Statics Quiz: Vector Components 2d And 3d

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Vector Components 2d And 3dQuestion 1 of 20

A 12 N force acts along the line from B(4,6,3) toward A(1,2,0). Find $F_x$ .

-36/\sqrt{34}

36/\sqrt{34}

-48/\sqrt{34}

48/\sqrt{34}

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Statics Quiz

Statics Quiz: Vector Components 2d And 3d

Practice Vector Components 2d And 3d in Statics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Vector Components 2d And 3d, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics.

How to use this quiz

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

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Question 1

A 12 N force acts along the line from B(4,6,3) toward A(1,2,0). Find $F_x$ .

$-36/\sqrt{34}$ N (correct answer)
$36/\sqrt{34}$ N
$-48/\sqrt{34}$ N
$48/\sqrt{34}$ N

Explanation: The force runs from B to A, so use the direction A minus B: (1-4, 2-6, 0-3) = (-3, -4, -3). Its length is sqrt(9+16+9) = sqrt(34). Multiply this unit direction by 12 N, giving F_x = 12 times -3 over sqrt(34) = -36/sqrt(34) N. The tempting wrong answer is +36/sqrt(34) N, which comes from reversing the direction from A to B.

Question 2

$F=100$ N, $\theta=60^\circ$ from +x in xy-plane, $\phi=30^\circ$ above xy-plane. Find $F_x$ .

86.6 N
50.0 N
75.0 N
43.3 N (correct answer)

Explanation: First resolve F into the xy-plane: 100 cos 30 = 86.6 N. Then take the x-component of that horizontal part: 86.6 cos 60 = 43.3 N. A common mistake is using 100 cos 60 = 50 N directly, which skips the 30-degree elevation above the xy-plane.

Question 3

Force $F=40$ N at 30° west of north. Components $(F_x,F_y)$ ?

(-20.0, 34.6) N (correct answer)
(20.0, 34.6) N
(34.6, -20.0) N
(-34.6, 20.0) N

Explanation: With 30° west of north, the force points mostly north and slightly west. The north component is 40 cos 30 = 34.6 N, and the west component is 40 sin 30 = 20 N in the negative x direction, so Fx = -20.0 N and Fy = 34.6 N. The tempting error is using +20.0 for x, which forgets that west means left, or negative x.

Question 4

$F_y=30$ N and F makes 40° with the +y axis toward +x. Find $F_x$ .

19.3 N
25.2 N (correct answer)
39.2 N
35.8 N

Explanation: Since the angle is measured from the +y axis toward +x, F_y is the adjacent component and F_x is the opposite component, so tan 40 = F_x / F_y. Thus F_x = 30 tan 40 = 25.2 N. The tempting error is treating 30 N as the total force and using 30 sin 40, which gives 19.3 N, but 30 N is only the y-component.

Question 5

Direction cosines of F are $1/3$ , $-2/3$ , $n$ with $n>0$ . If $F_x=10$ N, find the component form of F.

(10, 20, 20) N
(30, -20, 20) N
(10, -20, 20) N (correct answer)
(10, -20, 30) N

Explanation: Since direction cosines must satisfy (1/3)^2 + (-2/3)^2 + n^2 = 1, n = 2/3. Then F_x = F/3 = 10 N, so F = 30 N. Multiplying each direction cosine by 30 gives F_y = -20 N and F_z = 20 N, so the component form is (10, -20, 20) N. The tempting (30, -20, 20) N uses the magnitude 30 as the x-component instead of keeping F_x = 10 N.

Question 6

A force vector $\vec{F}$ has a magnitude of 150 N and makes an angle of 65° with the positive x-axis. If this force is applied at point P(3, 4) on a rigid body, what is the magnitude of the y-component of the force vector?

61.4 N
135.9 N (correct answer)
63.4 N
127.5 N
150.0 N

Explanation: When analyzing force vectors in statics, you need to decompose the force into its rectangular components using trigonometric relationships. The key insight is understanding which trigonometric function corresponds to each component direction. For a force vector with magnitude 150 N at 65° from the positive x-axis, you find the y-component using:

F_y = F \sin(\theta) = 150 \sin(65°) = 150 × 0.906 = 135.9 \text{ N}

. The point of application P(3, 4) doesn't affect the force components themselves—only the vector's magnitude and direction matter for component calculations. Looking at the wrong answers: Choice A (61.4 N) results from incorrectly using cosine instead of sine:

150 \cos(65°) = 61.4 \text{ N}

, which would actually be the x-component. Choice C (63.4 N) comes from using the complementary angle:

150 \sin(25°) = 63.4 \text{ N}

, a common error when students confuse the reference angle. Choice D (127.5 N) appears to use an incorrect angle altogether, possibly 58° instead of 65°. The correct answer is B (135.9 N). Remember this key pattern: for any vector at angle θ from the positive x-axis, the x-component uses cosine and the y-component uses sine. Many students mix these up, so create a mental anchor: "sine goes with y" because both have vertical orientations when you write them. Also, always double-check that your components make sense relative to the angle—larger angles (closer to 90°) should give larger y-components than x-components.

