Find Sum of Two Vectors - Precalculus

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.)
Answer: $\sqrt{200}$ 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.
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)?
Answer: Magnitude $13$ at $\tan^{-1}\!\left(\frac{12}{5}\right)^\circ$ 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.
For the vectors described, vector $\mathbf{u}$ has magnitude $10$ at direction $30^\circ$ and vector $\mathbf{v}$ has magnitude $10$ at direction $150^\circ$ , where directions are measured counterclockwise from the positive $x$ -axis. What is the magnitude of $\mathbf{u}+\mathbf{v}$ (direction not needed)?
Answer: $10$ 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 10 at direction 30°, so its components are ⟨10·cos(30°), 10·sin(30°)⟩ = ⟨5√3, 5⟩. Vector v has magnitude 10 at direction 150°, so its components are ⟨10·cos(150°), 10·sin(150°)⟩ = ⟨-5√3, 5⟩. Adding these gives u + v = ⟨0, 10⟩, and the magnitude is √(0² + 10²) = 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. 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.
For the vectors described, vector $\mathbf{u}$ has magnitude $10$ and direction $0^\circ$ , and vector $\mathbf{v}$ has magnitude $6$ and direction $180^\circ$ (angles from the positive $x$ -axis). What are the magnitude and direction of $\mathbf{u}+\mathbf{v}$ ?
Answer: Magnitude $4$ at $0^\circ$ 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 0°, so its components are ⟨10, 0⟩. Vector v has magnitude 6 at direction 180°, so its components are ⟨-6, 0⟩. Adding these gives u + v = ⟨4, 0⟩, the magnitude is √(16 + 0) =4, and the direction is 0° since it's along the positive x-axis. Choice A is correct because it properly converts to components, adds correctly, and determines both magnitude and direction accurately. Choice C makes an arithmetic error in the calculation, computing 10 - 6 =4 but placing it at 180° instead of 0°, confusing the direction of the resultant. 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.
A force $\mathbf{F}_1$ of $12$ N acts at direction $0^\circ$ (along the positive $x$ -axis). A force $\mathbf{F}_2$ of $5$ N acts at direction $90^\circ$ (along the positive $y$ -axis). For the vectors described, what is the magnitude of the resultant force $\mathbf{F}_1+\mathbf{F}_2$ (magnitude only)?
Answer: $13$ N 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 east with magnitude 12 and the other points north with magnitude 5, these are perpendicular, forming a right triangle. The magnitude of the sum is the hypotenuse: √(12² + 5²) = √(144 + 25) = √169 = 13. Choice B is correct because it properly applies the Pythagorean theorem for perpendicular vectors. Choice A incorrectly adds the magnitudes directly (12 + 5 = 17), 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. To verify your answer, check that the magnitude of the sum is between |r₁ - r₂| and r₁ + r₂ (triangle inequality), and indeed 7 < 13 < 17.
In an $xy$ -coordinate system where $0^\circ$ is along the positive $x$ -axis and angles increase counterclockwise, vector $\mathbf{u}$ has magnitude $5$ at direction $0^\circ$ , and vector $\mathbf{v}$ has magnitude $5$ at direction $180^\circ$ . Given the magnitudes and directions, what is $|\mathbf{u}+\mathbf{v}|$ ?
Answer: $0$ 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 5 at direction 0°, so its components are ⟨5·cos(0°), 5·sin(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. Choice B is correct because the vectors point in exactly opposite directions with equal magnitudes, so they cancel completely. Choice A incorrectly gives magnitude 10, which would be the result of adding the magnitudes directly without considering direction—but these vectors point in opposite directions, so their effects cancel. 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). When vectors have equal magnitudes and opposite directions (180° apart), their sum is always the zero vector.
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.
Answer: magnitude $5$ at $\tan^{-1}\!\left(\frac{4}{3}\right)$ 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.
