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AP Precalculus Quiz

AP Precalculus Quiz: Vectors

Practice Vectors in AP Precalculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 20

0 of 20 answered

Let v⃗=⟨−3,6⟩\vec{v} = \langle -3, 6 \ranglev=⟨−3,6⟩ and w⃗=⟨2,4⟩\vec{w} = \langle 2, 4 \ranglew=⟨2,4⟩. What is the dot product v⃗⋅w⃗\vec{v} \cdot \vec{w}v⋅w?

Select an answer to continue

What this quiz covers

This quiz focuses on Vectors, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Precalculus.

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

Let v⃗=⟨−3,6⟩\vec{v} = \langle -3, 6 \ranglev=⟨−3,6⟩ and w⃗=⟨2,4⟩\vec{w} = \langle 2, 4 \ranglew=⟨2,4⟩. What is the dot product v⃗⋅w⃗\vec{v} \cdot \vec{w}v⋅w?

  1. ⟨−6,24⟩\langle -6, 24 \rangle⟨−6,24⟩
  2. ⟨−1,10⟩\langle -1, 10 \rangle⟨−1,10⟩
  3. 181818 (correct answer)
  4. 303030

Explanation: The dot product of two vectors ⟨a,b⟩\langle a, b \rangle⟨a,b⟩ and ⟨c,d⟩\langle c, d \rangle⟨c,d⟩ is the scalar quantity ac+bdac + bdac+bd. Therefore, v⃗⋅w⃗=(−3)(2)+(6)(4)=−6+24=18\vec{v} \cdot \vec{w} = (-3)(2) + (6)(4) = -6 + 24 = 18v⋅w=(−3)(2)+(6)(4)=−6+24=18.

Question 2

Let a⃗=⟨2,2⟩\vec{a} = \langle 2, 2 \ranglea=⟨2,2⟩ and b⃗=⟨0,3⟩\vec{b} = \langle 0, 3 \rangleb=⟨0,3⟩. What is the measure of the angle θ\thetaθ, in radians, between vectors a⃗\vec{a}a and b⃗\vec{b}b?

  1. π6\frac{\pi}{6}6π​
  2. π4\frac{\pi}{4}4π​ (correct answer)
  3. π3\frac{\pi}{3}3π​
  4. π2\frac{\pi}{2}2π​

Explanation: The angle θ\thetaθ between two vectors is found using the formula cos⁡(θ)=a⃗⋅b⃗∥a⃗∥∥b⃗∥\cos(\theta) = \frac{\vec{a} \cdot \vec{b}}{\Vert\vec{a}\Vert \Vert\vec{b}\Vert}cos(θ)=∥a∥∥b∥a⋅b​. The dot product is a⃗⋅b⃗=(2)(0)+(2)(3)=6\vec{a} \cdot \vec{b} = (2)(0) + (2)(3) = 6a⋅b=(2)(0)+(2)(3)=6. The magnitudes are ∥a⃗∥=22+22=8=22\Vert\vec{a}\Vert = \sqrt{2^2+2^2} = \sqrt{8} = 2\sqrt{2}∥a∥=22+22​=8​=22​ and ∥b⃗∥=02+32=3\Vert\vec{b}\Vert = \sqrt{0^2+3^2} = 3∥b∥=02+32​=3. So, cos⁡(θ)=6(22)(3)=662=12\cos(\theta) = \frac{6}{(2\sqrt{2})(3)} = \frac{6}{6\sqrt{2}} = \frac{1}{\sqrt{2}}cos(θ)=(22​)(3)6​=62​6​=2​1​. The angle whose cosine is 12\frac{1}{\sqrt{2}}2​1​ is θ=π4\theta = \frac{\pi}{4}θ=4π​ radians.

Question 3

The vector w⃗\vec{w}w has initial point (1,9)(1, 9)(1,9) and terminal point (−4,3)(-4, 3)(−4,3). Which of the following expresses w⃗\vec{w}w as a linear combination of the standard unit vectors i⃗\vec{i}i and j⃗\vec{j}j​?

  1. 5i⃗+6j⃗5\vec{i} + 6\vec{j}5i+6j​
  2. −3i⃗+12j⃗-3\vec{i} + 12\vec{j}−3i+12j​
  3. −5i⃗−6j⃗-5\vec{i} - 6\vec{j}−5i−6j​ (correct answer)
  4. 3i⃗−12j⃗3\vec{i} - 12\vec{j}3i−12j​

Explanation: First, find the component form of the vector: w⃗=⟨−4−1,3−9⟩=⟨−5,−6⟩\vec{w} = \langle -4 - 1, 3 - 9 \rangle = \langle -5, -6 \ranglew=⟨−4−1,3−9⟩=⟨−5,−6⟩. A vector ⟨a,b⟩\langle a, b \rangle⟨a,b⟩ can be written as the linear combination ai⃗+bj⃗a\vec{i} + b\vec{j}ai+bj​. Therefore, w⃗=−5i⃗−6j⃗\vec{w} = -5\vec{i} - 6\vec{j}w=−5i−6j​.

