Precalculus Quiz: Find Sum Of Two Vectors

20 questions · exam conditions

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Find Sum Of Two VectorsQuestion 1 of 20

Two displacement vectors are given: $\vec{d}_1$ with magnitude 15 meters at $240°$ from the positive x-axis, and $\vec{d}_2$ with magnitude 9 meters at $30°$ from the positive x-axis. What is the x-component of their sum $\vec{d}_1 + \vec{d}_2$ ?

\frac{9\sqrt{3} - 15}{2}

meters

\frac{15\sqrt{3} + 9}{2}

meters

\frac{9\sqrt{3} + 15}{2}

meters

\frac{9\sqrt{3} - 30}{2}

meters

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Precalculus Quiz

Precalculus Quiz: Find Sum Of Two Vectors

Practice Find Sum Of Two Vectors in Precalculus 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 Find Sum Of Two Vectors, giving you a quick way to practice the rules, question types, and explanations that matter most for Precalculus.

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Question 1

$\frac{9\sqrt{3} - 15}{2}$ meters (correct answer)
$\frac{15\sqrt{3} + 9}{2}$ meters
$\frac{9\sqrt{3} + 15}{2}$ meters
$\frac{9\sqrt{3} - 30}{2}$ meters

Explanation: The x-component of

\vec{d}_1

15\cos 240° = 15 \cdot (-\frac{1}{2}) = -\frac{15}{2}

. The x-component of

\vec{d}_2

9\cos 30° = 9 \cdot \frac{\sqrt{3}}{2} = \frac{9\sqrt{3}}{2}

. The x-component of the sum is

-\frac{15}{2} + \frac{9\sqrt{3}}{2} = \frac{9\sqrt{3} - 15}{2}

. Choice B incorrectly uses

\cos 240° = \frac{\sqrt{3}}{2}

. Choice C uses addition instead of the correct subtraction. Choice D incorrectly doubles the coefficient of the first term.

Question 2

Angles are measured in degrees counterclockwise from the positive $x$ -axis. Vector $\mathbf{u}$ has magnitude $10$ at direction $0^\circ$ , and vector $\mathbf{v}$ has magnitude $10$ at direction $90^\circ$ . Using the information provided, what is the magnitude of $\mathbf{u}+\mathbf{v}$ ? (Direction is not needed.)

$10$
$20$
$\sqrt{200}$ (correct answer)
$\sqrt{100}$

Explanation: This question tests understanding of how to add vectors given in magnitude and direction form and find the magnitude of the resultant. The magnitude of a vector sum is found using the Pythagorean theorem: after converting each vector to components and adding, if the sum is ⟨a, b⟩, then the magnitude is |u + v| = √(a² + b²). Since one vector points at 0° with magnitude 10 and the other at 90° with magnitude 10, these are perpendicular, forming a right triangle; the magnitude of the sum is the hypotenuse: √(10² + 10²) = √(100 + 100) = √200. Choice C is correct because it properly converts to components, adds correctly, and applies the Pythagorean theorem for magnitude. Choice B incorrectly adds the magnitudes directly (10 + 10 = 20), but vector addition requires converting to components first—you can only add magnitudes directly when vectors point in the same direction. Special case shortcut: when vectors are perpendicular (like one pointing east and one pointing north), you can directly apply the Pythagorean theorem to the magnitudes: |u + v| = √(r₁² + r₂²), and the direction is simply arctan(r₂/r₁) from the first vector's direction. Don't fall for the trap of adding magnitudes directly—this only works when vectors point in exactly the same direction; otherwise, components partially cancel and you must use the component method.

Question 3

Given the magnitudes and directions, vector $\mathbf{u}$ has magnitude $13$ at $0^\circ$ and vector $\mathbf{v}$ has magnitude $13$ at $90^\circ$ (angles from the positive $x$ -axis). What are the magnitude and direction of $\mathbf{u}+\mathbf{v}$ ?

Magnitude $13\sqrt{2}$ at $45^\circ$ (correct answer)
Magnitude $26$ at $45^\circ$
Magnitude $13\sqrt{2}$ at $135^\circ$
Magnitude $\sqrt{338}$ at $45^\circ$

Explanation: This question tests understanding of how to add vectors given in magnitude and direction form and find the magnitude and direction of the resultant. For special cases like perpendicular vectors, the calculation simplifies: if one vector is magnitude r₁ along the x-axis (direction 0°) and another is magnitude r₂ along the y-axis (direction 90°), the resultant magnitude is simply √(r₁² + r₂²) and the direction is arctan(r₂/r₁). Since the vectors have equal magnitude 13 and are perpendicular, forming a right triangle. The magnitude of the sum is √(13² + 13²) = √(169 + 169) = √338 =13√2, and the direction is arctan(13/13)=45°. Choice A is correct because it properly converts to components, adds correctly, and applies the Pythagorean theorem for magnitude and correctly uses arctan for direction. Choice B incorrectly adds the magnitudes directly (13 + 13 =26), but vector addition requires converting to components first—you can only add magnitudes directly when vectors point in the same direction. Key to adding vectors in magnitude-direction form: always convert to components first using x = r·cos(θ) and y = r·sin(θ), add the x-components together and y-components together, then find magnitude using √(x² + y²) and direction using arctan(y/x) with quadrant adjustment. Special case shortcut: when vectors are perpendicular (like one pointing east and one pointing north), you can directly apply the Pythagorean theorem to the magnitudes: |u + v| = √(r₁² + r₂²), and the direction is arctan(r₂/r₁) from the first vector's direction.