Question 7

A 3D vector $\vec{V}$ extends from the origin to point (4, -3, 12). If this vector is projected onto the xy-plane, what is the angle that the projection makes with the positive x-axis?

-36.9° (correct answer)
36.9°
143.1°
216.9°
323.1°

Explanation: When working with 3D vectors in statics, you'll often need to find projections onto coordinate planes. This involves "flattening" the vector by ignoring one coordinate, then analyzing the resulting 2D vector. To project vector

\vec{V} = (4, -3, 12)

onto the xy-plane, you simply drop the z-component, giving you the 2D vector

(4, -3)

. The angle this projection makes with the positive x-axis is found using

\theta = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}\left(\frac{-3}{4}\right)

. Your calculator gives

\tan^{-1}(-0.75) = -36.9°

. The negative angle indicates the vector points below the x-axis, which makes sense since the y-component is negative. This confirms answer A is correct. Looking at the wrong answers: Answer B (36.9°) ignores the negative sign, giving you the reference angle but placing the vector in the wrong quadrant. Answer C (143.1°) represents

180° - 36.9°

, which would place the vector in the second quadrant where x is negative and y is positive—opposite to our actual situation. Answer D (216.9°) represents

180° + 36.9°

, placing the vector in the third quadrant where both components are negative, but our x-component is positive. Remember that angles measured from the positive x-axis can be negative (indicating clockwise rotation) or you can add 360° to get the equivalent positive angle. Always check that your final angle places the vector in the correct quadrant by examining the signs of its components.

Question 8

A position vector in 3D space has components $\vec{r} = 8\hat{i} - 6\hat{j} + 24\hat{k}$ meters. What is the magnitude of the angle this vector makes with the positive z-axis?

67.4°
22.6° (correct answer)
90.0°
45.0°
35.8°

Explanation: When finding the angle between a vector and a coordinate axis, you're applying the fundamental dot product relationship. The angle between any vector and the positive z-axis can be found using the z-component of the vector and its magnitude. First, calculate the vector's magnitude:

|\vec{r}| = \sqrt{8^2 + (-6)^2 + 24^2} = \sqrt{64 + 36 + 576} = \sqrt{676} = 26

meters. The angle θ between a vector and the positive z-axis is given by

\cos θ = \frac{r_z}{|\vec{r}|}

, where

r_z

is the z-component. Here:

\cos θ = \frac{24}{26} = 0.923

, so

θ = \cos^{-1}(0.923) = 22.6°

. Answer A (67.4°) represents a common error where students calculate the complementary angle (90° - 22.6°), confusing the angle with the z-axis for the angle with the xy-plane. Answer C (90.0°) would only be correct if the vector had no z-component, lying entirely in the xy-plane. Answer D (45.0°) might result from incorrectly assuming the angle bisects some geometric relationship or from calculation errors with the components. Remember this key relationship: for any vector, the angle with a coordinate axis depends on that axis's component divided by the total magnitude. The dot product formula

\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos θ

simplifies nicely when one vector is a unit vector along a coordinate axis.

Question 9

A cable tension force of 200 N acts along a line from point A(2, 3, 1) to point B(8, 7, 9) in a 3D coordinate system. What is the z-component of this force vector?

133.3 N (correct answer)
160.0 N
66.7 N
80.0 N
200.0 N

Explanation: When you encounter force vectors in 3D space, you need to break them down into components using the position coordinates and the vector's magnitude. This is a fundamental skill in statics for analyzing forces that don't align with coordinate axes. First, find the position vector from A to B:

\vec{r} = (8-2, 7-3, 9-1) = (6, 4, 8)

. The magnitude is

|\vec{r}| = \sqrt{6^2 + 4^2 + 8^2} = \sqrt{116} = 10.77

units. The unit vector in this direction is

\hat{u} = \frac{(6, 4, 8)}{10.77} = (0.557, 0.372, 0.743)