Question 4

For what value of kkk are the vectors u⃗=⟨k,−4⟩\vec{u} = \langle k, -4 \rangleu=⟨k,−4⟩ and v⃗=⟨3,6⟩\vec{v} = \langle 3, 6 \ranglev=⟨3,6⟩ perpendicular?

  1. −8-8−8
  2. −2-2−2
  3. 222
  4. 888 (correct answer)

Explanation: Two vectors are perpendicular if and only if their dot product is zero. The dot product is u⃗⋅v⃗=(k)(3)+(−4)(6)=3k−24\vec{u} \cdot \vec{v} = (k)(3) + (-4)(6) = 3k - 24u⋅v=(k)(3)+(−4)(6)=3k−24. Setting the dot product to zero gives the equation 3k−24=03k - 24 = 03k−24=0. Solving for kkk, we get 3k=243k = 243k=24, so k=8k = 8k=8.

Question 5

A crate rests in equilibrium: weight W=−50j-50\mathbf{j}−50j N, horizontal push P=20i20\mathbf{i}20i N, and rope tension T balances them. Vectors have magnitude and direction. Refer to the vectors described in the passage. Which vector represents the equilibrium in the system described?​

  1. T=−20i+50j\mathbf{T}=-20\mathbf{i}+50\mathbf{j}T=−20i+50j N (correct answer)
  2. T=20i−50j\mathbf{T}=20\mathbf{i}-50\mathbf{j}T=20i−50j N
  3. T=70i+0j\mathbf{T}=70\mathbf{i}+0\mathbf{j}T=70i+0j N
  4. T=−20i−50j\mathbf{T}=-20\mathbf{i}-50\mathbf{j}T=−20i−50j N

Explanation: This question tests AP Precalculus skills in vectors, specifically equilibrium conditions and vector addition. Vectors are quantities having both magnitude and direction, essential in physics for analyzing forces in static equilibrium. In this scenario, three forces act on a crate at rest, requiring the tension vector T to balance the weight and push forces so their sum equals zero. Choice A is correct because for equilibrium, W + P + T = 0, so T = -(W + P) = -(-50j + 20i) = -20i + 50j N, which exactly balances the other forces. Choice C is incorrect because it only considers magnitudes without proper vector addition, failing to account for the vertical component needed to balance the weight. To help students: Emphasize that equilibrium means the vector sum equals zero, not just balancing magnitudes. Draw free body diagrams showing all forces and practice setting up equilibrium equations component by component.

Question 6

A drone flies in a steady wind, modeled with vectors (magnitude and direction). The drone’s air-velocity is d⃗=6i+1j+0k\vec{d}=6\mathbf{i}+1\mathbf{j}+0\mathbf{k}d=6i+1j+0k (m/s), and the wind velocity is w⃗=−2i+3j+0k\vec{w}=-2\mathbf{i}+3\mathbf{j}+0\mathbf{k}w=−2i+3j+0k (m/s). The drone’s ground velocity is the vector sum g⃗=d⃗+w⃗\vec{g}=\vec{d}+\vec{w}g​=d+w because both effects act simultaneously. Subtraction represents “removing” an effect: if you know g⃗\vec{g}g​ and w⃗\vec{w}w, then d⃗=g⃗−w⃗\vec{d}=\vec{g}-\vec{w}d=g​−w. The coordinate axes use xxx east and yyy north; arrows indicate direction. Refer to the vectors described in the passage. Determine the magnitude of the resultant vector g⃗\vec{g}g​.

  1. 20\sqrt{20}20​ m/s
  2. 32\sqrt{32}32​ m/s (correct answer)
  3. 52\sqrt{52}52​ m/s
  4. 8\sqrt{8}8​ m/s

Explanation: This question tests AP Precalculus skills in vectors, specifically vector addition and magnitude calculation using the Pythagorean theorem. Vectors combine effects acting simultaneously, such as a drone's air velocity and wind velocity producing a ground velocity. In this scenario, the ground velocity g = d + w = (6-2)i + (1+3)j = 4i + 4j represents the combined effect of drone motion and wind. Choice B is correct because the magnitude of g = 4i + 4j is calculated as √(4² + 4²) = √(16 + 16) = √32, using the formula |v| = √(x² + y²) for a 2D vector. Choice C is incorrect because it appears to calculate √(6² + 4²) = √52, possibly using incorrect components or confusing which vector's magnitude to find. To help students: Emphasize performing vector addition first, then calculating magnitude. Practice identifying when to add vectors versus when to find magnitudes, as these operations occur in different orders.