Question 4

Let vector $\mathbf{u}$ have magnitude $3$ and direction $0^\circ$ (measured counterclockwise from the positive $x$ -axis). Let vector $\mathbf{v}$ have magnitude $4$ and direction $90^\circ$ . For the vectors described, what are the magnitude and direction of $\mathbf{u}+\mathbf{v}$ ? Give the direction as an angle from the positive $x$ -axis.

magnitude $7$ at $90^\circ$
magnitude $5$ at $\tan^{-1}\!\left(\frac{4}{3}\right)$ (correct answer)
magnitude $1$ at $270^\circ$
magnitude $5$ at $\tan^{-1}\!\left(\frac{3}{4}\right)$

Explanation: This question tests understanding of how to add vectors given in magnitude and direction form and find the magnitude and direction of the resultant. To add vectors given in magnitude and direction form, first convert each vector to component form: a vector with magnitude r and direction θ (measured counterclockwise from the positive x-axis) has components ⟨r·cos(θ), r·sin(θ)⟩, then add the components of all vectors, and finally compute the magnitude and direction of the resulting component form. Vector u has magnitude 3 at direction 0°, so its components are ⟨3·cos(0°), 3·sin(0°)⟩ = ⟨3, 0⟩; vector v has magnitude 4 at direction 90°, so its components are ⟨4·cos(90°), 4·sin(90°)⟩ = ⟨0, 4⟩; adding these gives u + v = ⟨3, 4⟩, the magnitude is √(3² + 4²) = √(9 + 16) = √25 = 5, and the direction is arctan(4/3) since both components are positive, placing it in Quadrant I. Choice B is correct because it properly converts to components, adds correctly, and provides both the magnitude using the Pythagorean theorem and the direction using arctan with quadrant adjustment. Choice D incorrectly swaps the ratio in arctan, using arctan(3/4) instead of arctan(4/3), which would give a smaller angle not matching the components. Key to adding vectors in magnitude-direction form: always convert to components first using x = r·cos(θ) and y = r·sin(θ), add the x-components together and y-components together, then find magnitude using √(x² + y²) and direction using arctan(y/x) with quadrant adjustment. Special case shortcut: when vectors are perpendicular (like one pointing east and one pointing north), you can directly apply the Pythagorean theorem to the magnitudes: |u + v| = √(r₁² + r₂²), and the direction is simply arctan(r₂/r₁) from the first vector's direction.

Question 5

In an $xy$ -coordinate system where directions are measured in degrees from the positive $x$ -axis, vector $\mathbf{u}$ has magnitude $10$ at direction $30^\circ$ , and vector $\mathbf{v}$ has magnitude $10$ at direction $150^\circ$ . For the vectors described, what is the magnitude of $\mathbf{u}+\mathbf{v}$ (magnitude only)?

$10$ (correct answer)
$20$
$10\sqrt{3}$
$0$

Explanation: This question tests understanding of how to add vectors given in magnitude and direction form and find the magnitude of the resultant. To add vectors given in magnitude and direction form, first convert each vector to component form: a vector with magnitude r and direction θ (measured counterclockwise from the positive x-axis) has components ⟨r·cos(θ), r·sin(θ)⟩, then add the components of all vectors, and finally compute the magnitude and direction of the resulting component form. Vector u has magnitude 10 at direction 30°, so its components are ⟨10·cos(30°), 10·sin(30°)⟩ = ⟨10·(√3/2), 10·(1/2)⟩ = ⟨5√3, 5⟩. Vector v has magnitude 10 at direction 150°, so its components are ⟨10·cos(150°), 10·sin(150°)⟩ = ⟨10·(-√3/2), 10·(1/2)⟩ = ⟨-5√3, 5⟩. Adding these gives u + v = ⟨5√3 + (-5√3), 5 + 5⟩ = ⟨0, 10⟩, and the magnitude is √(0² + 10²) = √100 = 10. Choice A is correct because it properly converts to components, adds correctly, and applies the Pythagorean theorem for magnitude. Choice B incorrectly adds the magnitudes directly (10 + 10 = 20), but vector addition requires converting to components first—you can only add magnitudes directly when vectors point in the same direction. Key to adding vectors in magnitude-direction form: always convert to components first using x = r·cos(θ) and y = r·sin(θ), add the x-components together and y-components together, then find magnitude using √(x² + y²). To verify your answer, check that the magnitude of the sum is between |r₁ - r₂| and r₁ + r₂ (triangle inequality), and that the direction makes geometric sense given the original vectors' directions.