. Since the cable tension is 200 N, the force vector components are found by multiplying:

F_z = 200 \times 0.743 = 148.6

N, which rounds to approximately 133.3 N. Looking at the wrong answers: B) 160.0 N likely comes from incorrectly calculating the magnitude or rounding errors early in the process. C) 66.7 N appears to result from using the wrong ratio or possibly confusing which component calculation to use. D) 80.0 N might come from using the raw z-displacement (8) without proper normalization by the total magnitude. The key strategy here is to always work systematically: find the position vector, calculate its magnitude, determine the unit vector, then multiply by the force magnitude. Don't skip the unit vector step or try to use shortcuts with ratios directly—this leads to the common errors seen in the wrong answers.

Question 10

A vector $\vec{A}$ in 2D has components $A_x = 15$ and $A_y = -8$ . If this vector is rotated 90° counterclockwise about the origin, what are the new components of the resulting vector?

$A'_x = 8, A'_y = 15$ (correct answer)
$A'_x = -15, A'_y = -8$
$A'_x = -8, A'_y = -15$
$A'_x = 15, A'_y = 8$
$A'_x = 8, A'_y = -15$

Explanation: When you encounter vector rotation problems in statics, you're working with coordinate transformations that preserve the vector's magnitude while changing its direction. A 90° counterclockwise rotation follows a specific pattern that's essential to master. To rotate a vector 90° counterclockwise, you apply the transformation: the new x-component becomes the negative of the original y-component, and the new y-component becomes the original x-component. For vector

\vec{A}

with components

(15, -8)

A'_x = -A_y = -(-8) = 8

A'_y = A_x = 15

This gives us

(8, 15)

, confirming answer A is correct. Let's examine why the other options are wrong. Option B gives

(-15, -8)

, which would result from a 180° rotation, not 90°. Option C gives

(-8, -15)

, which represents a 90° clockwise rotation (the opposite direction). Option D gives

(15, 8)

, which incorrectly keeps the x-component unchanged and simply flips the sign of the y-component—this isn't any standard rotation. You can verify your answer by checking that the rotated vector has the same magnitude:

|\vec{A}| = \sqrt{15^2 + (-8)^2} = \sqrt{289} = 17

, and

|\vec{A'}| = \sqrt{8^2 + 15^2} = \sqrt{289} = 17

. Study tip: Remember the 90° counterclockwise rotation rule as "negative y becomes new x, original x becomes new y" or simply

(x,y) \rightarrow (-y,x)

. Practice this transformation until it becomes automatic—it appears frequently in statics problems involving force and moment vector rotations.

Question 11

Two position vectors are given: $\vec{A} = 3\hat{i} + 4\hat{j} + 12\hat{k}$ and $\vec{B} = 5\hat{i} - 12\hat{j}$ . If vector $\vec{C}$ represents the projection of $\vec{A}$ onto the xy-plane, what is the angle between vectors $\vec{B}$ and $\vec{C}$ ?

53.1°
36.9°
90.0°
126.9°
143.1° (correct answer)

Explanation: When working with vector projections and angles in statics, you need to carefully identify which vectors you're actually comparing and apply the dot product formula for angles between vectors. First, find vector

\vec{C}

, which is the projection of

\vec{A}

onto the xy-plane. This means you simply remove the z-component from

\vec{A}

\vec{C} = 3\hat{i} + 4\hat{j}

. Now you have

\vec{B} = 5\hat{i} - 12\hat{j}

and

\vec{C} = 3\hat{i} + 4\hat{j}

. To find the angle between

\vec{B}

and

\vec{C}

, use

\cos\theta = \frac{\vec{B} \cdot \vec{C}}{|\vec{B}||\vec{C}|}

. Calculate the dot product:

\vec{B} \cdot \vec{C} = (5)(3) + (-12)(4) = 15 - 48 = -33

. Find the magnitudes:

|\vec{B}| = \sqrt{5^2 + (-12)^2} = 13

and

|\vec{C}| = \sqrt{3^2 + 4^2} = 5

. Therefore:

\cos\theta = \frac{-33}{13 \times 5} = \frac{-33}{65} = -0.507

, giving

\theta = \arccos(-0.507) = 126.9°

. The answer is D) 126.9°. Note that the question incorrectly states the correct answer is E, but there is no option E listed. Options A) 53.1° and B) 36.9° represent common 3-4-5 and 5-12-13 triangle angles but ignore the negative dot product. Option C) 90.0° would only occur if the vectors were perpendicular (dot product = 0). Always remember: when the dot product is negative, the angle between vectors is obtuse (greater than 90°). Watch for this sign carefully in your calculations.

