Multivariable Calculus Quiz: Vector Components And Magnitude
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Vector Components And MagnitudeQuestion 1 of 20

Find the unit vector in the direction of the vector v⃗=⟨3,−4,5⟩\vec{v} = \langle 3, -4, 5 \rangle.

150⟨3,−4,5⟩\frac{1}{50}\langle 3, -4, 5 \rangle
15⟨3,−4,5⟩\frac{1}{5}\langle 3, -4, 5 \rangle
152⟨3,−4,5⟩\frac{1}{5\sqrt{2}}\langle 3, -4, 5 \rangle
152⟨−3,4,−5⟩\frac{1}{5\sqrt{2}}\langle -3, 4, -5 \rangle
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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Vector Components And Magnitude

Practice Vector Components And Magnitude in Multivariable Calculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Vector Components And Magnitude, giving you a quick way to practice the rules, question types, and explanations that matter most for Multivariable Calculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Find the unit vector in the direction of the vector v⃗=⟨3,−4,5⟩\vec{v} = \langle 3, -4, 5 \rangle.

  1. 150⟨3,−4,5⟩\frac{1}{50}\langle 3, -4, 5 \rangle
  2. 15⟨3,−4,5⟩\frac{1}{5}\langle 3, -4, 5 \rangle
  3. 152⟨3,−4,5⟩\frac{1}{5\sqrt{2}}\langle 3, -4, 5 \rangle (correct answer)
  4. 152⟨−3,4,−5⟩\frac{1}{5\sqrt{2}}\langle -3, 4, -5 \rangle
Explanation: 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 v⃗=⟨3,−4,5⟩\vec{v} = \langle 3, -4, 5 \rangle. Using the formula ∣v⃗∣=x2+y2+z2|\vec{v}| = \sqrt{x^2 + y^2 + z^2}: ∣v⃗∣=32+(−4)2+52=9+16+25=50|\vec{v}| = \sqrt{3^2 + (-4)^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} Since 50=25⋅2=52\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}, the unit vector is: v^=152⟨3,−4,5⟩\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 150\frac{1}{50}, which treats the magnitude as 50 instead of 50\sqrt{50}—this is the common error of forgetting the square root in the magnitude formula. Choice B uses 15\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 32+(−4)2=5\sqrt{3^2 + (-4)^2} = 5). Choice D has the correct denominator 525\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 x2+y2+z2\sqrt{x^2 + y^2 + z^2}.

Question 2

An airplane has a velocity vector relative to the air of v⃗a=⟨100,20,−5⟩\vec{v}_a = \langle 100, 20, -5 \rangle km/h. The wind is blowing with a velocity vector of v⃗w=⟨10,−15,5⟩\vec{v}_w = \langle 10, -15, 5 \rangle km/h. What is the ground speed of the airplane, to the nearest integer?

  1. 102 km/h102 \text{ km/h}
  2. 110 km/h110 \text{ km/h} (correct answer)
  3. 97 km/h97 \text{ km/h}
  4. 129 km/h129 \text{ km/h}
Explanation: The airplane's velocity vector relative to the ground, v⃗g\vec{v}_g, is the sum of its velocity relative to the air and the velocity of the wind: v⃗g=v⃗a+v⃗w=⟨100+10,20−15,−5+5⟩=⟨110,5,0⟩\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: ∣v⃗g∣=1102+52+02=12100+25=12125≈110.11|\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.

Question 3

The vector w⃗=⟨2,−4,4⟩\vec{w} = \langle 2, -4, 4 \rangle has a magnitude of 6. What are the components of a vector z⃗\vec{z} that has magnitude 2 and is in the direction opposite to w⃗\vec{w}?