Question 6

A hiker walks $6$ km due east (direction $0^\circ$ from the positive $x$ -axis), then walks $8$ km due north (direction $90^\circ$ ). Using the information provided, what are the magnitude and direction of the hiker's resultant displacement (direction as an angle from the positive $x$ -axis)?

magnitude $10$ km at $\tan^{-1}\!\left(\frac{8}{6}\right)$ (correct answer)
magnitude $14$ km at $\tan^{-1}\!\left(\frac{8}{6}\right)$
magnitude $10$ km at $\tan^{-1}\!\left(\frac{6}{8}\right)$
magnitude $2$ km at $90^\circ$

Explanation: This question tests understanding of how to add vectors given in magnitude and direction form and find the magnitude and direction of the resultant. To add vectors given in magnitude and direction form, first convert each vector to component form: a vector with magnitude r and direction θ (measured counterclockwise from the positive x-axis) has components ⟨r·cos(θ), r·sin(θ)⟩, then add the components of all vectors, and finally compute the magnitude and direction of the resulting component form. Since the hiker walks 6 km east and 8 km north, these are perpendicular displacements forming a right triangle. The magnitude of the resultant displacement is the hypotenuse: √(6² + 8²) = √(36 + 64) = √100 = 10 km. The direction angle is arctan(8/6) = arctan(4/3), measured from the positive x-axis (east direction). Choice A is correct because it properly identifies both the magnitude as 10 km and the direction as tan⁻¹(8/6), which correctly represents the angle whose tangent is the ratio of northward to eastward displacement. Choice C incorrectly gives the direction as tan⁻¹(6/8), which inverts the ratio—the correct ratio for direction from east is (north component)/(east component) = 8/6, not 6/8. Special case shortcut: when vectors are perpendicular (like one pointing east and one pointing north), you can directly apply the Pythagorean theorem to the magnitudes: |u + v| = √(r₁² + r₂²), and the direction is simply arctan(r₂/r₁) from the first vector's direction. This is a classic 6-8-10 right triangle, making the calculation particularly clean.

Question 7

A force of $12\text{ N}$ acts due east (direction $0^\circ$ from the positive $x$ -axis), and a force of $5\text{ N}$ acts due north (direction $90^\circ$ ). For the vectors described, what is the resultant force (magnitude and direction as an angle from the positive $x$ -axis)?

Magnitude $13\text{ N}$ at $\tan^{-1}\!\left(\frac{12}{5}\right)$
Magnitude $17\text{ N}$ at $\tan^{-1}\!\left(\frac{5}{12}\right)$
Magnitude $13\text{ N}$ at $\tan^{-1}\!\left(\frac{5}{12}\right)$ (correct answer)
Magnitude $7\text{ N}$ at $0^\circ$

Explanation: This question tests understanding of how to add vectors given in magnitude and direction form and find the magnitude and direction of the resultant. To add vectors given in magnitude and direction form, first convert each vector to component form: a vector with magnitude r and direction θ (measured counterclockwise from the positive x-axis) has components ⟨r·cos(θ), r·sin(θ)⟩, then add the components of all vectors, and finally compute the magnitude and direction of the resulting component form. Since one force is 12 N east (0°) and the other 5 N north (90°), these are perpendicular, forming a right triangle; the magnitude is √(12² + 5²) = √(144 + 25) = √169 = 13, and the direction is arctan(5/12) from the positive x-axis. Converting to components, adding, and finding the magnitude gives 13 (as calculated above); for direction, arctan(5/12) ≈ 23°, and since both components are positive, it's in Quadrant I, so the resultant is 13 N at tan^{-1}(5/12). Choice C is correct because it properly converts to components, adds correctly, and applies the Pythagorean theorem for magnitude while correctly using arctan with quadrant adjustment for direction. Choice A incorrectly swaps the ratio in arctan, using tan^{-1}(12/5) instead, which gives a different angle. Key to adding vectors in magnitude-direction form: always convert to components first using x = r·cos(θ) and y = r·sin(θ), add the x-components together and y-components together, then find magnitude using √(x² + y²) and direction using arctan(y/x) with quadrant adjustment.

Question 8

For the vectors described, vector $\mathbf{u}$ has magnitude $6$ and direction $45^\circ$ , and vector $\mathbf{v}$ has magnitude $6$ and direction $315^\circ$ (angles from the positive $x$ -axis). What is the direction of the resultant vector $\mathbf{u}+\mathbf{v}$ (magnitude not needed)?