Question 7

An engineering test cable is pulled by a force F⃗=⟨3,−4,0⟩=3i−4j+0k\vec{F}=\langle 3,-4,0\rangle=3\mathbf{i}-4\mathbf{j}+0\mathbf{k}F=⟨3,−4,0⟩=3i−4j+0k kN, where magnitude is the pull strength and direction is along the cable. The technician doubles the load to model a stronger test, using scalar multiplication ∗k∗F⃗*k*\vec{F}∗k∗F with k = 2. Refer to the vectors described in the passage. How does the scalar multiplication of F⃗\vec{F}F affect its magnitude?

  1. It halves the magnitude and reverses direction.
  2. It doubles the magnitude and keeps direction. (correct answer)
  3. It keeps magnitude and rotates 90∘90^\circ90∘.
  4. It triples the magnitude and keeps direction.

Explanation: This question tests AP Precalculus skills in vectors, specifically scalar multiplication and its effect on magnitude and direction. Scalar multiplication of a vector changes its magnitude by the absolute value of the scalar and reverses direction if the scalar is negative. In this scenario, multiplying force vector F⃗=⟨3,−4,0⟩\vec{F}=\langle 3,-4,0\rangleF=⟨3,−4,0⟩ by scalar k=2k=2k=2 gives 2F⃗=⟨6,−8,0⟩2\vec{F}=\langle 6,-8,0\rangle2F=⟨6,−8,0⟩. Choice B is correct because scalar multiplication by 2 doubles the magnitude while keeping the same direction (since 2 is positive). The original magnitude is ∣F⃗∣=9+16=5|\vec{F}|=\sqrt{9+16}=5∣F∣=9+16​=5 kN, and after multiplication, ∣2F⃗∣=36+64=10|2\vec{F}|=\sqrt{36+64}=10∣2F∣=36+64​=10 kN, confirming the doubling. Choice A is incorrect because it describes multiplication by -0.5, not 2. To help students: Emphasize that positive scalars preserve direction while changing magnitude proportionally. Practice calculating magnitudes before and after scalar multiplication, and watch for confusion between the effects of positive versus negative scalars.

Question 8

A robot arm moves in 3D with displacement a=2i−1j+3k2\mathbf{i}-1\mathbf{j}+3\mathbf{k}2i−1j+3k cm and correction b=−4i+5j−1k-4\mathbf{i}+5\mathbf{j}-1\mathbf{k}−4i+5j−1k cm; vectors have magnitude and direction. Refer to the vectors described in the passage. What is the resultant vector when a and b are added?​

  1. −2i+4j+2k-2\mathbf{i}+4\mathbf{j}+2\mathbf{k}−2i+4j+2k cm (correct answer)
  2. 2i+4j+2k2\mathbf{i}+4\mathbf{j}+2\mathbf{k}2i+4j+2k cm
  3. −2i+6j−4k-2\mathbf{i}+6\mathbf{j}-4\mathbf{k}−2i+6j−4k cm
  4. 6i−6j+4k6\mathbf{i}-6\mathbf{j}+4\mathbf{k}6i−6j+4k cm

Explanation: This question tests AP Precalculus skills in vectors, specifically three-dimensional vector addition. Vectors are quantities having both magnitude and direction, extended to 3D space for robotics and spatial applications. In this scenario, a robot arm's displacement is corrected by adding another vector, requiring component-wise addition in three dimensions. Choice A is correct because vector addition works component-wise in 3D: (2i - 1j + 3k) + (-4i + 5j - 1k) = (2-4)i + (-1+5)j + (3-1)k = -2i + 4j + 2k cm. Choice B is incorrect because it appears to have sign errors in the i-component, possibly forgetting the negative sign on -4i. To help students: Extend 2D vector addition rules to 3D by treating each component independently. Use organized layouts showing i, j, and k components separately to avoid errors.

Question 9

A bridge joint is modeled with two tension forces: T⃗1=⟨8,0,0⟩=8i+0j+0k\vec{T}_1=\langle 8,0,0\rangle=8\mathbf{i}+0\mathbf{j}+0\mathbf{k}T1​=⟨8,0,0⟩=8i+0j+0k kN and T⃗2=⟨3,4,0⟩=3i+4j+0k\vec{T}_2=\langle 3,4,0\rangle=3\mathbf{i}+4\mathbf{j}+0\mathbf{k}T2​=⟨3,4,0⟩=3i+4j+0k kN. Vectors have magnitude and direction; the difference T⃗2−T⃗1\vec{T}_2-\vec{T}_1T2​−T1​ compares their effects. Refer to the vectors described in the passage. What is the resultant vector when T⃗2\vec{T}_2T2​ and T⃗1\vec{T}_1T1​ are subtracted?