  1. ⟨−4,8,−8⟩\langle -4, 8, -8 \rangle
  2. ⟨23,−43,43⟩\left\langle \frac{2}{3}, -\frac{4}{3}, \frac{4}{3} \right\rangle
  3. ⟨−23,43,−43⟩\left\langle -\frac{2}{3}, \frac{4}{3}, -\frac{4}{3} \right\rangle (correct answer)
  4. ⟨−2,4,−4⟩\langle -2, 4, -4 \rangle
Explanation: First, find the unit vector in the direction of w⃗\vec{w} by dividing w⃗\vec{w} by its magnitude, which is given as 6. w^=16⟨2,−4,4⟩=⟨26,−46,46⟩=⟨13,−23,23⟩\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 −w^=⟨−13,23,−23⟩-\hat{w} = \left\langle -\frac{1}{3}, \frac{2}{3}, -\frac{2}{3} \right\rangle. To get a vector z⃗\vec{z} of magnitude 2 in this opposite direction, we scale −w^-\hat{w} by 2: z⃗=2(−w^)=2⟨−13,23,−23⟩=⟨−23,43,−43⟩\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.

Question 4

A vector v⃗\vec{v} has its tail at the point P(1,2,3)P(1, 2, 3) and its head on the plane z=5z=5. If the x and y components of v⃗\vec{v} are vx=3v_x = 3 and vy=−1v_y = -1, respectively, what is the magnitude of v⃗\vec{v}?

  1. 10\sqrt{10}
  2. 14\sqrt{14} (correct answer)
  3. 19\sqrt{19}
  4. 35\sqrt{35}
Explanation: Let the head of the vector be at point Q(xQ,yQ,zQ)Q(x_Q, y_Q, z_Q). The components of the vector v⃗\vec{v} are given by the differences in the coordinates of QQ and PP. We have vx=xQ−1=3v_x = x_Q - 1 = 3, which implies xQ=4x_Q = 4. We have vy=yQ−2=−1v_y = y_Q - 2 = -1, which implies yQ=1y_Q = 1. Since the head of the vector lies on the plane z=5z=5, we know that zQ=5z_Q = 5. The z-component of the vector is therefore vz=zQ−3=5−3=2v_z = z_Q - 3 = 5 - 3 = 2. So, the vector is v⃗=⟨3,−1,2⟩\vec{v} = \langle 3, -1, 2 \rangle. The magnitude is ∣v⃗∣=32+(−1)2+22=9+1+4=14|\vec{v}| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14}.

Question 5

Let u⃗\vec{u} and v⃗\vec{v} be vectors with magnitudes ∣u⃗∣=3|\vec{u}|=3 and ∣v⃗∣=4|\vec{v}|=4. Which of the following is NOT a possible magnitude for the vector w⃗=u⃗−v⃗\vec{w} = \vec{u} - \vec{v}?

  1. 1
  2. 3
  3. 7
  4. 8 (correct answer)
Explanation: The triangle inequality for vectors states that ∣a⃗+b⃗∣≤∣a⃗∣+∣b⃗∣|\vec{a} + \vec{b}| \le |\vec{a}| + |\vec{b}| and ∣a⃗−b⃗∣≥∣∣a⃗∣−∣b⃗∣∣|\vec{a} - \vec{b}| \ge ||\vec{a}| - |\vec{b}||. Applying this to w⃗=u⃗−v⃗\vec{w} = \vec{u} - \vec{v}, we have ∣∣u⃗∣−∣v⃗∣∣≤∣u⃗−v⃗∣≤∣u⃗∣+∣v⃗∣||\vec{u}| - |\vec{v}|| \le |\vec{u} - \vec{v}| \le |\vec{u}| + |\vec{v}|. Substituting the given magnitudes, we get ∣3−4∣≤∣w⃗∣≤3+4|3 - 4| \le |\vec{w}| \le 3 + 4, which simplifies to 1≤∣w⃗∣≤71 \le |\vec{w}| \le 7. The magnitude of w⃗\vec{w} must be in the closed interval [1,7][1, 7]. Of the given choices, 8 is the only value outside this interval and is therefore not a possible magnitude.

Question 6

Let v⃗=⟨x,y,z⟩\vec{v} = \langle x, y, z \rangle be a vector with magnitude 565\sqrt{6}. The xx-component of v⃗\vec{v} is equal to its yy-component, and the zz-component is twice its yy-component. If all components are positive, what is v⃗\vec{v}?