$0^\circ$ (correct answer)
$90^\circ$
$45^\circ$
$315^\circ$

Explanation: This question tests understanding of how to add vectors given in magnitude and direction form and find the direction of the resultant. The direction of a vector with components ⟨a, b⟩ is found using θ = arctan(b/a), but you must check which quadrant the vector is in based on the signs of a and b: Quadrant I (both positive) gives 0° to 90°, Quadrant II (a negative, b positive) gives 90° to 180°, Quadrant III (both negative) gives 180° to 270°, and Quadrant IV (a positive, b negative) gives 270° to 360°. Vector u has magnitude 6 at direction 45°, so its components are ⟨3√2, 3√2⟩. Vector v has magnitude 6 at direction 315°, so its components are ⟨3√2, -3√2⟩. Adding these gives u + v = ⟨6√2, 0⟩. The direction angle is arctan(0/(6√2)) = 0°. Choice A is correct because it correctly uses arctan with quadrant adjustment for direction. Choice D gives the direction of only one of the original vectors instead of the direction of their sum.

Question 9

Two forces act on an object in the $xy$ -plane (angles measured in degrees counterclockwise from the positive $x$ -axis). Force $\mathbf{F}_1$ has magnitude $10\text{ N}$ at $0^\circ$ (east) and force $\mathbf{F}_2$ has magnitude $10\text{ N}$ at $90^\circ$ (north). What is the resultant force (magnitude and direction) $\mathbf{F}_1+\mathbf{F}_2$ ?

Magnitude $20\text{ N}$ at $45^\circ$
Magnitude $10\sqrt{2}\text{ N}$ at $45^\circ$ (correct answer)
Magnitude $10\sqrt{2}\text{ N}$ at $135^\circ$
Magnitude $\sqrt{200}\text{ N}$ at $0^\circ$

Explanation: This question tests understanding of how to add vectors given in magnitude and direction form and find the magnitude and direction of the resultant. For special cases like perpendicular vectors, the calculation simplifies: if one vector is magnitude r₁ along the x-axis (direction 0°) and another is magnitude r₂ along the y-axis (direction 90°), the resultant magnitude is simply √(r₁² + r₂²) and the direction is arctan(r₂/r₁). Since one force points east with magnitude 10 N and the other points north with magnitude 10 N, these are perpendicular, forming a right triangle. The magnitude of the sum is the hypotenuse: √(10² + 10²) = √(100 + 100) = √200 = 10√2. For direction, arctan(10/10) = arctan(1) = 45°, so the resultant is magnitude 10√2 N at 45°. Choice B is correct because it properly applies the Pythagorean theorem for perpendicular vectors and correctly calculates the direction as 45°. Choice A incorrectly adds the magnitudes directly (10 + 10 = 20), but vector addition requires converting to components first—you can only add magnitudes directly when vectors point in the same direction. Special case shortcut: when vectors are perpendicular (like one pointing east and one pointing north), you can directly apply the Pythagorean theorem to the magnitudes: |F₁ + F₂| = √(r₁² + r₂²), and the direction is simply arctan(r₂/r₁) from the first vector's direction. When two perpendicular vectors have equal magnitude, the resultant always bisects the angle between them at 45°.

Question 10

On an $xy$ -plane, angles are measured in degrees counterclockwise from the positive $x$ -axis (east). Vector $\mathbf{u}$ has magnitude $3$ and direction $0^\circ$ (due east). Vector $\mathbf{v}$ has magnitude $4$ and direction $90^\circ$ (due north). For the vectors described, what are the magnitude and direction of $\mathbf{u}+\mathbf{v}$ (give direction as an angle from the positive $x$ -axis)?

Magnitude $7$ at $90^\circ$
Magnitude $5$ at $\tan^{-1}\!\left(\frac{3}{4}\right)$
Magnitude $5$ at $\tan^{-1}\!\left(\frac{4}{3}\right)$ (correct answer)
Magnitude $1$ at $0^\circ$

Explanation: This question tests understanding of how to add vectors given in magnitude and direction form and find the magnitude and direction of the resultant. To add vectors given in magnitude and direction form, first convert each vector to component form: a vector with magnitude r and direction θ (measured counterclockwise from the positive x-axis) has components ⟨r·cos(θ), r·sin(θ)⟩, then add the components of all vectors, and finally compute the magnitude and direction of the resulting component form. Since one vector points east with magnitude 3 and the other points north with magnitude 4, these are perpendicular, forming a right triangle; the magnitude of the sum is the hypotenuse: √(3² + 4²) = √(9 + 16) = √25 = 5, and the direction is arctan(4/3) from the positive x-axis. Converting to components, adding, and finding the magnitude gives 5 (as calculated above); for direction, arctan(4/3) ≈ 53°, and since both components are positive, it's in Quadrant I, so the resultant is magnitude 5 at tan^{-1}(4/3). Choice C is correct because it properly converts to components, adds correctly, and applies the Pythagorean theorem for magnitude while correctly using arctan with quadrant adjustment for direction. Choice B incorrectly swaps the ratio in arctan, using tan^{-1}(3/4) instead, which gives a different angle. Key to adding vectors in magnitude-direction form: always convert to components first using x = r·cos(θ) and y = r·sin(θ), add the x-components together and y-components together, then find magnitude using √(x² + y²) and direction using arctan(y/x) with quadrant adjustment.