  1. ⟨11,4,0⟩\langle 11,4,0\rangle⟨11,4,0⟩
  2. ⟨−5,4,0⟩\langle -5,4,0\rangle⟨−5,4,0⟩ (correct answer)
  3. ⟨5,−4,0⟩\langle 5,-4,0\rangle⟨5,−4,0⟩
  4. ⟨−11,−4,0⟩\langle -11,-4,0\rangle⟨−11,−4,0⟩

Explanation: This question tests AP Precalculus skills in vectors, specifically vector subtraction in structural analysis. Vector subtraction T⃗2−T⃗1\vec{T}_2-\vec{T}_1T2​−T1​ finds the difference between two vectors, showing how one vector must change to become the other. In this scenario, T⃗2−T⃗1=⟨3,4,0⟩−⟨8,0,0⟩=⟨3−8,4−0,0−0⟩=⟨−5,4,0⟩\vec{T}_2-\vec{T}_1 = \langle 3,4,0\rangle - \langle 8,0,0\rangle = \langle 3-8, 4-0, 0-0\rangle = \langle -5,4,0\rangleT2​−T1​=⟨3,4,0⟩−⟨8,0,0⟩=⟨3−8,4−0,0−0⟩=⟨−5,4,0⟩. Choice B is correct because it properly subtracts each component: x-component is 3-8=-5, y-component is 4-0=4, z-component is 0-0=0. Choice A is incorrect because it appears to add the vectors instead of subtracting them. To help students: Remember that vector subtraction means subtracting corresponding components, and be careful with the order (which vector comes first matters). Practice rewriting a⃗−b⃗\vec{a}-\vec{b}a−b as a⃗+(−b⃗)\vec{a}+(-\vec{b})a+(−b) to reinforce the concept.

Question 10

Refer to the coordinate-plane diagram: a sailboat’s intended displacement is a⃗=⟨5,1,0⟩=5i+1j+0k\vec{a}=\langle 5,1,0\rangle=5\mathbf{i}+1\mathbf{j}+0\mathbf{k}a=⟨5,1,0⟩=5i+1j+0k km and wind drift is b⃗=⟨−2,3,0⟩=−2i+3j+0k\vec{b}=\langle -2,3,0\rangle=-2\mathbf{i}+3\mathbf{j}+0\mathbf{k}b=⟨−2,3,0⟩=−2i+3j+0k km. Vectors have magnitude and direction; the actual displacement is a⃗+b⃗\vec{a}+\vec{b}a+b. Refer to the vectors described in the passage. What is the resultant vector when a⃗\vec{a}a and b⃗\vec{b}b are added?

  1. ⟨3,4,0⟩\langle 3,4,0\rangle⟨3,4,0⟩ (correct answer)
  2. ⟨7,−2,0⟩\langle 7,-2,0\rangle⟨7,−2,0⟩
  3. ⟨−3,4,0⟩\langle -3,4,0\rangle⟨−3,4,0⟩
  4. ⟨3,−2,0⟩\langle 3,-2,0\rangle⟨3,−2,0⟩

Explanation: This question tests AP Precalculus skills in vectors, specifically vector addition for navigation problems involving drift. Vectors represent displacements with both magnitude and direction, and actual displacement equals intended displacement plus drift. In this scenario, the actual displacement is a⃗+b⃗=⟨5,1,0⟩+⟨−2,3,0⟩=⟨5+(−2),1+3,0+0⟩=⟨3,4,0⟩\vec{a}+\vec{b} = \langle 5,1,0\rangle + \langle -2,3,0\rangle = \langle 5+(-2), 1+3, 0+0\rangle = \langle 3,4,0\ranglea+b=⟨5,1,0⟩+⟨−2,3,0⟩=⟨5+(−2),1+3,0+0⟩=⟨3,4,0⟩. Choice A is correct because it properly adds each component: x-component is 5+(-2)=3, y-component is 1+3=4, z-component is 0+0=0. Choice B is incorrect because it shows ⟨7,−2,0⟩\langle 7,-2,0\rangle⟨7,−2,0⟩, which would require subtracting the y-components instead of adding them. To help students: Draw vector diagrams showing how intended path and drift combine to give actual path. Practice systematic component addition, and watch for sign errors when adding negative components.

Question 11

A projectile launches with initial velocity v⃗0=⟨12,16,0⟩=12i+16j+0k\vec{v}_0=\langle 12,16,0\rangle=12\mathbf{i}+16\mathbf{j}+0\mathbf{k}v0​=⟨12,16,0⟩=12i+16j+0k m/s. After 1 s, gravity contributes Δv⃗=⟨0,−9.8,0⟩=0i−9.8j+0k\Delta\vec{v}=\langle 0,-9.8,0\rangle=0\mathbf{i}-9.8\mathbf{j}+0\mathbf{k}Δv=⟨0,−9.8,0⟩=0i−9.8j+0k m/s. Vectors have magnitude and direction; new velocity is v⃗0+Δv⃗\vec{v}_0+\Delta\vec{v}v0​+Δv. Refer to the vectors described in the passage. What is the resultant vector when v⃗0\vec{v}_0v0​ and Δv⃗\Delta\vec{v}Δv are added?

