Vector Magnitude & Direction - Statics

A vector has magnitude 13 N and x-component 12 N; its y-component is negative. Find its direction from +x.
Answer: 22.6° below +x With magnitude 13 and x-component 12, the angle from the +x axis has cosine 12/13, so it is 22.6 degrees. Since the y-component is negative, the vector points below the +x axis, giving 22.6° below +x. The tempting 67.4° below +x is the complement, measured from the y-axis instead of from +x.
A force vector $\vec{F}$ makes angles of 60° with the positive x-axis and 45° with the positive y-axis. If the magnitude of this force is 100 N, what angle does it make with the positive z-axis?
Answer: 60.0° Using direction cosines: cos²α + cos²β + cos²γ = 1, where α, β, γ are angles with x, y, z axes respectively. cos(60°) = 0.5, cos(45°) = √2/2 = 0.707. So: (0.5)² + (0.707)² + cos²γ = 1, which gives 0.25 + 0.5 + cos²γ = 1, therefore cos²γ = 0.25, so cos(γ) = 0.5, and γ = 60°. Choice B assumes wrong calculation of remaining angle, C confuses with complement, D uses supplementary angle.
Two 100-N forces act at 60° and 120° from the +x-axis. Resultant magnitude and direction are:
Answer: 173 N at 90° Resolve each force into components. The x-components are 100 cos 60 = 50 N and 100 cos 120 = -50 N, so they cancel. The y-components are 100 sin 60 = 86.6 N and 100 sin 120 = 86.6 N, adding to 173.2 N. The resultant is thus 173 N straight up along the +y-axis. The tempting error is 200 N at 90°, which just adds the magnitudes without considering that the forces point in different directions.
Which vector has magnitude 10 and direction 143.1° from +x?
Answer: -8i + 6j For a vector of magnitude 10 at 143.1° from +x, the components are 10 cos 143.1° = -8 and 10 sin 143.1° = 6. So the vector is -8i + 6j. The tempting -6i + 8j also has magnitude 10 but its direction is about 126.9°, not 143.1°.
A force vector has a magnitude of 200 N and makes an angle of 150° with the positive x-axis. What are the x and y components of this force?
Answer: $F_x = -173.2$ N, $F_y = 100$ N When you encounter force component problems in statics, you're applying vector decomposition using trigonometry. The key is remembering that force components are found using $F_x = F \cos \theta$ and $F_y = F \sin \theta$ , where $\theta$ is measured counterclockwise from the positive x-axis. For this 200 N force at 150°, let's calculate each component: $F_x = 200 \cos(150°) = 200 \times (-0.866) = -173.2 \text{ N}$ $F_y = 200 \sin(150°) = 200 \times (0.5) = 100 \text{ N}$ The negative x-component makes sense because 150° places the force in the second quadrant, where x-components are negative and y-components are positive. Looking at the wrong answers: Choice B gives positive $F_x$ and negative $F_y$ , which would correspond to a force in the fourth quadrant (around 330°), not 150°. Choice C switches the magnitudes entirely—this happens when students confuse the sine and cosine values or mix up which trigonometric function goes with which component. Choice D makes both components positive, placing the force in the first quadrant, which contradicts the 150° angle. The correct answer is A: $F_x = -173.2$ N, $F_y = 100$ N. Study tip: Always sketch the force vector first to check your component signs. In the second quadrant (90° to 180°), x-components are always negative and y-components are always positive. This visual check will catch sign errors before you submit your answer.
Two perpendicular vectors $\vec{F_1}$ and $\vec{F_2}$ have magnitudes of 120 N and 160 N respectively. If $\vec{F_1}$ points in the positive x-direction, what is the direction of their resultant vector $\vec{R} = \vec{F_1} + \vec{F_2}$ ?
Answer: $53.13°$ above the positive x-axis When you encounter vector addition problems in statics, you're applying the fundamental principle that forces combine according to vector mathematics, not simple arithmetic. Since these vectors are perpendicular, this becomes a right triangle problem where you can use basic trigonometry. Given that $\vec{F_1} = 120\text{ N}$ points in the positive x-direction and $\vec{F_2} = 160\text{ N}$ is perpendicular to it, you can set up a coordinate system. The resultant vector forms the hypotenuse of a right triangle with legs of 120 N and 160 N. To find the direction, you need the angle $\theta$ that the resultant makes with the positive x-axis. Using trigonometry: $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{160}{120} = \frac{4}{3} = 1.333$ Taking the inverse tangent: $\theta = \arctan(1.333) = 53.13°$ Choice A (53.13°) is correct because it properly applies the arctangent of the ratio of perpendicular components. Choice B (36.87°) represents a common error where students accidentally use the reciprocal ratio (120/160 instead of 160/120). Choice C (45°) would only be correct if both force magnitudes were equal, creating an isosceles right triangle. Choice D (126.87°) appears to use the supplementary angle, which would apply if $\vec{F_2}$ pointed in the negative y-direction. Remember: when finding vector directions, always carefully identify which component is opposite and which is adjacent to your angle of interest, and double-check that your angle makes physical sense given the problem setup.
A force has $F_x=-6.0$ N and $F_y=8.0$ N. Its angle counterclockwise from the + $x$ -axis is:
Answer: 127° The vector points left and up (quadrant II). Its reference angle is arctan(8.0 / 6.0) = 53° from the negative x-axis, so the angle counterclockwise from the +x-axis is 180° - 53° = 127°. The tempting 53° comes from using that reference angle directly without accounting for the negative $F_x$ .
A force F = 24i + 7j kN. Find its direction angle from the +y axis.