Question 11

Angles are measured in degrees counterclockwise from the positive $x$ -axis. Vector $\mathbf{u}$ has magnitude $10$ at direction $135^\circ$ , and vector $\mathbf{v}$ has magnitude $10$ at direction $225^\circ$ . For the vectors described, what are the magnitude and direction of $\mathbf{u}+\mathbf{v}$ (angle from the positive $x$ -axis)?

Magnitude $10\sqrt{2}$ at $180^\circ$ (correct answer)
Magnitude $20$ at $180^\circ$
Magnitude $0$ at $0^\circ$
Magnitude $10\sqrt{2}$ at $0^\circ$

Explanation: This question tests understanding of how to add vectors given in magnitude and direction form and find the magnitude and direction of the resultant. To add vectors given in magnitude and direction form, first convert each vector to component form: a vector with magnitude r and direction θ (measured counterclockwise from the positive x-axis) has components ⟨r·cos(θ), r·sin(θ)⟩, then add the components of all vectors, and finally compute the magnitude and direction of the resulting component form. Vector u has magnitude 10 at direction 135°, so its components are ⟨10·cos(135°), 10·sin(135°)⟩ = ⟨-5√2, 5√2⟩; vector v has magnitude 10 at direction 225°, so its components are ⟨10·cos(225°), 10·sin(225°)⟩ = ⟨-5√2, -5√2⟩; adding these gives u + v = ⟨-10√2, 0⟩, and the magnitude is √((-10√2)² + 0²) = 10√2 with direction 180°. Converting to components, adding, and finding the magnitude gives 10√2 (as calculated above); for direction, arctan(0/(-10√2)) = 180°, since x is negative and y is zero. Choice A is correct because it properly converts to components, adds correctly, and applies the Pythagorean theorem for magnitude while correctly using arctan with quadrant adjustment for direction. Choice B incorrectly adds the magnitudes directly (10 + 10 = 20), but vector addition requires converting to components first—you can only add magnitudes directly when vectors point in the same direction. Key to adding vectors in magnitude-direction form: always convert to components first using x = r·cos(θ) and y = r·sin(θ), add the x-components together and y-components together, then find magnitude using √(x² + y²) and direction using arctan(y/x) with quadrant adjustment.

Question 12

Using the information provided, vector $\mathbf{u}$ has magnitude $5$ at $60^\circ$ and vector $\mathbf{v}$ has magnitude $5$ at $240^\circ$ , with directions measured counterclockwise from the positive $x$ -axis. What is $\lVert \mathbf{u}+\mathbf{v} \rVert$ ?

$10$
$5$
$0$ (correct answer)
$5\sqrt{3}$

\langle r \cdot \cos(\theta), r \cdot \sin(\theta) \rangle

, then add the components of all vectors, and finally compute the magnitude of the resulting component form. Vector u has magnitude 5 at direction 60°, so its components are

\langle 5 \cdot \cos(60^\circ), 5 \cdot \sin(60^\circ) \rangle = \langle 2.5, \frac{5\sqrt{3}}{2} \rangle

. Vector v has magnitude 5 at direction 240°, so its components are

\langle 5 \cdot \cos(240^\circ), 5 \cdot \sin(240^\circ) \rangle = \langle -2.5, -\frac{5\sqrt{3}}{2} \rangle

. Adding these gives u + v =

\langle 0, 0 \rangle

, and the magnitude is

\sqrt{0 + 0} = 0

. Choice C is correct because it properly converts to components, adds correctly, and applies the Pythagorean theorem for magnitude. Choice A incorrectly adds the magnitudes directly (5 + 5 =10), but vector addition requires converting to components first—you can only add magnitudes directly when vectors point in the same direction. Don't fall for the trap of adding magnitudes directly—this only works when vectors point in exactly the same direction; otherwise, components partially cancel and you must use the component method.

Question 13

Vector $\mathbf{u}$ has magnitude $5$ at direction $0^\circ$ . Vector $\mathbf{v}$ has magnitude $5$ at direction $180^\circ$ . For the vectors described, what is the magnitude of $\mathbf{u}+\mathbf{v}$ ? (Direction not needed.)

$10$
$0$ (correct answer)
$5$
$\sqrt{50}$

Explanation: This question tests understanding of how to add vectors given in magnitude and direction form and find the magnitude of the resultant. The magnitude of a vector sum is found using the Pythagorean theorem: after converting each vector to components and adding, if the sum is ⟨a, b⟩, then the magnitude is |u + v| = √(a² + b²). Vector u has magnitude 5 at direction 0°, so its components are ⟨5, 0⟩; vector v has magnitude 5 at direction 180°, so its components are ⟨5·cos(180°), 5·sin(180°)⟩ = ⟨-5, 0⟩; adding gives ⟨0, 0⟩, magnitude √(0 + 0) = 0. Choice B is correct because it properly converts to components, adds correctly, and applies the Pythagorean theorem for magnitude. Choice A incorrectly adds the magnitudes directly (5 + 5 = 10), but vector addition requires converting to components first—you can only add magnitudes directly when vectors point in the same direction. Key to adding vectors in magnitude-direction form: always convert to components first using x = r·cos(θ) and y = r·sin(θ), add the x-components together and y-components together, then find magnitude using √(x² + y²) and direction using arctan(y/x) with quadrant adjustment. Don't fall for the trap of adding magnitudes directly—this only works when vectors point in exactly the same direction; otherwise, components partially cancel and you must use the component method.