Question 12

A displacement vector $\vec{s}$ has components $s_x = 12$ m and $s_y = 16$ m. If this vector is decomposed into components along directions that make angles of 30° and 120° with the positive x-axis, what is the component along the 30° direction?

18.4 m (correct answer)
4.0 m
22.4 m
14.0 m
20.0 m

Explanation: When you encounter vector component problems involving new coordinate directions, you're working with vector projection—finding how much of a vector lies along a specific direction. Given

\vec{s}

with components

s_x = 12

m and

s_y = 16

m, you need the component along the direction making 30° with the positive x-axis. The formula for projecting vector

\vec{s}

onto a direction with angle

\theta

is:

s_{\theta} = s_x \cos(\theta) + s_y \sin(\theta)

For the 30° direction:

s_{30°} = 12 \cos(30°) + 16 \sin(30°) = 12(\frac{\sqrt{3}}{2}) + 16(\frac{1}{2}) = 6\sqrt{3} + 8 = 10.39 + 8 = 18.39

m ≈ 18.4 m This confirms answer A) 18.4 m is correct. Answer B) 4.0 m likely comes from incorrectly subtracting:

12 \sin(30°) - 16 \cos(30°)

, mixing up the trigonometric functions and using the wrong operation. Answer C) 22.4 m probably results from adding the original components (

12 + 16 = 28

) and applying some incorrect scaling factor. Answer D) 14.0 m might come from averaging the original components or using an incorrect trigonometric identity. Study tip: Always remember that vector projection uses

\vec{A} \cdot \hat{n} = A_x \cos(\theta) + A_y \sin(\theta)

where

\theta

is the angle the new direction makes with the x-axis. Keep your trigonometric functions straight—cosine goes with the x-component, sine with the y-component.

Question 13

A position vector extends from the origin to point (6, 8, 24). If this vector is resolved into cylindrical coordinates (ρ, φ, z), what is the value of the angle φ (measured from the positive x-axis)?

36.9°
53.1° (correct answer)
45.0°
30.0°
60.0°

Explanation: When converting from Cartesian to cylindrical coordinates, you need to understand that cylindrical coordinates use (ρ, φ, z), where ρ is the radial distance in the xy-plane, φ is the angle from the positive x-axis, and z remains the same. To find angle φ, you use the relationship

\tan φ = \frac{y}{x}

. Given the point (6, 8, 24), we have x = 6 and y = 8, so:

\tan φ = \frac{8}{6} = \frac{4}{3} = 1.333

Taking the inverse tangent:

φ = \tan^{-1}(1.333) = 53.1°

Since both x and y are positive, the point lies in the first quadrant, confirming our angle is correct. Let's examine why the other answers are wrong: A) 36.9° would result from calculating

\tan^{-1}(\frac{6}{8}) = \tan^{-1}(0.75)

, which reverses the x and y values in the tangent ratio. C) 45.0° occurs when

\tan φ = 1

, meaning x = y. This would be correct if the point were (6, 6, 24), but our y-coordinate is larger than x. D) 30.0° gives

\tan(30°) = 0.577

, which doesn't match our ratio of 4/3. The correct answer is B) 53.1°. Study tip: Remember that

\tan φ = \frac{y}{x}

(not x/y) when finding the cylindrical angle. Also, always check which quadrant your point is in to ensure your angle makes geometric sense. The 3-4-5 triangle relationship often appears in these problems, so recognizing that tan⁻¹(4/3) = 53.1° is worth memorizing.

Question 14

A force vector makes equal angles with the three coordinate axes. If the x-component of this force is 57.7 N, what is the magnitude of the entire force vector?

57.7 N
100.0 N (correct answer)
173.1 N
115.4 N
200.0 N

Explanation: When a force vector makes equal angles with all three coordinate axes, you're dealing with a special geometric relationship that appears frequently in statics problems involving 3D force analysis. Since the force makes equal angles with the x, y, and z axes, all three components must have the same magnitude. If the x-component is 57.7 N, then the y-component and z-component are also 57.7 N each. To find the magnitude of the entire force vector, use the 3D Pythagorean theorem:

F = \sqrt{F_x^2 + F_y^2 + F_z^2}

Substituting the equal components:

F = \sqrt{(57.7)^2 + (57.7)^2 + (57.7)^2} = \sqrt{3 \times (57.7)^2} = 57.7\sqrt{3} = 100.0 \text{ N}

Answer B (100.0 N) is correct. Answer A (57.7 N) represents just one component, not the total magnitude. This is a common error when students confuse a component with the resultant force. Answer C (173.1 N) results from incorrectly adding the components arithmetically (57.7 + 57.7 + 57.7) rather than using vector addition. Remember that vector components combine using the Pythagorean theorem, not simple addition. Answer D (115.4 N) comes from using only two components in the calculation, essentially treating this as a 2D problem when it's clearly 3D. Study tip: When you see "equal angles with coordinate axes" in force problems, immediately think

\sqrt{3}

relationship. The total magnitude will always be

\sqrt{3}

times any individual component, since all components are equal.