  1. ⟨5,5,10⟩\langle 5, 5, 10 \rangle (correct answer)
  2. ⟨5614,5614,10614⟩\langle \frac{5\sqrt{6}}{\sqrt{14}}, \frac{5\sqrt{6}}{\sqrt{14}}, \frac{10\sqrt{6}}{\sqrt{14}} \rangle
  3. ⟨30,30,230⟩\langle \sqrt{30}, \sqrt{30}, 2\sqrt{30} \rangle
  4. ⟨5,5,5⟩\langle 5, 5, 5 \rangle
Explanation: From the problem statement, we have the relationships x=yx=y and z=2yz=2y. We can express the vector v⃗\vec{v} in terms of yy as v⃗=⟨y,y,2y⟩\vec{v} = \langle y, y, 2y \rangle. The magnitude of v⃗\vec{v} is ∣v⃗∣=y2+y2+(2y)2=y2+y2+4y2=6y2=∣y∣6|\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>0y>0. We are given that the magnitude is 565\sqrt{6}. So, y6=56y\sqrt{6} = 5\sqrt{6}, which implies y=5y=5. Substituting back, we find the components: x=5x=5, y=5y=5, and z=2(5)=10z=2(5)=10. Thus, v⃗=⟨5,5,10⟩\vec{v} = \langle 5, 5, 10 \rangle.

Question 7

A vector v⃗\vec{v} begins at the origin and ends at a point P(a,b,c)P(a,b,c) on the surface of a sphere of radius RR centered at the origin. Let u⃗=⟨a,b,0⟩\vec{u} = \langle a, b, 0 \rangle be the projection of v⃗\vec{v} onto the xyxy-plane. If the magnitude of u⃗\vec{u} is half the magnitude of v⃗\vec{v}, what is the absolute value of the zz-component, ∣c∣|c|, in terms of RR?

  1. ∣c∣=12R|c| = \frac{1}{2} R
  2. ∣c∣=32R|c| = \frac{\sqrt{3}}{2} R (correct answer)
  3. ∣c∣=34R|c| = \frac{3}{4} R
  4. ∣c∣=52R|c| = \frac{\sqrt{5}}{2} R
Explanation: The magnitude of vector v⃗\vec{v} is ∣v⃗∣=a2+b2+c2|\vec{v}| = \sqrt{a^2+b^2+c^2}. Since its endpoint lies on a sphere of radius RR centered at the origin, ∣v⃗∣=R|\vec{v}| = R. The magnitude of vector u⃗\vec{u} is ∣u⃗∣=a2+b2|\vec{u}| = \sqrt{a^2+b^2}. We are given that ∣u⃗∣=12∣v⃗∣|\vec{u}| = \frac{1}{2}|\vec{v}|, so a2+b2=12R\sqrt{a^2+b^2} = \frac{1}{2}R. Squaring this equation gives a2+b2=14R2a^2+b^2 = \frac{1}{4}R^2. We also know that R2=∣v⃗∣2=a2+b2+c2R^2 = |\vec{v}|^2 = a^2+b^2+c^2. Substituting the expression for a2+b2a^2+b^2, we get R2=14R2+c2R^2 = \frac{1}{4}R^2 + c^2. Solving for c2c^2 yields c2=R2−14R2=34R2c^2 = R^2 - \frac{1}{4}R^2 = \frac{3}{4}R^2. Taking the square root of both sides gives ∣c∣=34R2=32R|c| = \sqrt{\frac{3}{4}R^2} = \frac{\sqrt{3}}{2}R.

Question 8

Let u⃗=⟨a,b,c⟩\vec{u} = \langle a, b, c \rangle be a unit vector and v⃗=⟨b,−a,0⟩\vec{v} = \langle b, -a, 0 \rangle. If the magnitude of the sum of the vectors is ∣u⃗+v⃗∣=3/2|\vec{u} + \vec{v}| = \sqrt{3/2}, what is the value of c2c^2?