Vector $\mathbf{u}$ has magnitude $8$ at direction $30^\circ$ (from the positive $x$ -axis). Vector $\mathbf{v}$ has magnitude $8$ at direction $150^\circ$ . Given the magnitudes and directions, what is the magnitude of $\mathbf{u}+\mathbf{v}$ ? (Direction not needed.)
Answer: $8$ 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 8 at direction 30°, so its components are ⟨8·cos(30°), 8·sin(30°)⟩ = ⟨4√3, 4⟩; vector v has magnitude 8 at direction 150°, so its components are ⟨8·cos(150°), 8·sin(150°)⟩ = ⟨-4√3, 4⟩; adding these gives u + v = ⟨0, 8⟩, and the magnitude is √(0² + 8²) = √64 = 8. 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 (8 + 8 = 16), 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.
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.)
Answer: Magnitude $0$ (direction undefined) 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.
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 $\vec{d}_1 + \vec{d}_2$ ?
Answer: $\frac{9\sqrt{3} - 15}{2}$ meters The x-component of $\vec{d}_1$ is $15\cos 240° = 15 \cdot (-\frac{1}{2}) = -\frac{15}{2}$ . The x-component of $\vec{d}_2$ is $9\cos 30° = 9 \cdot \frac{\sqrt{3}}{2} = \frac{9\sqrt{3}}{2}$ . 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.
In an $xy$ -coordinate system where directions are measured from the positive $x$ -axis, vector $\mathbf{u}$ has magnitude $10$ at direction $0^\circ$ , and vector $\mathbf{v}$ has magnitude $10$ at direction $90^\circ$ . For the vectors described, what is the direction of the resultant vector $\mathbf{u}+\mathbf{v}$ ? (Direction only; magnitude not needed.)
Answer: $45^\circ$ 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°. Since one vector points east with magnitude 10 and the other points north with magnitude 10, these are perpendicular vectors along the coordinate axes. The sum has components ⟨10, 10⟩. The direction angle is arctan(10/10) = arctan(1) = 45°. Since both components are positive, the vector is in Quadrant I, so the direction is 45° from the positive x-axis. Choice B is correct because it correctly identifies that equal perpendicular vectors at 0° and 90° produce a resultant at 45°, which is the angle that bisects the right angle between them. Choice C incorrectly gives 90°, which is the direction of only one of the original vectors, not their sum. For direction, remember that arctan alone gives angles between -90° and 90°, so you must check the signs of the components to determine the correct quadrant. When two perpendicular vectors have equal magnitudes, their sum always points at 45° from the first vector's direction.
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) $\mathbf{F}_1+\mathbf{F}_2$ ?
Answer: Magnitude $10\sqrt{2}\text{ N}$ at $45^\circ$ 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°.
Using the information provided (angles measured in degrees counterclockwise from the positive $x$ -axis), vector $\mathbf{u}$ has magnitude $8$ at $0^\circ$ and vector $\mathbf{v}$ has magnitude $6$ at $120^\circ$ . What is $\lvert \mathbf{u}+\mathbf{v} \rvert$ ?
Answer: $\sqrt{52}$ 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 8 at direction 0°, so its components are ⟨8·cos(0°), 8·sin(0°)⟩ = ⟨8·1, 8·0⟩ = ⟨8, 0⟩. Vector v has magnitude 6 at direction 120°, so its components are ⟨6·cos(120°), 6·sin(120°)⟩ = ⟨6·(-1/2), 6·(√3/2)⟩ = ⟨-3, 3√3⟩. Adding these gives u + v = ⟨8 + (-3), 0 + 3√3⟩ = ⟨5, 3√3⟩, and the magnitude is √(5² + (3√3)²) = √(25 + 27) = √52. 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 (8 + 6 = 14), 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). 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.
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.
Answer: magnitude $\sqrt{42500}\text{ km/h}$ at $\tan^{-1}\!\left(\frac{1}{4}\right)$ 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.