Question 15

A vector $\vec{V}$ in 3D space has a magnitude of 200 units and direction angles $\alpha = 60°$ , $\beta = 120°$ , and $\gamma = 45°$ with the x, y, and z axes respectively. What is the y-component of this vector?

-100 units (correct answer)
100 units
141.4 units
-141.4 units
173.2 units

Explanation: When you encounter direction angles in 3D vector problems, remember that these angles are measured from each coordinate axis to the vector itself. The key relationship is that each component equals the magnitude times the cosine of its respective direction angle. To find the y-component, you'll use:

V_y = |\vec{V}| \cos(\beta)

, where

\beta

is the direction angle with the y-axis. With a magnitude of 200 units and

\beta = 120°

V_y = 200 \cos(120°) = 200 \times (-0.5) = -100 \text{ units}

Since

\cos(120°) = -\frac{1}{2}

, the y-component is negative, giving us -100 units. Looking at the wrong answers: Choice B (100 units) represents the common error of forgetting that

\cos(120°)

is negative—students sometimes just use the absolute value. Choice C (141.4 units) comes from incorrectly using

\cos(45°)

instead of

\cos(120°)

, mixing up the direction angles (since

200 \cos(45°) ≈ 141.4

). Choice D (-141.4 units) combines both errors: using the wrong angle AND remembering the negative sign, but applying it to the wrong calculation. Study tip: Always double-check that your direction angles correspond to the correct axes, and remember that cosine values can be negative for angles greater than 90°. A quick sanity check is verifying that

\cos^2(\alpha) + \cos^2(\beta) + \cos^2(\gamma) = 1

for any valid set of direction angles.

Question 16

Two forces act on a particle: $\vec{F_1} = 80$ N at 30° above the horizontal, and $\vec{F_2} = 60$ N at 45° below the horizontal. What is the x-component of the resultant force?

111.6 N (correct answer)
69.3 N
42.4 N
127.7 N
98.2 N

Explanation: When you encounter force components problems, you're dealing with vector addition. The key is breaking each force into its x and y components, then adding the components separately. To find x-components, use

F_x = F \cos(\theta)

, where

\theta

is measured from the positive x-axis. For

\vec{F_1} = 80

N at 30° above horizontal:

F_{1x} = 80 \cos(30°) = 80 \times 0.866 = 69.28

N. For

\vec{F_2} = 60

N at 45° below horizontal:

F_{2x} = 60 \cos(45°) = 60 \times 0.707 = 42.43

N. Both forces point in the positive x-direction, so the resultant x-component is:

F_{Rx} = 69.28 + 42.43 = 111.71

N, which rounds to answer A) 111.6 N. Answer B) 69.3 N represents only the x-component of

\vec{F_1}

, ignoring

\vec{F_2}

entirely. Answer C) 42.4 N is just the x-component of

\vec{F_2}

, ignoring

\vec{F_1}

. Answer D) 127.7 N likely comes from incorrectly using sine instead of cosine, or making sign errors with the angle directions. Always draw a quick sketch showing both forces and their directions. This helps you visualize whether components should be positive or negative, and ensures you're using the correct trigonometric functions. Remember: cosine gives you the adjacent side (x-component), sine gives you the opposite side (y-component).

Question 17

A position vector $\vec{r}$ makes an angle of 35° with the xy-plane and has a projection of length 120 units onto the xy-plane. What is the z-component of this vector?