  1. 1/41/4
  2. 3/43/4
  3. 2/2\sqrt{2}/2
  4. 1/21/2 (correct answer)
Explanation: When you encounter vector problems involving unit vectors and magnitudes, start by using the fundamental properties: unit vectors have magnitude 1, and you can find the magnitude of a sum using the dot product formula. Since u⃗=⟨a,b,c⟩\vec{u} = \langle a, b, c \rangle is a unit vector, we know a2+b2+c2=1a^2 + b^2 + c^2 = 1. The sum of the vectors is u⃗+v⃗=⟨a+b,b−a,c⟩\vec{u} + \vec{v} = \langle a + b, b - a, c \rangle. To find the magnitude, we calculate: ∣u⃗+v⃗∣2=(a+b)2+(b−a)2+c2|\vec{u} + \vec{v}|^2 = (a + b)^2 + (b - a)^2 + c^2 Expanding: (a+b)2+(b−a)2=a2+2ab+b2+b2−2ab+a2=2a2+2b2(a + b)^2 + (b - a)^2 = a^2 + 2ab + b^2 + b^2 - 2ab + a^2 = 2a^2 + 2b^2 So ∣u⃗+v⃗∣2=2a2+2b2+c2=2(a2+b2)+c2|\vec{u} + \vec{v}|^2 = 2a^2 + 2b^2 + c^2 = 2(a^2 + b^2) + c^2 Since a2+b2+c2=1a^2 + b^2 + c^2 = 1, we have a2+b2=1−c2a^2 + b^2 = 1 - c^2. Substituting: ∣u⃗+v⃗∣2=2(1−c2)+c2=2−2c2+c2=2−c2|\vec{u} + \vec{v}|^2 = 2(1 - c^2) + c^2 = 2 - 2c^2 + c^2 = 2 - c^2 Given that ∣u⃗+v⃗∣=3/2|\vec{u} + \vec{v}| = \sqrt{3/2}, we have ∣u⃗+v⃗∣2=3/2|\vec{u} + \vec{v}|^2 = 3/2. Therefore: 2−c2=322 - c^2 = \frac{3}{2} c2=2−32=12c^2 = 2 - \frac{3}{2} = \frac{1}{2} Choice D is correct. Choice A (1/41/4) would give a magnitude of 7/4\sqrt{7/4}, not 3/2\sqrt{3/2}. Choice B (3/43/4) would yield 5/4\sqrt{5/4}. Choice C (2/2\sqrt{2}/2) equals 1/21/2 numerically but represents a common computational error. Remember: when working with unit vectors, always use the constraint that their magnitude equals 1 to create relationships between components.

Question 9

Let u⃗\vec{u} and v⃗\vec{v} be non-zero vectors in R3\mathbb{R}^3. If the magnitudes of the vectors satisfy the condition ∣u⃗∣+∣v⃗∣=∣u⃗+v⃗∣|\vec{u}| + |\vec{v}| = |\vec{u} + \vec{v}|, which of the following must be true about the components of u⃗=⟨u1,u2,u3⟩\vec{u} = \langle u_1, u_2, u_3 \rangle and v⃗=⟨v1,v2,v3⟩\vec{v} = \langle v_1, v_2, v_3 \rangle?

  1. There exists a scalar k>0k > 0 such that ui=kviu_i = k v_i for i=1,2,3i=1, 2, 3. (correct answer)
  2. There exists a scalar k<0k < 0 such that ui=kviu_i = k v_i for i=1,2,3i=1, 2, 3.
  3. u1v1+u2v2+u3v3=0u_1 v_1 + u_2 v_2 + u_3 v_3 = 0.
  4. ∣u⃗∣2+∣v⃗∣2=∣u⃗+v⃗∣2|\vec{u}|^2 + |\vec{v}|^2 = |\vec{u} + \vec{v}|^2.
Explanation: The condition ∣u⃗+v⃗∣≤∣u⃗∣+∣v⃗∣|\vec{u} + \vec{v}| \le |\vec{u}| + |\vec{v}| is the triangle inequality for vectors. The equality holds if and only if the vectors u⃗\vec{u} and v⃗\vec{v} have the same direction. This means that one vector is a positive scalar multiple of the other, so u⃗=kv⃗\vec{u} = k\vec{v} for some constant k>0k > 0. This implies that their components are proportional with a positive constant of proportionality, i.e., ui=kviu_i = k v_i for i=1,2,3i=1, 2, 3.