  1. ⟨12,6.2,0⟩\langle 12,6.2,0\rangle⟨12,6.2,0⟩ (correct answer)
  2. ⟨12,25.8,0⟩\langle 12,25.8,0\rangle⟨12,25.8,0⟩
  3. ⟨−12,6.2,0⟩\langle -12,6.2,0\rangle⟨−12,6.2,0⟩
  4. ⟨2.2,16,0⟩\langle 2.2,16,0\rangle⟨2.2,16,0⟩

Explanation: This question tests AP Precalculus skills in vectors, specifically vector addition in projectile motion with gravity effects. Vectors can represent velocities with magnitude (speed) and direction, and velocity changes are found by vector addition. In this scenario, the new velocity after 1 second is v⃗0+Δv⃗=⟨12,16,0⟩+⟨0,−9.8,0⟩=⟨12+0,16+(−9.8),0+0⟩=⟨12,6.2,0⟩\vec{v}_0+\Delta\vec{v} = \langle 12,16,0\rangle + \langle 0,-9.8,0\rangle = \langle 12+0, 16+(-9.8), 0+0\rangle = \langle 12,6.2,0\ranglev0​+Δv=⟨12,16,0⟩+⟨0,−9.8,0⟩=⟨12+0,16+(−9.8),0+0⟩=⟨12,6.2,0⟩. Choice A is correct because it properly adds the initial velocity and the change due to gravity: the x-component remains 12 (no horizontal acceleration), the y-component becomes 16-9.8=6.2 (gravity reduces upward velocity), and z remains 0. Choice B is incorrect because it appears to add 9.8 instead of subtracting it, ignoring that gravity acts downward. To help students: Emphasize that gravity contributes negative y-velocity in standard coordinates. Practice problems involving multiple time steps, and watch for sign errors with gravity.

Question 12

An AC circuit uses phasors: voltage V=6i+8j6\mathbf{i}+8\mathbf{j}6i+8j V and current I=8i−6j8\mathbf{i}-6\mathbf{j}8i−6j A; vectors have magnitude and direction. Refer to the vectors described in the passage. Calculate the angle between V and I.​

  1. 0∘0^\circ0∘
  2. 90∘90^\circ90∘ (correct answer)
  3. 45∘45^\circ45∘
  4. 180∘180^\circ180∘

Explanation: This question tests AP Precalculus skills in vectors, specifically calculating angles between vectors using the dot product. Vectors are quantities having both magnitude and direction, used in AC circuit analysis where phase relationships are crucial. In this scenario, voltage and current phasors require finding their angular relationship through the dot product formula. Choice B is correct because the dot product V·I = (6)(8) + (8)(-6) = 48 - 48 = 0, and when the dot product equals zero, vectors are perpendicular (90° angle). Choice C is incorrect as 45° would require the dot product to equal |V||I|/√2, not zero. To help students: Remember that perpendicular vectors have zero dot product, parallel vectors have maximum dot product. Practice using cos θ = (u·v)/(|u||v|) and recognizing special cases like perpendicular and parallel vectors.

Question 13

A robot controller stores a 3D velocity vector v⃗=⟨2,−1,3⟩=2i−1j+3k\vec{v}=\langle 2,-1,3\rangle=2\mathbf{i}-1\mathbf{j}+3\mathbf{k}v=⟨2,−1,3⟩=2i−1j+3k m/s, where magnitude is speed and direction is heading. It applies scalar multiplication with k = −2-2−2 for a reverse-and-scale maneuver. Refer to the vectors described in the passage. How does the scalar multiplication of v⃗\vec{v}v affect its magnitude?

  1. Magnitude doubles, and direction reverses. (correct answer)
  2. Magnitude halves, and direction reverses.
  3. Magnitude doubles, and direction stays same.
  4. Magnitude stays same, and direction reverses.

Explanation: This question tests AP Precalculus skills in vectors, specifically scalar multiplication with negative scalars and its effects on magnitude and direction. When a vector is multiplied by a negative scalar, the magnitude changes by the absolute value of the scalar and the direction reverses. In this scenario, multiplying v⃗=⟨2,−1,3⟩\vec{v}=\langle 2,-1,3\ranglev=⟨2,−1,3⟩ by k=−2k=-2k=−2 gives −2v⃗=⟨−4,2,−6⟩-2\vec{v}=\langle -4,2,-6\rangle−2v=⟨−4,2,−6⟩. Choice A is correct because the magnitude doubles (from ∣v⃗∣=4+1+9=14|\vec{v}|=\sqrt{4+1+9}=\sqrt{14}∣v∣=4+1+9​=14​ to ∣−2v⃗∣=16+4+36=56=214|-2\vec{v}|=\sqrt{16+4+36}=\sqrt{56}=2\sqrt{14}∣−2v∣=16+4+36​=56​=214​) and the direction reverses (due to the negative scalar). Choice D is incorrect because it ignores the magnitude change that occurs with scalar multiplication. To help students: Emphasize that the absolute value of the scalar determines magnitude change, while its sign determines direction. Practice with both positive and negative scalars, and watch for students who forget that magnitude always remains positive.

