\langle \frac{2}{6}, -\frac{4}{6}, \frac{4}{6} \right\rangle = \left\langle \frac{1}{3}, -\frac{2}{3}, \frac{2}{3} \right\rangle$. A unit vector in the opposite direction is $-\hat{w} = \left\langle -\frac{1}{3}, \frac{2}{3}, -\frac{2}{3} \right\rangle$. To get a vector $\vec{z}$ of magnitude 2 in this opposite direction, we scale $-\hat{w}$ by 2: $\vec{z} = 2(-\hat{w}) = 2\left\langle -\frac{1}{3}, \frac{2}{3}, -\frac{2}{3} \right\rangle = \left\langle -\frac{2}{3}, \frac{4}{3}, -\frac{4}{3} \right\rangle$.
6. Given the vectors $\vec{a} = \langle 1, 2, 2 \rangle$ and $\vec{b} = \langle 3, 0, -4 \rangle$, find the magnitude of the vector $\vec{w} = |\vec{b}|\vec{a} - |\vec{a}|\vec{b}$.
   Answer: $10\sqrt{6}$ First, calculate the magnitudes of $\vec{a}$ and $\vec{b}$. $|\vec{a}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1+4+4} = \sqrt{9} = 3$. $|\vec{b}| = \sqrt{3^2 + 0^2 + (-4)^2} = \sqrt{9+0+16} = \sqrt{25} = 5$. Now, substitute these scalar values into the expression for $\vec{w}$: $\vec{w} = 5\vec{a} - 3\vec{b} = 5\langle 1, 2, 2 \rangle - 3\langle 3, 0, -4 \rangle$. Perform the scalar multiplication and vector subtraction: $\vec{w} = \langle 5, 10, 10 \rangle - \langle 9, 0, -12 \rangle = \langle 5-9, 10-0, 10-(-12) \rangle = \langle -4, 10, 22 \rangle$. Finally, calculate the magnitude of $\vec{w}$: $|\vec{w}| = \sqrt{(-4)^2 + 10^2 + 22^2} = \sqrt{16 + 100 + 484} = \sqrt{600} = \sqrt{100 \cdot 6} = 10\sqrt{6}$.
7. An airplane has a velocity vector relative to the air of $\vec{v}_a = \langle 100, 20, -5 \rangle$ km/h. The wind is blowing with a velocity vector of $\vec{v}_w = \langle 10, -15, 5 \rangle$ km/h. What is the ground speed of the airplane, to the nearest integer?
   Answer: $110 \text{ km/h}$ The airplane's velocity vector relative to the ground, $\vec{v}_g$, is the sum of its velocity relative to the air and the velocity of the wind: $\vec{v}_g = \vec{v}_a + \vec{v}_w = \langle 100+10, 20-15, -5+5 \rangle = \langle 110, 5, 0 \rangle$. The ground speed is the magnitude of this vector: $|\vec{v}_g| = \sqrt{110^2 + 5^2 + 0^2} = \sqrt{12100 + 25} = \sqrt{12125} \approx 110.11$ km/h. To the nearest integer, the ground speed is 110 km/h.
8. A vector $\vec{v}$ has initial point $P(2, -1, 3)$ and terminal point $Q(5, 3, -2)$. If $\vec{w}$ is a unit vector in the same direction as $\vec{v}$, what is the $z$-component of $\vec{w}$?
   Answer: $-\frac{1}{\sqrt{2}}$ First find $\vec{v} = Q - P = (5-2, 3-(-1), -2-3) = (3, 4, -5)$. The magnitude is $|\vec{v}| = \sqrt{3^2 + 4^2 + (-5)^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2}$. The unit vector is $\vec{w} = \frac{\vec{v}}{|\vec{v}|} = \frac{(3, 4, -5)}{5\sqrt{2}}$. The z-component is $\frac{-5}{5\sqrt{2}} = -\frac{1}{\sqrt{2}}$. Choice A uses the wrong magnitude ($\sqrt{50}$ instead of $5\sqrt{2}$). Choice B is correct algebraically but doesn't simplify. Choice C has the wrong sign.
9. Two forces, $\vec{F}_1 = \langle 2, -3, 1 \rangle$ and $\vec{F}_2 = \langle -1, 5, 3 \rangle$, are applied to an object. What is the magnitude of the resultant force, $\vec{F}_1 + \vec{F}_2$?