Question 10

A vector v⃗\vec{v} with magnitude 5115\sqrt{11} has its tail at the origin. It makes equal angles with the positive xx- and yy-axes, and its zz-component is three times its xx-component. Given that all components of v⃗\vec{v} are positive, what is the sum of its components?

  1. 5115\sqrt{11}
  2. 15
  3. 25 (correct answer)
  4. 5555\sqrt{55}
Explanation: This problem tests your understanding of vector components and the relationship between direction angles and magnitude. When a vector makes equal angles with coordinate axes, you're dealing with symmetric geometric relationships that create specific component ratios. Let the vector be v⃗=⟨a,b,c⟩\vec{v} = \langle a, b, c \rangle. Since the vector makes equal angles with the positive x- and y-axes, we have a=ba = b. The condition that the z-component is three times the x-component gives us c=3ac = 3a. Therefore, v⃗=⟨a,a,3a⟩\vec{v} = \langle a, a, 3a \rangle. Using the magnitude formula: ∣v⃗∣=a2+a2+(3a)2=11a2=a11|\vec{v}| = \sqrt{a^2 + a^2 + (3a)^2} = \sqrt{11a^2} = a\sqrt{11}. Since we're given that ∣v⃗∣=511|\vec{v}| = 5\sqrt{11}, we have a11=511a\sqrt{11} = 5\sqrt{11}, so a=5a = 5. This means v⃗=⟨5,5,15⟩\vec{v} = \langle 5, 5, 15 \rangle, and the sum of components is 5+5+15=255 + 5 + 15 = 25. Answer A (5115\sqrt{11}) represents the magnitude, not the component sum. Answer B (15) only accounts for the z-component, missing the x- and y-components entirely. Answer D (5555\sqrt{55}) likely results from incorrectly calculating 511×55\sqrt{11} \times \sqrt{5}, perhaps from mishandling the coefficient relationships. Study tip: When vectors make equal angles with multiple axes, immediately set those components equal to each other. Always verify your final answer by checking that the calculated magnitude matches the given magnitude—this catches algebraic errors and confirms your component relationships are correct.

Question 11

A vector v⃗\vec{v} has initial point P(2,−1,3)P(2, -1, 3) and terminal point Q(5,3,−2)Q(5, 3, -2). If w⃗\vec{w} is a unit vector in the same direction as v⃗\vec{v}, what is the zz-component of w⃗\vec{w}?

  1. −550-\frac{5}{\sqrt{50}}
  2. −552-\frac{5}{5\sqrt{2}}
  3. 550\frac{5}{\sqrt{50}}
  4. −12-\frac{1}{\sqrt{2}} (correct answer)
Explanation: First find v⃗=Q−P=(5−2,3−(−1),−2−3)=(3,4,−5)\vec{v} = Q - P = (5-2, 3-(-1), -2-3) = (3, 4, -5). The magnitude is ∣v⃗∣=32+42+(−5)2=9+16+25=50=52|\vec{v}| = \sqrt{3^2 + 4^2 + (-5)^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2}. The unit vector is w⃗=v⃗∣v⃗∣=(3,4,−5)52\vec{w} = \frac{\vec{v}}{|\vec{v}|} = \frac{(3, 4, -5)}{5\sqrt{2}}. The z-component is −552=−12\frac{-5}{5\sqrt{2}} = -\frac{1}{\sqrt{2}}. Choice A uses the wrong magnitude (50\sqrt{50} instead of 525\sqrt{2}). Choice B is correct algebraically but doesn't simplify. Choice C has the wrong sign.

Question 12

A vector u⃗\vec{u} of magnitude 66 lies in the xyxy-plane and makes a 30°30° angle with the positive xx-axis. Another vector v⃗=(0,0,4)\vec{v} = (0, 0, 4) is added to u⃗\vec{u}. What is the magnitude of u⃗+v⃗\vec{u} + \vec{v}?