Question 14

A bridge cable tension is T=9i+12j9\mathbf{i}+12\mathbf{j}9i+12j kN and a support reaction is R=4i−5j4\mathbf{i}-5\mathbf{j}4i−5j kN; vectors have magnitude and direction. Refer to the vectors described in the passage. What is the resultant vector when T and R are added?​

  1. 13i+7j13\mathbf{i}+7\mathbf{j}13i+7j kN (correct answer)
  2. 5i+17j5\mathbf{i}+17\mathbf{j}5i+17j kN
  3. 13i+17j13\mathbf{i}+17\mathbf{j}13i+17j kN
  4. −13i−7j-13\mathbf{i}-7\mathbf{j}−13i−7j kN

Explanation: This question tests AP Precalculus skills in vectors, specifically vector addition with engineering applications. Vectors are quantities having both magnitude and direction, crucial for analyzing structural forces in engineering. In this scenario, cable tension and support reaction vectors combine to find the total force effect on the bridge structure. Choice A is correct because vector addition is performed component-wise: (9i + 12j) + (4i - 5j) = (9+4)i + (12-5)j = 13i + 7j kN. Choice C is incorrect because it fails to properly subtract the negative j-component of R, adding 12 + 5 instead of 12 + (-5), a common sign error. To help students: Emphasize careful attention to signs when adding vectors, especially with negative components. Use color coding or separate calculations for i and j components to avoid mixing them.

Question 15

A navigation app models two 2D vectors in a coordinate plane: a⃗=⟨4,1,0⟩=4i+1j+0k\vec{a}=\langle 4,1,0\rangle=4\mathbf{i}+1\mathbf{j}+0\mathbf{k}a=⟨4,1,0⟩=4i+1j+0k and b⃗=⟨1,5,0⟩=1i+5j+0k\vec{b}=\langle 1,5,0\rangle=1\mathbf{i}+5\mathbf{j}+0\mathbf{k}b=⟨1,5,0⟩=1i+5j+0k. Vectors have magnitude and direction; the angle between them depends on the dot product. Refer to the vectors described in the passage. Calculate the angle between a⃗\vec{a}a and b⃗\vec{b}b.

  1. cos⁡−1 ⁣(91726)\cos^{-1}\!\left(\frac{9}{\sqrt{17}\sqrt{26}}\right)cos−1(17​26​9​) (correct answer)
  2. cos⁡−1 ⁣(211726)\cos^{-1}\!\left(\frac{21}{\sqrt{17}\sqrt{26}}\right)cos−1(17​26​21​)
  3. cos⁡−1 ⁣(917+26)\cos^{-1}\!\left(\frac{9}{17+26}\right)cos−1(17+269​)
  4. cos⁡−1 ⁣(201726)\cos^{-1}\!\left(\frac{20}{\sqrt{17}\sqrt{26}}\right)cos−1(17​26​20​)

Explanation: This question tests AP Precalculus skills in vectors, specifically finding the angle between vectors using the dot product formula. The angle between vectors is found using cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}cosθ=∣a∣∣b∣a⋅b​, where the dot product measures how aligned the vectors are. In this scenario, a⃗⋅b⃗=(4)(1)+(1)(5)+(0)(0)=4+5=9\vec{a}\cdot\vec{b} = (4)(1)+(1)(5)+(0)(0) = 4+5 = 9a⋅b=(4)(1)+(1)(5)+(0)(0)=4+5=9, ∣a⃗∣=16+1=17|\vec{a}| = \sqrt{16+1} = \sqrt{17}∣a∣=16+1​=17​, and ∣b⃗∣=1+25=26|\vec{b}| = \sqrt{1+25} = \sqrt{26}∣b∣=1+25​=26​. Choice A is correct because it properly applies the formula: θ=cos⁡−1(91726)\theta = \cos^{-1}\left(\frac{9}{\sqrt{17}\sqrt{26}}\right)θ=cos−1(17​26​9​). Choice B is incorrect because it uses 21 instead of 9 for the dot product, likely from an arithmetic error. To help students: Break down the angle formula into steps - calculate dot product, find magnitudes, then divide. Practice computing dot products carefully, and watch for arithmetic errors in the numerator.

Question 16

A delivery drone corrects its position using displacement vectors (magnitude and direction). Its planned displacement is p⃗=10i+2j+0k\vec{p}=10\mathbf{i}+2\mathbf{j}+0\mathbf{k}p​=10i+2j+0k meters, but a gust causes an unintended drift of d⃗=−3i+5j+0k\vec{d}=-3\mathbf{i}+5\mathbf{j}+0\mathbf{k}d=−3i+5j+0k meters. The drone’s actual displacement is a⃗=p⃗+d⃗\vec{a}=\vec{p}+\vec{d}a=p​+d. To compute the correction needed to return to the planned endpoint, the drone uses vector subtraction: c⃗=p⃗−a⃗\vec{c}=\vec{p}-\vec{a}c=p​−a. On the coordinate plane, xxx is east and yyy is north; arrows show each displacement from the origin. Refer to the vectors described in the passage. What is the resultant vector when p⃗−a⃗\vec{p}-\vec{a}p​−a is computed?

