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Vectors - AP Precalculus

  1. 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} 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} km. Vectors have magnitude and direction; the actual displacement is a⃗+b⃗\vec{a}+\vec{b}. Refer to the vectors described in the passage. What is the resultant vector when a⃗\vec{a} and b⃗\vec{b} are added?
    Answer: ⟨3,4,0⟩\langle 3,4,0\rangle 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\rangle. 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, 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.
  2. A robot's displacement is d=6i−8j6\mathbf{i}-8\mathbf{j} m (magnitude and direction). The program scales motion by *k*=12\tfrac12. Refer to the vectors described in the passage. How does the scalar multiplication of d affect its magnitude?​
    Answer: Magnitude halves; direction unchanged 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.
  3. The vector w⃗\vec{w} has initial point (1,9)(1, 9) and terminal point (−4,3)(-4, 3). Which of the following expresses w⃗\vec{w} as a linear combination of the standard unit vectors i⃗\vec{i} and j⃗\vec{j}?
    Answer: −5i⃗−6j⃗-5\vec{i} - 6\vec{j} 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 \rangle. A vector ⟨a,b⟩\langle a, b \rangle can be written as the linear combination ai⃗+bj⃗a\vec{i} + b\vec{j}. Therefore, w⃗=−5i⃗−6j⃗\vec{w} = -5\vec{i} - 6\vec{j}.
  4. 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} notation. The applied force is F⃗=2i−1j+3k\vec{F}=2\mathbf{i}-1\mathbf{j}+3\mathbf{k} N. A calibration routine scales the force by the scalar k = 3, producing G⃗=kF⃗\vec{G}=k\vec{F}. Scalar multiplication multiplies each component by the same scalar, changing magnitude while keeping direction if the scalar is positive. The coordinate axes correspond to xx, yy, and zz directions. Refer to the vectors described in the passage. What is the resultant vector when G⃗=3F⃗\vec{G}=3\vec{F} is computed?
    Answer: G⃗=6i−3j+9k\vec{G}=6\mathbf{i}-3\mathbf{j}+9\mathbf{k} 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.
  5. 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} 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} kN. Vectors have magnitude and direction; the difference T⃗2−T⃗1\vec{T}_2-\vec{T}_1 compares their effects. Refer to the vectors described in the passage. What is the resultant vector when T⃗2\vec{T}_2 and T⃗1\vec{T}_1 are subtracted?
    Answer: ⟨−5,4,0⟩\langle -5,4,0\rangle This question tests AP Precalculus skills in vectors, specifically vector subtraction in structural analysis. Vector subtraction T⃗2−T⃗1\vec{T}_2-\vec{T}_1 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\rangle. 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} as a⃗+(−b⃗)\vec{a}+(-\vec{b}) to reinforce the concept.
  6. A vector v⃗\vec{v} has its initial point at P1=(2,−5)P_1 = (2, -5) and its terminal point at P2=(−3,7)P_2 = (-3, 7). Which of the following represents the component form of v⃗\vec{v}?
    Answer: ⟨−5,12⟩\langle -5, 12 \rangle The component form of a vector from initial point (x1,y1)(x_1, y_1) to terminal point (x2,y2)(x_2, y_2) is given by ⟨x2−x1,y2−y1⟩\langle x_2 - x_1, y_2 - y_1 \rangle. For the given points, this is ⟨−3−2,7−(−5)⟩=⟨−5,12⟩\langle -3 - 2, 7 - (-5) \rangle = \langle -5, 12 \rangle.
  7. 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} 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} 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}_b and v⃗c\vec{v}_c are added?
    Answer: ⟨5,5,0⟩\langle 5,5,0\rangle 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\rangle to the current vector v⃗c=⟨−1,3,0⟩\vec{v}_c=\langle -1,3,0\rangle. 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\rangle. Choice B is incorrect because it appears to subtract components incorrectly, getting ⟨7,−1,0⟩\langle 7,-1,0\rangle 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.