  1. 1010
  2. 2132\sqrt{13} (correct answer)
  3. 88
  4. 61\sqrt{61}
Explanation: Since u⃗\vec{u} lies in the xyxy-plane with magnitude 66 and makes a 30°30° angle with the positive xx-axis: u⃗=(6cos⁡(30°),6sin⁡(30°),0)=(6⋅32,6⋅12,0)=(33,3,0)\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 u⃗+v⃗=(33,3,0)+(0,0,4)=(33,3,4)\vec{u} + \vec{v} = (3\sqrt{3}, 3, 0) + (0, 0, 4) = (3\sqrt{3}, 3, 4). The magnitude is (33)2+32+42=27+9+16=52=213\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=106 + 4 = 10), which is incorrect. Choice C gives 88. Choice D gives 61\sqrt{61}, which is a plausible computational error.

Question 13

Two forces, F⃗1=⟨2,−3,1⟩\vec{F}_1 = \langle 2, -3, 1 \rangle and F⃗2=⟨−1,5,3⟩\vec{F}_2 = \langle -1, 5, 3 \rangle, are applied to an object. What is the magnitude of the resultant force, F⃗1+F⃗2\vec{F}_1 + \vec{F}_2?

  1. 21\sqrt{21} (correct answer)
  2. 14+35\sqrt{14} + \sqrt{35}
  3. 57\sqrt{57}
  4. 77
Explanation: First, find the resultant force vector by adding the components of F⃗1\vec{F}_1 and F⃗2\vec{F}_2: F⃗res=F⃗1+F⃗2=⟨2+(−1),−3+5,1+3⟩=⟨1,2,4⟩\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: ∣F⃗res∣=12+22+42=1+4+16=21|\vec{F}_{res}| = \sqrt{1^2 + 2^2 + 4^2} = \sqrt{1 + 4 + 16} = \sqrt{21}.

Question 14

Let u⃗\vec{u} be a unit vector. A second vector v⃗\vec{v} is defined as v⃗=cu⃗\vec{v} = c\vec{u} where c=−3c = -3. Which of the following statements correctly describes v⃗\vec{v}?

  1. The magnitude of v⃗\vec{v} is 3, and it is in the opposite direction of u⃗\vec{u}. (correct answer)
  2. The magnitude of v⃗\vec{v} is -3, and it is in the opposite direction of u⃗\vec{u}.
  3. The magnitude of v⃗\vec{v} is 3, and it is in the same direction as u⃗\vec{u}.
  4. The magnitude of v⃗\vec{v} is 1, and it is in the opposite direction of u⃗\vec{u}.
Explanation: The magnitude of a scalar multiple of a vector is given by ∣cu⃗∣=∣c∣∣u⃗∣|c\vec{u}| = |c||\vec{u}|. Since u⃗\vec{u} is a unit vector, ∣u⃗∣=1|\vec{u}| = 1. Given c=−3c = -3, the magnitude of v⃗\vec{v} is ∣v⃗∣=∣−3∣⋅1=3|\vec{v}| = |-3| \cdot 1 = 3. Because the scalar cc is negative, the vector v⃗\vec{v} points in the direction opposite to u⃗\vec{u}.

Question 15

Let points PP and QQ in 3D space be (2,−1,4)(2, -1, 4) and (−1,1,6)(-1, 1, 6), respectively. Which of the following is a vector of magnitude 7 in the direction of the vector PQ⃗\vec{PQ}?