  1. c⃗=−3i+5j+0k\vec{c}=-3\mathbf{i}+5\mathbf{j}+0\mathbf{k}c=−3i+5j+0k
  2. c⃗=3i−5j+0k\vec{c}=3\mathbf{i}-5\mathbf{j}+0\mathbf{k}c=3i−5j+0k (correct answer)
  3. c⃗=7i+7j+0k\vec{c}=7\mathbf{i}+7\mathbf{j}+0\mathbf{k}c=7i+7j+0k
  4. c⃗=13i−3j+0k\vec{c}=13\mathbf{i}-3\mathbf{j}+0\mathbf{k}c=13i−3j+0k

Explanation: This question tests AP Precalculus skills in vectors, specifically vector subtraction in correction calculations. Vectors model displacements, and subtraction finds the difference between planned and actual positions. In this scenario, the actual displacement is a = p + d = (10-3)i + (2+5)j = 7i + 7j, so the correction needed is c = p - a = (10-7)i + (2-7)j = 3i - 5j. Choice B is correct because it properly calculates the correction vector as 3i - 5j + 0k, representing 3 m east and 5 m south to return to the planned position. Choice A is incorrect because it has the wrong signs (-3i + 5j), which would move the drone further from the planned position rather than correcting it. To help students: Emphasize that vector subtraction finds the displacement from one position to another. Practice problems involving corrections and adjustments where the direction of the correction matters.

Question 17

A robot arm applies a force in 3D, where vectors have magnitude and direction and are written in i,j,k\mathbf{i},\mathbf{j},\mathbf{k}i,j,k notation. The applied force is F⃗=2i−1j+3k\vec{F}=2\mathbf{i}-1\mathbf{j}+3\mathbf{k}F=2i−1j+3k N. A calibration routine scales the force by the scalar k = 3, producing G⃗=kF⃗\vec{G}=k\vec{F}G=kF. Scalar multiplication multiplies each component by the same scalar, changing magnitude while keeping direction if the scalar is positive. The coordinate axes correspond to xxx, yyy, and zzz directions. Refer to the vectors described in the passage. What is the resultant vector when G⃗=3F⃗\vec{G}=3\vec{F}G=3F is computed?

  1. G⃗=6i−3j+9k\vec{G}=6\mathbf{i}-3\mathbf{j}+9\mathbf{k}G=6i−3j+9k (correct answer)
  2. G⃗=5i−4j+6k\vec{G}=5\mathbf{i}-4\mathbf{j}+6\mathbf{k}G=5i−4j+6k
  3. G⃗=23i−13j+k\vec{G}=\tfrac{2}{3}\mathbf{i}-\tfrac{1}{3}\mathbf{j}+\mathbf{k}G=32​i−31​j+k
  4. G⃗=6i+3j−9k\vec{G}=6\mathbf{i}+3\mathbf{j}-9\mathbf{k}G=6i+3j−9k

Explanation: This question tests AP Precalculus skills in vectors, specifically scalar multiplication in three dimensions. Vectors in 3D use i, j, k notation for x, y, z components, and scalar multiplication scales all components equally. In this scenario, multiplying F = 2i - 1j + 3k by scalar k = 3 gives G = 3(2i - 1j + 3k) = 6i - 3j + 9k. Choice A is correct because it accurately multiplies each component by 3: 3(2) = 6 for i, 3(-1) = -3 for j, and 3(3) = 9 for k, yielding 6i - 3j + 9k. Choice D is incorrect because it reverses the signs of the j and k components, giving 6i + 3j - 9k, possibly confusing scalar multiplication with some other operation. To help students: Emphasize that scalar multiplication affects magnitude but preserves direction relationships between components. Practice 3D vector operations to build comfort with i, j, k notation.

Question 18

A boat’s velocity relative to water is v⃗b=⟨6,2,0⟩=6i+2j+0k\vec{v}_b=\langle 6,2,0\rangle=6\mathbf{i}+2\mathbf{j}+0\mathbf{k}vb​=⟨6,2,0⟩=6i+2j+0k m/s and the river current is v⃗c=⟨−1,3,0⟩=−1i+3j+0k\vec{v}_c=\langle -1,3,0\rangle=-1\mathbf{i}+3\mathbf{j}+0\mathbf{k}vc​=⟨−1,3,0⟩=−1i+3j+0k m/s. Vectors have magnitude and direction, and ground velocity is vector addition. Refer to the vectors described in the passage. What is the resultant vector when v⃗b\vec{v}_bvb​ and v⃗c\vec{v}_cvc​ are added?