   Answer: $\sqrt{21}$ First, find the resultant force vector by adding the components of $\vec{F}_1$ and $\vec{F}_2$: $\vec{F}_{res} = \vec{F}_1 + \vec{F}_2 = \langle 2+(-1), -3+5, 1+3 \rangle = \langle 1, 2, 4 \rangle$. Next, calculate the magnitude of this resultant vector: $|\vec{F}_{res}| = \sqrt{1^2 + 2^2 + 4^2} = \sqrt{1 + 4 + 16} = \sqrt{21}$.
10. A force is described by the vector $\vec{F} = \langle 2, a, 3 \rangle$. The magnitude of the force is 7, and its $y$-component is positive. What is the unit vector in the direction of $\vec{F}$?
    Answer: $\langle \frac{2}{7}, \frac{6}{7}, \frac{3}{7} \rangle$ The magnitude of $\vec{F}$ is given by $|\vec{F}| = \sqrt{2^2 + a^2 + 3^2} = \sqrt{4+a^2+9} = \sqrt{13+a^2}$. We are given that the magnitude is 7, so $\sqrt{13+a^2} = 7$. Squaring both sides gives $13+a^2 = 49$, which leads to $a^2 = 36$. Since the $y$-component is positive, we take $a=6$. The force vector is $\vec{F} = \langle 2, 6, 3 \rangle$. The unit vector is found by dividing the vector by its magnitude: $\hat{F} = \frac{\vec{F}}{|\vec{F}|} = \frac{\langle 2, 6, 3 \rangle}{7} = \langle \frac{2}{7}, \frac{6}{7}, \frac{3}{7} \rangle$.
11. A vector $\vec{v}$ has the form $\langle a, 2a, 3a \rangle$ for some real number $a$. If the magnitude of $\vec{v}$ is $\sqrt{56}$, what is a possible value of $a$?
    Answer: $2$ The magnitude of $\vec{v}$ is given by $|\vec{v}| = \sqrt{a^2 + (2a)^2 + (3a)^2}$. We are given that $|\vec{v}| = \sqrt{56}$. Squaring both sides, we get $|\vec{v}|^2 = 56$. Calculating the squared magnitude from the components: $a^2 + 4a^2 + 9a^2 = 14a^2$. Setting this equal to 56 gives $14a^2 = 56$, which simplifies to $a^2 = 4$. Therefore, the possible values for $a$ are $a=2$ and $a=-2$. Of the choices provided, $2$ is a possible value.
12. Let $\vec{v}$ be a vector that starts at the origin and ends at a point on the surface of the sphere defined by $x^2 + y^2 + z^2 = 16$. Which of the following statements about $\vec{v}$ must be true?
    Answer: The magnitude of $\vec{v}$ is 4. Let the endpoint of the vector $\vec{v}$ be the point $(x, y, z)$. Since the vector starts at the origin, its components are $\vec{v} = \langle x, y, z \rangle$. The magnitude of $\vec{v}$ is $|\vec{v}| = \sqrt{x^2 + y^2 + z^2}$. Because the endpoint $(x, y, z)$ lies on the sphere $x^2 + y^2 + z^2 = 16$, we can substitute 16 for $x^2 + y^2 + z^2$ in the magnitude formula. This gives $|\vec{v}| = \sqrt{16} = 4$.
13. A vector $\vec{v}$ has its tail at the point $P(1, 2, 3)$ and its head on the plane $z=5$. If the x and y components of $\vec{v}$ are $v_x = 3$ and $v_y = -1$, respectively, what is the magnitude of $\vec{v}$?
    Answer: $\sqrt{14}$ Let the head of the vector be at point $Q(x_Q, y_Q, z_Q)$. The components of the vector $\vec{v}$ are given by the differences in the coordinates of $Q$ and $P$. We have $v_x = x_Q - 1 = 3$, which implies $x_Q = 4$. We have $v_y = y_Q - 2 = -1$, which implies $y_Q = 1$. Since the head of the vector lies on the plane $z=5$, we know that $z_Q = 5$. The z-component of the vector is therefore $v_z = z_Q - 3 = 5 - 3 = 2$. So, the vector is $\vec{v} = \langle 3, -1, 2 \rangle$. The magnitude is $|\vec{v}| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$.