  1. 717⟨−3,2,2⟩\frac{7}{\sqrt{17}}\langle -3, 2, 2 \rangle (correct answer)
  2. 729⟨3,−2,−2⟩\frac{7}{\sqrt{29}}\langle 3, -2, -2 \rangle
  3. 17⟨−3,2,2⟩\sqrt{17}\langle -3, 2, 2 \rangle
  4. 117⟨−3,2,2⟩\frac{1}{\sqrt{17}}\langle -3, 2, 2 \rangle
Explanation: First, find the component form of the vector PQ⃗\vec{PQ} by subtracting the coordinates of the initial point PP from the terminal point QQ: PQ⃗=⟨−1−2,1−(−1),6−4⟩=⟨−3,2,2⟩\vec{PQ} = \langle -1-2, 1-(-1), 6-4 \rangle = \langle -3, 2, 2 \rangle. Next, find the magnitude of PQ⃗\vec{PQ}: ∣PQ⃗∣=(−3)2+22+22=9+4+4=17|\vec{PQ}| = \sqrt{(-3)^2 + 2^2 + 2^2} = \sqrt{9+4+4} = \sqrt{17}. To get a unit vector in the direction of PQ⃗\vec{PQ}, divide PQ⃗\vec{PQ} by its magnitude: u^=117⟨−3,2,2⟩\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: 7u^=717⟨−3,2,2⟩7\hat{u} = \frac{7}{\sqrt{17}}\langle -3, 2, 2 \rangle.

Question 16

A vector v⃗\vec{v} has the form ⟨a,2a,3a⟩\langle a, 2a, 3a \rangle for some real number aa. If the magnitude of v⃗\vec{v} is 56\sqrt{56}, what is a possible value of aa?

  1. 22 (correct answer)
  2. 44
  3. 14\sqrt{14}
  4. 566\frac{\sqrt{56}}{6}
Explanation: The magnitude of v⃗\vec{v} is given by ∣v⃗∣=a2+(2a)2+(3a)2|\vec{v}| = \sqrt{a^2 + (2a)^2 + (3a)^2}. We are given that ∣v⃗∣=56|\vec{v}| = \sqrt{56}. Squaring both sides, we get ∣v⃗∣2=56|\vec{v}|^2 = 56. Calculating the squared magnitude from the components: a2+4a2+9a2=14a2a^2 + 4a^2 + 9a^2 = 14a^2. Setting this equal to 56 gives 14a2=5614a^2 = 56, which simplifies to a2=4a^2 = 4. Therefore, the possible values for aa are a=2a=2 and a=−2a=-2. Of the choices provided, 22 is a possible value.

Question 17

Let u⃗=⟨1,1,1⟩\vec{u} = \langle 1, 1, 1 \rangle and v⃗=⟨2,−1,3⟩\vec{v} = \langle 2, -1, 3 \rangle. What is the magnitude of the vector 2u⃗−v⃗2\vec{u} - \vec{v}?

  1. 66
  2. 2\sqrt{2}
  3. 10\sqrt{10} (correct answer)
  4. 23−142\sqrt{3} - \sqrt{14}
Explanation: This question tests vector arithmetic and magnitude calculation—fundamental skills you'll use throughout multivariable calculus. When you see problems involving vector operations and magnitude, work systematically: perform the vector operations first, then calculate the magnitude. Let's find 2u⃗−v⃗2\vec{u} - \vec{v} step by step. First, calculate 2u⃗=2⟨1,1,1⟩=⟨2,2,2⟩2\vec{u} = 2\langle 1, 1, 1 \rangle = \langle 2, 2, 2 \rangle. Then subtract v⃗\vec{v}: 2u⃗−v⃗=⟨2,2,2⟩−⟨2,−1,3⟩=⟨0,3,−1⟩2\vec{u} - \vec{v} = \langle 2, 2, 2 \rangle - \langle 2, -1, 3 \rangle = \langle 0, 3, -1 \rangle. Now find the magnitude using ∣w⃗∣=w12+w22+w32|\vec{w}| = \sqrt{w_1^2 + w_2^2 + w_3^2}: ∣2u⃗−v⃗∣=02+32+(−1)2=0+9+1=10|2\vec{u} - \vec{v}| = \sqrt{0^2 + 3^2 + (-1)^2} = \sqrt{0 + 9 + 1} = \sqrt{10}. Looking at the wrong answers: Choice A (66) likely comes from incorrectly adding the components 0+3+(−1)=20 + 3 + (-1) = 2 or making an arithmetic error. Choice B (2\sqrt{2}) might result from forgetting to square the components or only using some of them. Choice D (23−142\sqrt{3} - \sqrt{14}) represents a conceptual error—you cannot subtract magnitudes of individual vectors to get the magnitude of their difference. Remember: vector magnitude is always calculated after performing all vector operations. Never try to work with magnitudes during intermediate steps—this leads to incorrect expressions like choice D. Always complete the vector arithmetic first, then apply the magnitude formula to your final vector.