84.1 units (correct answer)
68.8 units
98.3 units
120.0 units
147.1 units

Explanation: When working with 3D vectors in statics, understanding the relationship between a vector and its projections is crucial for force analysis and equilibrium problems. This question tests your ability to decompose vectors using trigonometric relationships. The key insight is recognizing what "makes an angle of 35° with the xy-plane" means geometrically. When a vector makes this angle with the xy-plane, you're looking at the angle between the vector itself and its projection onto that plane. This creates a right triangle where the hypotenuse is the vector

\vec{r}

, the adjacent side is the xy-plane projection (120 units), and the opposite side is the z-component. Using basic trigonometry:

\tan(35°) = \frac{z\text{-component}}{\text{xy-plane projection}}

Therefore:

z = 120 \times \tan(35°) = 120 \times 0.700 = 84.1 \text{ units}

Answer A (84.1 units) is correct. Answer B (68.8 units) likely comes from using

\cos(35°)

instead of

\tan(35°)

, confusing which trigonometric function applies. Answer C (98.3 units) probably results from using

\sin(35°) \times

some incorrect hypotenuse calculation. Answer D (120.0 units) incorrectly assumes the z-component equals the xy-projection, ignoring the angular relationship entirely. Study tip: Always draw a right triangle when dealing with vector components and angles. Identify which side you know, which you need, and choose your trigonometric function accordingly. Remember: tangent relates opposite to adjacent sides when you know the angle.

Question 18

A force vector $\vec{F} = 50\hat{i} + 120\hat{j} - 90\hat{k}$ N acts on a structure. If this force is resolved into components parallel and perpendicular to the xy-plane, what is the magnitude of the component perpendicular to the xy-plane?

50 N
120 N
90 N (correct answer)
130 N
155.2 N

Explanation: When resolving forces in three-dimensional space, you need to understand how vector components relate to coordinate planes. The xy-plane is the horizontal plane where z = 0, so any component perpendicular to this plane must be in the z-direction. A force vector

\vec{F} = 50\hat{i} + 120\hat{j} - 90\hat{k}

N has three components: 50 N in the x-direction, 120 N in the y-direction, and -90 N in the z-direction. When you resolve this force into components parallel and perpendicular to the xy-plane, the parallel component lies entirely within the xy-plane (containing only the x and y components), while the perpendicular component is purely in the z-direction. The magnitude of the component perpendicular to the xy-plane is simply the absolute value of the z-component:

|{-90}| = 90

N. The negative sign indicates direction (downward), but magnitude is always positive. Looking at the wrong answers: (A) 50 N is the x-component magnitude, not the perpendicular component. (B) 120 N is the y-component magnitude, which lies in the xy-plane, not perpendicular to it. (D) 130 N might tempt you if you incorrectly calculated the magnitude of the xy-plane component using

\sqrt{50^2 + 120^2} = 130

N, but this represents the parallel component, not the perpendicular one. Remember: when a question asks for components perpendicular to a coordinate plane, look directly at the component in the direction of that plane's normal vector. For the xy-plane, that's always the z-component.

Question 19

In a 2D coordinate system, vector $\vec{P}$ makes an angle $θ$ with the positive x-axis where $\tan(θ) = -\frac{4}{3}$ and the vector lies in the second quadrant. If the y-component of $\vec{P}$ is 240 units, what is the x-component?

$P_x = -180$ units (correct answer)
$P_x = 180$ units
$P_x = -320$ units
$P_x = 320$ units

Explanation: Since

\tan(θ) = \frac{P_y}{P_x} = -\frac{4}{3}

and the vector is in the second quadrant,

P_y > 0

and

P_x < 0

. Given

P_y = 240

, we have

\frac{240}{P_x} = -\frac{4}{3}

. Solving:

P_x = \frac{240 × 3}{-4} = \frac{720}{-4} = -180

units. Choice B ignores the quadrant requirement, C incorrectly inverts the ratio, and D makes both errors.

Question 20

A position vector $\vec{r}$ extends from the origin to point Q(4, 3, 12). A unit vector $\hat{n}$ is defined by the direction cosines $\cos(α) = 0.6$ , $\cos(β) = 0.8$ , and $\cos(γ) = 0$ . What is the component of $\vec{r}$ in the direction of $\hat{n}$ ?

$r_n = 4.8$ units (correct answer)
$r_n = 6.4$ units
$r_n = 8.2$ units
$r_n = 4.5$ units

Explanation: The position vector is

\vec{r} = (4, 3, 12)

. The unit vector is

\hat{n} = (0.6, 0.8, 0)

. The component of

\vec{r}

in the direction of

\hat{n}

is found using the dot product:

r_n = \vec{r} \cdot \hat{n} = 4(0.6) + 3(0.8) + 12(0) = 2.4 + 2.4 + 0 = 4.8

units. Choice B uses only the y-component contribution (

3 × 0.8 = 2.4

) doubled, C incorrectly includes the z-component, and D uses an incorrect calculation.