14. Let points $P$ and $Q$ in 3D space be $(2, -1, 4)$ and $(-1, 1, 6)$, respectively. Which of the following is a vector of magnitude 7 in the direction of the vector $\vec{PQ}$?
    Answer: $\frac{7}{\sqrt{17}}\langle -3, 2, 2 \rangle$ First, find the component form of the vector $\vec{PQ}$ by subtracting the coordinates of the initial point $P$ from the terminal point $Q$: $\vec{PQ} = \langle -1-2, 1-(-1), 6-4 \rangle = \langle -3, 2, 2 \rangle$. Next, find the magnitude of $\vec{PQ}$: $|\vec{PQ}| = \sqrt{(-3)^2 + 2^2 + 2^2} = \sqrt{9+4+4} = \sqrt{17}$. To get a unit vector in the direction of $\vec{PQ}$, divide $\vec{PQ}$ by its magnitude: $\hat{u} = \frac{1}{\sqrt{17}}\langle -3, 2, 2 \rangle$. Finally, to create a vector of magnitude 7 in this direction, multiply the unit vector by 7: $7\hat{u} = \frac{7}{\sqrt{17}}\langle -3, 2, 2 \rangle$.
15. A vector $\vec{u}$ of magnitude $6$ lies in the $xy$-plane and makes a $30°$ angle with the positive $x$-axis. Another vector $\vec{v} = (0, 0, 4)$ is added to $\vec{u}$. What is the magnitude of $\vec{u} + \vec{v}$?
    Answer: $2\sqrt{13}$ Since $\vec{u}$ lies in the $xy$-plane with magnitude $6$ and makes a $30°$ angle with the positive $x$-axis: $\vec{u} = (6\cos(30°), 6\sin(30°), 0) = (6 \cdot \frac{\sqrt{3}}{2}, 6 \cdot \frac{1}{2}, 0) = (3\sqrt{3}, 3, 0)$. Then $\vec{u} + \vec{v} = (3\sqrt{3}, 3, 0) + (0, 0, 4) = (3\sqrt{3}, 3, 4)$. The magnitude is $\sqrt{(3\sqrt{3})^2 + 3^2 + 4^2} = \sqrt{27 + 9 + 16} = \sqrt{52} = 2\sqrt{13}$. Choice A would be correct if we simply added the magnitudes ($6 + 4 = 10$), which is incorrect. Choice C gives $8$. Choice D gives $\sqrt{61}$, which is a plausible computational error.
16. Let $\vec{v} = \langle x, y, z \rangle$ be a vector with magnitude $5\sqrt{6}$. The $x$-component of $\vec{v}$ is equal to its $y$-component, and the $z$-component is twice its $y$-component. If all components are positive, what is $\vec{v}$?
    Answer: $\langle 5, 5, 10 \rangle$ From the problem statement, we have the relationships $x=y$ and $z=2y$. We can express the vector $\vec{v}$ in terms of $y$ as $\vec{v} = \langle y, y, 2y \rangle$. The magnitude of $\vec{v}$ is $|\vec{v}| = \sqrt{y^2 + y^2 + (2y)^2} = \sqrt{y^2 + y^2 + 4y^2} = \sqrt{6y^2} = |y|\sqrt{6}$. Since all components are positive, $y>0$. We are given that the magnitude is $5\sqrt{6}$. So, $y\sqrt{6} = 5\sqrt{6}$, which implies $y=5$. Substituting back, we find the components: $x=5$, $y=5$, and $z=2(5)=10$. Thus, $\vec{v} = \langle 5, 5, 10 \rangle$.