Angles are measured in degrees counterclockwise from the positive $x$ -axis. Vector $\mathbf{u}$ has magnitude $6$ at direction $0^\circ$ , and vector $\mathbf{v}$ has magnitude $6$ at direction $60^\circ$ . Using the information provided, what is the magnitude of $\mathbf{u}+\mathbf{v}$ ? (Direction is not needed.)
Answer: $6\sqrt{3}$ 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 6 at direction 0°, so its components are ⟨6, 0⟩; vector v has magnitude 6 at direction 60°, so its components are ⟨6·cos(60°), 6·sin(60°)⟩ = ⟨3, 3√3⟩; adding these gives u + v = ⟨9, 3√3⟩, and the magnitude is √(9² + (3√3)²) = √(81 + 27) = √108 = 6√3. 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 (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. 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.
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 $12$ at $90^\circ$ . What are the magnitude and direction of $\mathbf{u}+\mathbf{v}$ ? (Give direction as an angle from the positive $x$ -axis.)
Answer: Magnitude $13$ at $\tan^{-1}\!\left(\frac{12}{5}\right)$ 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. For direction, arctan(12/5) gives the angle from the positive x-axis, so the resultant is magnitude 13 at tan⁻¹(12/5). Choice C is correct because it properly applies the Pythagorean theorem for perpendicular vectors and correctly calculates the direction as arctan(12/5). Choice A incorrectly inverts the ratio in the arctangent, using tan⁻¹(5/12) instead of tan⁻¹(12/5), which would give a much smaller angle. 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.
Angles are measured in degrees counterclockwise from the positive $x$ -axis. Vector $\mathbf{u}$ has magnitude $13$ at direction $0^\circ$ , and vector $\mathbf{v}$ has magnitude $5$ at direction $180^\circ$ . Given the magnitudes and directions, what are the magnitude and direction of $\mathbf{u}+\mathbf{v}$ (angle from the positive $x$ -axis)?
Answer: Magnitude $8$ at $0^\circ$ 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 13 at direction 0°, so its components are ⟨13, 0⟩; vector v has magnitude 5 at direction 180°, so its components are ⟨-5, 0⟩; adding these gives u + v = ⟨8, 0⟩, and the magnitude is √(8² + 0²) = 8 with direction 0°. Converting to components, adding, and finding the magnitude gives 8 (as calculated above); for direction, arctan(0/8) = 0°, and since x is positive and y is zero, the direction is 0°. Choice B 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 adds the magnitudes directly (13 + 5 = 18), 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.
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)?
Answer: magnitude $10$ km at $\tan^{-1}\!\left(\frac{8}{6}\right)$ 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.
In an $xy$ -coordinate system where $0^\circ$ is along the positive $x$ -axis and angles increase counterclockwise, vector $\mathbf{u}$ has magnitude $13$ at direction $0^\circ$ , and vector $\mathbf{v}$ has magnitude $13$ at direction $90^\circ$ . Using the information provided, what are the magnitude and direction of $\mathbf{u}+\mathbf{v}$ (direction as an angle from the positive $x$ -axis)?
Answer: magnitude $13\sqrt{2}$ at $45^\circ$ 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 13 and the other points north with magnitude 13, these are perpendicular, forming a right triangle. The magnitude of the sum is the hypotenuse: √(13² + 13²) = √(169 + 169) = √338 = √(169·2) = 13√2. For direction, since the vectors have equal magnitudes and are perpendicular, the resultant bisects the angle between them at 45°. Choice B is correct because it properly identifies the magnitude as 13√2 and the direction as 45°. Choice A incorrectly gives magnitude 26, which would be 13 + 13, incorrectly adding magnitudes directly—but vector addition requires the Pythagorean theorem for perpendicular vectors. 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 when the magnitudes are equal, the direction is always 45° from either original vector.
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.)
Answer: $0$ 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.

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