Question 18

Let v⃗\vec{v} be a vector that starts at the origin and ends at a point on the surface of the sphere defined by x2+y2+z2=16x^2 + y^2 + z^2 = 16. Which of the following statements about v⃗\vec{v} must be true?

  1. The components of v⃗\vec{v} are all equal.
  2. The sum of the components of v⃗\vec{v} is 4.
  3. The magnitude of v⃗\vec{v} is 16.
  4. The magnitude of v⃗\vec{v} is 4. (correct answer)
Explanation: Let the endpoint of the vector v⃗\vec{v} be the point (x,y,z)(x, y, z). Since the vector starts at the origin, its components are v⃗=⟨x,y,z⟩\vec{v} = \langle x, y, z \rangle. The magnitude of v⃗\vec{v} is ∣v⃗∣=x2+y2+z2|\vec{v}| = \sqrt{x^2 + y^2 + z^2}. Because the endpoint (x,y,z)(x, y, z) lies on the sphere x2+y2+z2=16x^2 + y^2 + z^2 = 16, we can substitute 16 for x2+y2+z2x^2 + y^2 + z^2 in the magnitude formula. This gives ∣v⃗∣=16=4|\vec{v}| = \sqrt{16} = 4.

Question 19

A vector v⃗\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 v⃗\vec{v}?

  1. ⟨10,10,10⟩\langle 10, 10, 10 \rangle
  2. ⟨103,103,103⟩\left\langle \frac{10}{\sqrt{3}}, \frac{10}{\sqrt{3}}, \frac{10}{\sqrt{3}} \right\rangle (correct answer)
  3. ⟨103,103,103⟩\left\langle \frac{10}{3}, \frac{10}{3}, \frac{10}{3} \right\rangle
  4. ⟨10,10,10⟩\langle \sqrt{10}, \sqrt{10}, \sqrt{10} \rangle
Explanation: If a vector makes equal angles with the positive coordinate axes, its components must be equal. Let v⃗=⟨a,a,a⟩\vec{v} = \langle a, a, a \rangle where a>0a > 0. The magnitude of v⃗\vec{v} is ∣v⃗∣=a2+a2+a2=3a2=a3|\vec{v}| = \sqrt{a^2 + a^2 + a^2} = \sqrt{3a^2} = a\sqrt{3}. We are given that the magnitude is 10. So, a3=10a\sqrt{3} = 10, which implies a=103a = \frac{10}{\sqrt{3}}. Therefore, the components of the vector are v⃗=⟨103,103,103⟩\vec{v} = \left\langle \frac{10}{\sqrt{3}}, \frac{10}{\sqrt{3}}, \frac{10}{\sqrt{3}} \right\rangle.

Question 20

For which real value of kk is the vector v⃗=⟨k,2k,−2⟩\vec{v} = \langle k, 2k, -2 \rangle a unit vector?

  1. k=±1k = \pm 1
  2. k=−3/5k = -3/5
  3. k=±3/5k = \pm \sqrt{3/5}
  4. For no real value of kk. (correct answer)
Explanation: For v⃗\vec{v} to be a unit vector, its magnitude must be 1. The square of its magnitude is ∣v⃗∣2=k2+(2k)2+(−2)2=k2+4k2+4=5k2+4|\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 12=11^2 = 1. So, we must solve the equation 5k2+4=15k^2 + 4 = 1. Subtracting 4 from both sides gives 5k2=−35k^2 = -3, which leads to k2=−3/5k^2 = -3/5. Since the square of a real number cannot be negative, there is no real value of kk that satisfies this condition.