  8. For what value of kk are the vectors u⃗=⟨k,−4⟩\vec{u} = \langle k, -4 \rangle and v⃗=⟨3,6⟩\vec{v} = \langle 3, 6 \rangle perpendicular?
    Answer: 88 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 - 24. Setting the dot product to zero gives the equation 3k−24=03k - 24 = 0. Solving for kk, we get 3k=243k = 24, so k=8k = 8.
  9. 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} (m/s), and the wind velocity is w⃗=−2i+3j+0k\vec{w}=-2\mathbf{i}+3\mathbf{j}+0\mathbf{k} (m/s). The drone's ground velocity is the vector sum g⃗=d⃗+w⃗\vec{g}=\vec{d}+\vec{w} because both effects act simultaneously. Subtraction represents "removing" an effect: if you know g⃗\vec{g} and w⃗\vec{w}, then d⃗=g⃗−w⃗\vec{d}=\vec{g}-\vec{w}. The coordinate axes use xx east and yy north; arrows indicate direction. Refer to the vectors described in the passage. Determine the magnitude of the resultant vector g⃗\vec{g}.
    Answer: 32\sqrt{32} m/s 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.
  10. 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} m/s, where magnitude is speed and direction is heading. It applies scalar multiplication with k = −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} affect its magnitude?
    Answer: Magnitude doubles, and direction reverses. 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\rangle by k=−2k=-2 gives −2v⃗=⟨−4,2,−6⟩-2\vec{v}=\langle -4,2,-6\rangle. Choice A is correct because the magnitude doubles (from ∣v⃗∣=4+1+9=14|\vec{v}|=\sqrt{4+1+9}=\sqrt{14} to ∣−2v⃗∣=16+4+36=56=214|-2\vec{v}|=\sqrt{16+4+36}=\sqrt{56}=2\sqrt{14}) 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.
  11. 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} meters, but a gust causes an unintended drift of d⃗=−3i+5j+0k\vec{d}=-3\mathbf{i}+5\mathbf{j}+0\mathbf{k} meters. The drone's actual displacement is a⃗=p⃗+d⃗\vec{a}=\vec{p}+\vec{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}. On the coordinate plane, xx is east and yy 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} is computed?
    Answer: c⃗=3i−5j+0k\vec{c}=3\mathbf{i}-5\mathbf{j}+0\mathbf{k} 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.
  12. Which of the following is a unit vector in the same direction as the vector w⃗=⟨5,−12⟩\vec{w} = \langle 5, -12 \rangle?
    Answer: ⟨513,−1213⟩\langle \frac{5}{13}, -\frac{12}{13} \rangle To find a unit vector in the same direction as w⃗\vec{w}, we divide w⃗\vec{w} by its magnitude. The magnitude is ∥w⃗∥=52+(−12)2=25+144=169=13\Vert\vec{w}\Vert = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13. The unit vector is w⃗∥w⃗∥=113⟨5,−12⟩=⟨513,−1213⟩\frac{\vec{w}}{\Vert\vec{w}\Vert} = \frac{1}{13}\langle 5, -12 \rangle = \langle \frac{5}{13}, -\frac{12}{13} \rangle.
  13. Let the vector u⃗\vec{u} be represented by ⟨−8,15⟩\langle -8, 15 \rangle. What is the magnitude of u⃗\vec{u}?
    Answer: 1717 The magnitude of a vector ⟨a,b⟩\langle a, b \rangle is given by the formula a2+b2\sqrt{a^2 + b^2}. For the vector u⃗=⟨−8,15⟩\vec{u} = \langle -8, 15 \rangle, the magnitude is (−8)2+152=64+225=289=17\sqrt{(-8)^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17.
  14. A robot arm moves in 3D with displacement a=2i−1j+3k2\mathbf{i}-1\mathbf{j}+3\mathbf{k} cm and correction b=−4i+5j−1k-4\mathbf{i}+5\mathbf{j}-1\mathbf{k} 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?​
    Answer: −2i+4j+2k-2\mathbf{i}+4\mathbf{j}+2\mathbf{k} cm 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.