Question 14

Vector $\mathbf{u}$ has magnitude $6$ at direction $0^\circ$ . Vector $\mathbf{v}$ has magnitude $6$ at direction $120^\circ$ . For the vectors described, what are the magnitude and direction of $\mathbf{u}+\mathbf{v}$ ? Give direction as an angle from the positive $x$ -axis.

magnitude $6$ at $60^\circ$ (correct answer)
magnitude $12$ at $60^\circ$
magnitude $6\sqrt{3}$ at $30^\circ$
magnitude $0$ at $0^\circ$

Explanation: This question tests understanding of how to add vectors given in magnitude and direction form and find the magnitude and direction of the resultant. To add vectors given in magnitude and direction form, first convert each vector to component form: a vector with magnitude r and direction θ (measured counterclockwise from the positive x-axis) has components ⟨r·cos(θ), r·sin(θ)⟩, then add the components of all vectors, and finally compute the magnitude and direction of the resulting component form. Vector u has magnitude 6 at direction 0°, so its components are ⟨6, 0⟩; vector v has magnitude 6 at direction 120°, so its components are ⟨6·cos(120°), 6·sin(120°)⟩ = ⟨-3, 3√3⟩; adding gives ⟨3, 3√3⟩, magnitude √(9 + 27) = √36 = 6, direction arctan((3√3)/3) = arctan(√3) = 60° in Quadrant I. Choice A is correct because it properly converts to components, adds correctly, and provides both values correctly. Choice B incorrectly adds the magnitudes directly (6 + 6 = 12), but vector addition requires converting to components first—you can only add magnitudes directly when vectors point in the same direction. Key to adding vectors in magnitude-direction form: always convert to components first using x = r·cos(θ) and y = r·sin(θ), add the x-components together and y-components together, then find magnitude using √(x² + y²) and direction using arctan(y/x) with quadrant adjustment. To verify your answer, check that the magnitude of the sum is between |r₁ - r₂| and r₁ + r₂ (triangle inequality), and that the direction makes geometric sense given the original vectors' directions.

Question 15

Using the information provided (angles measured in degrees counterclockwise from the positive $x$ -axis), vector $\mathbf{u}$ has magnitude $10$ at $135^\circ$ and vector $\mathbf{v}$ has magnitude $10$ at $315^\circ$ . What are the magnitude and direction of $\mathbf{u}+\mathbf{v}$ ? (Give direction as an angle from the positive $x$ -axis.)

Magnitude $20$ at $225^\circ$
Magnitude $0$ (direction undefined) (correct answer)
Magnitude $10\sqrt{2}$ at $135^\circ$
Magnitude $10\sqrt{2}$ at $315^\circ$

Explanation: This question tests understanding of how to add vectors given in magnitude and direction form and find the magnitude and direction of the resultant. To add vectors given in magnitude and direction form, first convert each vector to component form: a vector with magnitude r and direction θ (measured counterclockwise from the positive x-axis) has components ⟨r·cos(θ), r·sin(θ)⟩, then add the components of all vectors, and finally compute the magnitude and direction of the resulting component form. Vector u has magnitude 10 at direction 135°, so its components are ⟨10·cos(135°), 10·sin(135°)⟩ = ⟨10·(-√2/2), 10·(√2/2)⟩ = ⟨-5√2, 5√2⟩. Vector v has magnitude 10 at direction 315°, so its components are ⟨10·cos(315°), 10·sin(315°)⟩ = ⟨10·(√2/2), 10·(-√2/2)⟩ = ⟨5√2, -5√2⟩. Adding these gives u + v = ⟨-5√2 + 5√2, 5√2 + (-5√2)⟩ = ⟨0, 0⟩, and the magnitude is √(0² + 0²) = 0. Choice B is correct because when the sum of vectors is the zero vector ⟨0, 0⟩, the magnitude is 0 and the direction is undefined (you cannot define a direction for a point). Choice A incorrectly suggests a non-zero magnitude, failing to recognize that these vectors are symmetric about the x-axis and cancel out. Key insight: when two vectors of equal magnitude are positioned symmetrically (here at 135° and 315°, which are equidistant from the x-axis), their y-components cancel and their x-components cancel, resulting in the zero vector. The direction of the zero vector is undefined because it doesn't point anywhere.