  1. ⟨5,5,0⟩\langle 5,5,0\rangle⟨5,5,0⟩ (correct answer)
  2. ⟨7,−1,0⟩\langle 7,-1,0\rangle⟨7,−1,0⟩
  3. ⟨−7,1,0⟩\langle -7,1,0\rangle⟨−7,1,0⟩
  4. ⟨6,6,0⟩\langle 6,6,0\rangle⟨6,6,0⟩

Explanation: This question tests AP Precalculus skills in vectors, specifically vector addition in the context of relative motion. Vectors represent quantities with both magnitude and direction, and when a boat moves through water with a current, the ground velocity is found by adding the boat's velocity relative to water and the water's velocity (current). In this scenario, we need to add the boat's velocity vector v⃗b=⟨6,2,0⟩\vec{v}_b=\langle 6,2,0\ranglevb​=⟨6,2,0⟩ to the current vector v⃗c=⟨−1,3,0⟩\vec{v}_c=\langle -1,3,0\ranglevc​=⟨−1,3,0⟩. Choice A is correct because v⃗b+v⃗c=⟨6,2,0⟩+⟨−1,3,0⟩=⟨6+(−1),2+3,0+0⟩=⟨5,5,0⟩\vec{v}_b + \vec{v}_c = \langle 6,2,0\rangle + \langle -1,3,0\rangle = \langle 6+(-1), 2+3, 0+0\rangle = \langle 5,5,0\ranglevb​+vc​=⟨6,2,0⟩+⟨−1,3,0⟩=⟨6+(−1),2+3,0+0⟩=⟨5,5,0⟩. Choice B is incorrect because it appears to subtract components incorrectly, getting ⟨7,−1,0⟩\langle 7,-1,0\rangle⟨7,−1,0⟩ instead of properly adding each component. To help students: Draw vector diagrams showing how velocities combine in relative motion problems. Practice component-wise addition systematically, and watch for sign errors when adding negative components.

Question 19

A drone experiences wind w=3i−2j3\mathbf{i}-2\mathbf{j}3i−2j m/s and thrust t=8i+5j8\mathbf{i}+5\mathbf{j}8i+5j m/s; vectors have magnitude and direction. Refer to the vectors described in the passage. What is the resultant vector when t and w are added?​

  1. 11i+3j11\mathbf{i}+3\mathbf{j}11i+3j m/s (correct answer)
  2. 5i+7j5\mathbf{i}+7\mathbf{j}5i+7j m/s
  3. 11i+7j11\mathbf{i}+7\mathbf{j}11i+7j m/s
  4. −11i−3j-11\mathbf{i}-3\mathbf{j}−11i−3j m/s

Explanation: This question tests AP Precalculus skills in vectors, specifically vector addition and understanding of vector components. Vectors are quantities having both magnitude and direction, essential in physics and engineering for representing forces and motions. In this scenario, vectors are used to model drone motion with wind and thrust forces, requiring precise calculation of the resultant vector through component-wise addition. Choice A is correct because it accurately calculates the resultant vector: (3i - 2j) + (8i + 5j) = (3+8)i + (-2+5)j = 11i + 3j m/s. Choice D is incorrect because it appears to subtract the vectors or apply incorrect signs, a common error when confusing vector addition with subtraction. To help students: Emphasize that vector addition is performed component-wise, adding i-components together and j-components together separately. Practice with physical interpretations like combining forces or velocities to solidify understanding.

Question 20

A robot’s displacement is d=6i−8j6\mathbf{i}-8\mathbf{j}6i−8j m (magnitude and direction). The program scales motion by k=12\tfrac1221​. Refer to the vectors described in the passage. How does the scalar multiplication of d affect its magnitude?​

  1. Magnitude doubles; direction unchanged
  2. Magnitude halves; direction unchanged (correct answer)
  3. Magnitude halves; direction reverses
  4. Magnitude unchanged; direction rotates 90∘90^\circ90∘

Explanation: This question tests AP Precalculus skills in vectors, specifically scalar multiplication and its effect on vector magnitude. Vectors are quantities having both magnitude and direction, where scalar multiplication scales the vector's length while preserving or reversing its direction. In this scenario, a robot's displacement vector is scaled by k = 1/2, requiring understanding of how scalar multiplication affects magnitude. Choice B is correct because scalar multiplication by 1/2 scales both components: (1/2)(6i - 8j) = 3i - 4j, and since magnitude scales linearly with positive scalars, the new magnitude is half the original. Choice A is incorrect because it reverses the scaling effect, possibly confusing multiplication by 1/2 with multiplication by 2. To help students: Emphasize that scalar multiplication by k scales magnitude by |k|, so multiplying by 1/2 halves the magnitude. Practice calculating magnitudes before and after scaling to verify the relationship.