  15. 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} 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} m/s. Vectors have magnitude and direction; new velocity is v⃗0+Δv⃗\vec{v}_0+\Delta\vec{v}. Refer to the vectors described in the passage. What is the resultant vector when v⃗0\vec{v}_0 and Δv⃗\Delta\vec{v} are added?
    Answer: ⟨12,6.2,0⟩\langle 12,6.2,0\rangle 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\rangle. 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.
  16. A projectile's initial velocity is v=15i+20j15\mathbf{i}+20\mathbf{j} m/s (magnitude and direction); gravity changes velocity by g=−10j-10\mathbf{j} m/s after 1 s. Refer to the vectors described in the passage. What is the resultant vector when v and g are added?​
    Answer: 15i+10j15\mathbf{i}+10\mathbf{j} m/s This question tests AP Precalculus skills in vectors, specifically vector addition in kinematics problems. Vectors are quantities having both magnitude and direction, essential for analyzing projectile motion where gravity affects velocity. In this scenario, the projectile's initial velocity is modified by gravity acting downward for 1 second, requiring vector addition to find the new velocity. Choice B is correct because adding the velocity change from gravity: (15i + 20j) + (0i - 10j) = 15i + (20-10)j = 15i + 10j m/s. Choice A is incorrect because it appears to add the gravity effect instead of subtracting it, misunderstanding that gravity acts downward (negative j direction). To help students: Emphasize that gravity always acts downward (negative j), reducing the upward component of velocity. Draw velocity vector diagrams showing how gravity changes the trajectory over time.
  17. Let v⃗=⟨−3,6⟩\vec{v} = \langle -3, 6 \rangle and w⃗=⟨2,4⟩\vec{w} = \langle 2, 4 \rangle. What is the dot product v⃗⋅w⃗\vec{v} \cdot \vec{w}?
    Answer: 1818 The dot product of two vectors ⟨a,b⟩\langle a, b \rangle and ⟨c,d⟩\langle c, d \rangle is the scalar quantity ac+bdac + bd. Therefore, v⃗⋅w⃗=(−3)(2)+(6)(4)=−6+24=18\vec{v} \cdot \vec{w} = (-3)(2) + (6)(4) = -6 + 24 = 18.
  18. 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} 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} with k = 2. Refer to the vectors described in the passage. How does the scalar multiplication of F⃗\vec{F} affect its magnitude?
    Answer: It doubles the magnitude and keeps direction. 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\rangle by scalar k=2k=2 gives 2F⃗=⟨6,−8,0⟩2\vec{F}=\langle 6,-8,0\rangle. 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 kN, and after multiplication, ∣2F⃗∣=36+64=10|2\vec{F}|=\sqrt{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.
  19. A bridge cable tension is T=9i+12j9\mathbf{i}+12\mathbf{j} kN and a support reaction is R=4i−5j4\mathbf{i}-5\mathbf{j} 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?​
    Answer: 13i+7j13\mathbf{i}+7\mathbf{j} kN 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.
  20. Let a⃗=⟨4,−1⟩\vec{a} = \langle 4, -1 \rangle and b⃗=⟨−2,5⟩\vec{b} = \langle -2, 5 \rangle. Which of the following is the vector 3a⃗−2b⃗3\vec{a} - 2\vec{b}?
    Answer: ⟨16,−13⟩\langle 16, -13 \rangle First, perform the scalar multiplications: 3a⃗=3⟨4,−1⟩=⟨12,−3⟩3\vec{a} = 3\langle 4, -1 \rangle = \langle 12, -3 \rangle and 2b⃗=2⟨−2,5⟩=⟨−4,10⟩2\vec{b} = 2\langle -2, 5 \rangle = \langle -4, 10 \rangle. Then, subtract the resulting vectors: ⟨12,−3⟩−⟨−4,10⟩=⟨12−(−4),−3−10⟩=⟨16,−13⟩\langle 12, -3 \rangle - \langle -4, 10 \rangle = \langle 12 - (-4), -3 - 10 \rangle = \langle 16, -13 \rangle.

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