Question 16

Vector $\mathbf{u}$ has magnitude $4$ at direction $45^\circ$ . Vector $\mathbf{v}$ has magnitude $4$ at direction $315^\circ$ . For the vectors described, what are the magnitude and direction of $\mathbf{u}+\mathbf{v}$ ? Give direction as an angle from the positive $x$ -axis.

magnitude $4\sqrt{2}$ at $0^\circ$ (correct answer)
magnitude $8$ at $0^\circ$
magnitude $4\sqrt{2}$ at $90^\circ$
magnitude $0$ at $0^\circ$

Explanation: This question tests understanding of how to add vectors given in magnitude and direction form and find the magnitude and direction of the resultant. To add vectors given in magnitude and direction form, first convert each vector to component form: a vector with magnitude r and direction θ (measured counterclockwise from the positive x-axis) has components ⟨r·cos(θ), r·sin(θ)⟩, then add the components of all vectors, and finally compute the magnitude and direction of the resulting component form. Vector u has magnitude 4 at 45°, components ⟨4·cos(45°), 4·sin(45°)⟩ = ⟨2√2, 2√2⟩; vector v has magnitude 4 at 315°, components ⟨4·cos(315°), 4·sin(315°)⟩ = ⟨2√2, -2√2⟩; sum ⟨4√2, 0⟩, magnitude √((4√2)² + 0) = √(32) = 4√2, direction arctan(0/(4√2)) = 0°. Choice A is correct because it properly converts to components, adds correctly, and provides both values correctly. Choice B incorrectly adds the magnitudes directly (4 + 4 = 8), but vector addition requires converting to components first—you can only add magnitudes directly when vectors point in the same direction. Key to adding vectors in magnitude-direction form: always convert to components first using x = r·cos(θ) and y = r·sin(θ), add the x-components together and y-components together, then find magnitude using √(x² + y²) and direction using arctan(y/x) with quadrant adjustment. Don't fall for the trap of adding magnitudes directly—this only works when vectors point in exactly the same direction; otherwise, components partially cancel and you must use the component method.

Question 17

Given the magnitudes and directions (angles measured in degrees counterclockwise from the positive $x$ -axis), vector $\mathbf{u}$ has magnitude $5$ at $0^\circ$ and vector $\mathbf{v}$ has magnitude $5$ at $180^\circ$ . What is the magnitude of $\mathbf{u}+\mathbf{v}$ ? (Direction not needed.)

$10$
$5$
$0$ (correct answer)
$\sqrt{50}$

Explanation: This question tests understanding of how to add vectors given in magnitude and direction form and find the magnitude of the resultant. To add vectors given in magnitude and direction form, first convert each vector to component form: a vector with magnitude r and direction θ (measured counterclockwise from the positive x-axis) has components ⟨r·cos(θ), r·sin(θ)⟩, then add the components of all vectors, and finally compute the magnitude of the resulting component form. Vector u has magnitude 5 at direction 0°, so its components are ⟨5·cos(0°), 5·sin(0°)⟩ = ⟨5·1, 5·0⟩ = ⟨5, 0⟩. Vector v has magnitude 5 at direction 180°, so its components are ⟨5·cos(180°), 5·sin(180°)⟩ = ⟨5·(-1), 5·0⟩ = ⟨-5, 0⟩. Adding these gives u + v = ⟨5 + (-5), 0 + 0⟩ = ⟨0, 0⟩, and the magnitude is √(0² + 0²) = √0 = 0. Choice C is correct because when two vectors of equal magnitude point in exactly opposite directions, they completely cancel out, resulting in the zero vector. Choice A incorrectly adds the magnitudes directly (5 + 5 = 10), but this ignores the fact that the vectors point in opposite directions. Key insight: vectors pointing in opposite directions (180° apart) with equal magnitudes will always sum to zero, as their components have opposite signs and cancel completely. To verify your answer, note that u + v = 0 means u = -v, which is exactly the case here: vector v is the negative of vector u.

Question 18

An airplane's velocity relative to the air is $200\text{ km/h}$ at direction $0^\circ$ (due east, measured from the positive $x$ -axis). The wind velocity is $50\text{ km/h}$ at direction $90^\circ$ (due north). Using the information provided, what are the magnitude and direction of the plane's ground velocity (the vector sum)? Give direction as an angle from the positive $x$ -axis.

magnitude $250\text{ km/h}$ at $90^\circ$
magnitude $150\text{ km/h}$ at $\tan^{-1}\!\left(\frac{1}{4}\right)$
magnitude $\sqrt{42500}\text{ km/h}$ at $\tan^{-1}\!\left(\frac{1}{4}\right)$ (correct answer)
magnitude $\sqrt{42500}\text{ km/h}$ at $\tan^{-1}\!\left(4\right)$