17. Find the unit vector in the direction of the vector $\vec{v} = \langle 3, -4, 5 \rangle$.
    Answer: $\frac{1}{5\sqrt{2}}\langle 3, -4, 5 \rangle$ When you see a question asking for a unit vector, you're being asked to find a vector with magnitude 1 that points in the same direction as the given vector. The process is straightforward: divide the original vector by its magnitude. First, let's find the magnitude of $\vec{v} = \langle 3, -4, 5 \rangle$. Using the formula $|\vec{v}| = \sqrt{x^2 + y^2 + z^2}$: $|\vec{v}| = \sqrt{3^2 + (-4)^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50}$ Since $\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}$, the unit vector is: $\hat{v} = \frac{1}{5\sqrt{2}}\langle 3, -4, 5 \rangle$ This confirms answer C is correct. Now let's examine why the other options fail. Choice A uses $\frac{1}{50}$, which treats the magnitude as 50 instead of $\sqrt{50}$—this is the common error of forgetting the square root in the magnitude formula. Choice B uses $\frac{1}{5}$, which would be correct if the magnitude were 5, but this ignores the z-component entirely (as if working in 2D where $\sqrt{3^2 + (-4)^2} = 5$). Choice D has the correct denominator $5\sqrt{2}$ but inexplicably changes all the vector components to their opposites, giving a unit vector in the opposite direction. Remember: unit vectors require dividing by the magnitude, which always involves a square root for the distance formula. Don't forget any components when calculating $\sqrt{x^2 + y^2 + z^2}$.
18. Let $\vec{u}$ and $\vec{v}$ be non-zero vectors in $\mathbb{R}^3$. If the magnitudes of the vectors satisfy the condition $|\vec{u}| + |\vec{v}| = |\vec{u} + \vec{v}|$, which of the following must be true about the components of $\vec{u} = \langle u_1, u_2, u_3 \rangle$ and $\vec{v} = \langle v_1, v_2, v_3 \rangle$?
    Answer: There exists a scalar $k > 0$ such that $u_i = k v_i$ for $i=1, 2, 3$. The condition $|\vec{u} + \vec{v}| \le |\vec{u}| + |\vec{v}|$ is the triangle inequality for vectors. The equality holds if and only if the vectors $\vec{u}$ and $\vec{v}$ have the same direction. This means that one vector is a positive scalar multiple of the other, so $\vec{u} = k\vec{v}$ for some constant $k > 0$. This implies that their components are proportional with a positive constant of proportionality, i.e., $u_i = k v_i$ for $i=1, 2, 3$.
19. Vector $\vec{v}$ has components $(3, -4, k)$ where $k > 0$. If the angle between $\vec{v}$ and the positive $z$-axis is $60°$, what is the value of $k$?
    Answer: $k = \frac{5}{\sqrt{3}}$ The angle between $\vec{v}$ and the positive $z$-axis is given by $\cos(60°) = \frac{\vec{v} \cdot \hat{k}}{|\vec{v}|} = \frac{k}{\sqrt{9 + 16 + k^2}} = \frac{k}{\sqrt{25 + k^2}}$. Since $\cos(60°) = \frac{1}{2}$: $\frac{k}{\sqrt{25 + k^2}} = \frac{1}{2}$. Cross-multiplying: $2k = \sqrt{25 + k^2}$. Squaring both sides: $4k^2 = 25 + k^2$, so $3k^2 = 25$ and $k = \frac{5}{\sqrt{3}}$. Choice A gives $k = 5$, which would make $\cos \theta = \frac{5}{\sqrt{50}} = \frac{1}{\sqrt{2}} \neq \frac{1}{2}$. Choice B is too large. Choice C equals $\frac{5\sqrt{3}}{3}$, which doesn't satisfy the equation.
20. A vector $\vec{v}$ in 3D space has a magnitude of 10 and makes equal angles with the positive x, y, and z axes. What are the components of $\vec{v}$?
    Answer: $\left\langle \frac{10}{\sqrt{3}}, \frac{10}{\sqrt{3}}, \frac{10}{\sqrt{3}} \right\rangle$ If a vector makes equal angles with the positive coordinate axes, its components must be equal. Let $\vec{v} = \langle a, a, a \rangle$ where $a > 0$. The magnitude of $\vec{v}$ is $|\vec{v}| = \sqrt{a^2 + a^2 + a^2} = \sqrt{3a^2} = a\sqrt{3}$. We are given that the magnitude is 10. So, $a\sqrt{3} = 10$, which implies $a = \frac{10}{\sqrt{3}}$. Therefore, the components of the vector are $\vec{v} = \left\langle \frac{10}{\sqrt{3}}, \frac{10}{\sqrt{3}}, \frac{10}{\sqrt{3}} \right\rangle$.

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