Explanation: This question tests understanding of how to add vectors given in magnitude and direction form and find the magnitude and direction of the resultant. To add vectors given in magnitude and direction form, first convert each vector to component form: a vector with magnitude r and direction θ (measured counterclockwise from the positive x-axis) has components ⟨r·cos(θ), r·sin(θ)⟩, then add the components of all vectors, and finally compute the magnitude and direction of the resulting component form. The airplane has magnitude 200 at 0°, components ⟨200, 0⟩; wind has magnitude 50 at 90°, components ⟨0, 50⟩; sum ⟨200, 50⟩, magnitude √(200² + 50²) = √(40000 + 2500) = √42500, direction arctan(50/200) = arctan(1/4) in Quadrant I. Choice C is correct because it properly converts to components, adds correctly, and provides both values correctly. Choice D uses the wrong trigonometric function, calculating arctan(4) instead of arctan(1/4), swapping the ratio for the direction. Key to adding vectors in magnitude-direction form: always convert to components first using x = r·cos(θ) and y = r·sin(θ), add the x-components together and y-components together, then find magnitude using √(x² + y²) and direction using arctan(y/x) with quadrant adjustment. Special case shortcut: when vectors are perpendicular (like one pointing east and one pointing north), you can directly apply the Pythagorean theorem to the magnitudes: |u + v| = √(r₁² + r₂²), and the direction is simply arctan(r₂/r₁) from the first vector's direction.

Question 19

Given the magnitudes and directions, a force $\mathbf{F}_1$ is $8\text{ N}$ at $0^\circ$ (due east) and a force $\mathbf{F}_2$ is $15\text{ N}$ at $90^\circ$ (due north), with angles measured from the positive $x$ -axis. What is the magnitude of the resultant force $\mathbf{F}_1+\mathbf{F}_2$ (direction not needed)?

$23\text{ N}$
$17\text{ N}$ (correct answer)
$\sqrt{161}\text{ N}$
$7\text{ N}$

Explanation: This question tests understanding of how to add vectors given in magnitude and direction form and find the magnitude of the resultant. For special cases like perpendicular vectors, the calculation simplifies: if one vector is magnitude r₁ along the x-axis (direction 0°) and another is magnitude r₂ along the y-axis (direction 90°), the resultant magnitude is simply √(r₁² + r₂²). Since one vector points east with magnitude 8 and the other points north with magnitude 15, these are perpendicular, forming a right triangle. The magnitude of the sum is the hypotenuse: √(8² + 15²) = √(64 + 225) = √289 =17. Choice B is correct because it properly applies the Pythagorean theorem for magnitude. Choice A incorrectly adds the magnitudes directly (8 + 15 = 23), but vector addition requires converting to components first—you can only add magnitudes directly when vectors point in the same direction. Special case shortcut: when vectors are perpendicular (like one pointing east and one pointing north), you can directly apply the Pythagorean theorem to the magnitudes: |u + v| = √(r₁² + r₂²), and the direction is arctan(r₂/r₁) from the first vector's direction. Don't fall for the trap of adding magnitudes directly—this only works when vectors point in exactly the same direction; otherwise, components partially cancel and you must use the component method.

Question 20

Using the information provided, let vector $\mathbf{u}$ have magnitude $5$ and direction $0^\circ$ (measured from the positive $x$ -axis), and let vector $\mathbf{v}$ have magnitude $12$ and direction $90^\circ$ . What are the magnitude and direction of $\mathbf{u}+\mathbf{v}$ (give direction as an angle from the positive $x$ -axis)?

Magnitude $13$ at $\tan^{-1}\!\left(\frac{12}{5}\right)^\circ$ (correct answer)
Magnitude $17$ at $\tan^{-1}\!\left(\frac{5}{12}\right)^\circ$
Magnitude $7$ at $90^\circ$
Magnitude $13$ at $\tan^{-1}\!\left(\frac{5}{12}\right)^\circ$

Explanation: This question tests understanding of how to add vectors given in magnitude and direction form and find the magnitude and direction of the resultant. For special cases like perpendicular vectors, the calculation simplifies: if one vector is magnitude r₁ along the x-axis (direction 0°) and another is magnitude r₂ along the y-axis (direction 90°), the resultant magnitude is simply √(r₁² + r₂²) and the direction is arctan(r₂/r₁). Since one vector points east with magnitude 5 and the other points north with magnitude 12, these are perpendicular, forming a right triangle. The magnitude of the sum is the hypotenuse: √(5² + 12²) = √(25 + 144) = √169 =13, and the direction is arctan(12/5) from the positive x-axis. Choice A is correct because it properly converts to components, adds correctly, and applies the Pythagorean theorem for magnitude and correctly uses arctan with the proper ratio for direction. Choice D incorrectly switches the ratio in the arctan, calculating arctan(5/12) instead of arctan(12/5), perhaps confusing the magnitudes of the vectors. Key to adding vectors in magnitude-direction form: always convert to components first using x = r·cos(θ) and y = r·sin(θ), add the x-components together and y-components together, then find magnitude using √(x² + y²) and direction using arctan(y/x) with quadrant adjustment. Special case shortcut: when vectors are perpendicular (like one pointing east and one pointing north), you can directly apply the Pythagorean theorem to the magnitudes: |u + v| = √(r₁² + r₂²), and the direction is arctan(r₂/r₁) from the first vector's direction.