Answer: 73.7° from +y For the angle from the +y axis, the y-component is the adjacent side and the x-component is the opposite side, so theta = arctan(24/7) = 73.7 degrees. The tempting wrong answer is 16.3 degrees from +x, which uses arctan(7/24) and measures from the +x axis instead of from +y.
A vector has components $F_x = -45$ N and $F_y = 60$ N. What is the magnitude of this vector and the angle it makes with the positive x-axis?
Answer: $75$ N, $126.9°$ When you encounter vector component problems in statics, you're working with two fundamental calculations: finding magnitude using the Pythagorean theorem and determining direction using inverse trigonometry. To find the magnitude, use $|\vec{F}| = \sqrt{F_x^2 + F_y^2}$ . With $F_x = -45$ N and $F_y = 60$ N, you get $|\vec{F}| = \sqrt{(-45)^2 + (60)^2} = \sqrt{2025 + 3600} = \sqrt{5625} = 75$ N. For the angle with the positive x-axis, use $\theta = \tan^{-1}\left(\frac{F_y}{F_x}\right) = \tan^{-1}\left(\frac{60}{-45}\right) = \tan^{-1}(-1.333) = -53.1°$ . However, since $F_x < 0$ and $F_y > 0$ , the vector lies in the second quadrant. The reference angle is $53.1°$ , so the actual angle is $180° - 53.1° = 126.9°$ . Choice A gives the correct magnitude and angle. Choice B has the right magnitude but uses $53.1°$ —this would place the vector in the first quadrant, ignoring the negative x-component. Choices C and D both show $105$ N, which comes from incorrectly adding the component magnitudes ( $45 + 60 = 105$ ) instead of using the Pythagorean theorem. Always remember: magnitude requires the Pythagorean theorem, and angle calculation must account for the proper quadrant based on component signs. Sketching the vector helps visualize which quadrant you're working in.
Two forces act on a particle: $\vec{F_1} = 40$ N at 150° from the positive x-axis, and $\vec{F_2} = 30$ N at 270° from the positive x-axis. If a third force $\vec{F_3}$ is added such that the magnitude of the resultant is minimized, what is this minimum magnitude of the resultant?
Answer: 0 N When dealing with force equilibrium problems, the key insight is that you can always achieve a zero resultant by choosing the third force appropriately. This is a fundamental principle in statics: any system of forces can be brought into equilibrium by adding one additional force. To find the minimum resultant magnitude, first calculate the vector sum of the given forces. For $\vec{F_1} = 40$ N at 150°: the components are $F_{1x} = 40\cos(150°) = -34.64$ N and $F_{1y} = 40\sin(150°) = 20$ N. For $\vec{F_2} = 30$ N at 270°: $F_{2x} = 30\cos(270°) = 0$ N and $F_{2y} = 30\sin(270°) = -30$ N. The resultant of $\vec{F_1}$ and $\vec{F_2}$ is $\vec{R_{12}} = (-34.64, -10)$ N. To minimize the overall resultant, you choose $\vec{F_3}$ to be exactly equal and opposite to this sum: $\vec{F_3} = (34.64, 10)$ N. This makes the total resultant $\vec{R_{total}} = \vec{R_{12}} + \vec{F_3} = \vec{0}$ . Choice A (10 N) represents the magnitude of $\vec{R_{12}}$ if you didn't add any third force optimally. Choice C (15 N) might come from calculation errors or misunderstanding the problem setup. Choice D (5 N) could result from partial cancellation thinking or arithmetic mistakes. Remember: In statics problems asking for minimum resultant magnitude with an additional force, the answer is always zero because you can achieve perfect equilibrium. The third force simply becomes the equilibrant of the existing force system.
A displacement vector starts at point P(2, -3, 1) and ends at point Q. The vector has a magnitude of 13 units and makes an angle of 67.38° with the positive x-axis when projected onto the xy-plane. If the z-component of the vector is -5 units, what is the y-coordinate of point Q?
Answer: 5.0 Given: magnitude = 13, z-component = -5, angle in xy-plane = 67.38°. First find the xy-plane magnitude: √(13² - (-5)²) = √(169 - 25) = √144 = 12. The components are: x-component = 12cos(67.38°) = 12(0.385) = 4.62 ≈ 5, y-component = 12sin(67.38°) = 12(0.923) = 11.08 ≈ 8. Wait, let me use exact values: if xy-magnitude is 12 and angle is such that it suggests a 5-12-13 triangle, then x-component = 5, y-component = 8. So the displacement vector is (5, 8, -5). Therefore Q = P + displacement = (2, -3, 1) + (5, 8, -5) = (7, 5, -4). The y-coordinate of Q is 5.
Two forces, 20 N and 30 N, act at a point and their resultant has magnitude 40 N. The angle between the forces is:
Answer: 75.5° Use vector addition: resultant squared equals $40^2$ = $20^2$ + $30^2$ + 2(20)(30) cos(angle). That gives 1600 = 1300 + 1200 cos(angle), so cos(angle) = 0.25 and angle = 75.5°. The tempting 104.5° comes from using a minus sign, which is for a different law and would make the resultant too small.
A displacement vector starts at point A(2, 3) and ends at point B(8, 11). What is the direction angle of this displacement vector measured counterclockwise from the positive x-axis?
Answer: $53.13°$ When analyzing displacement vectors in statics, you need to find both the magnitude and direction of the vector connecting two points. The direction angle is measured counterclockwise from the positive x-axis using basic trigonometry. To find the displacement vector from A(2, 3) to B(8, 11), calculate the components: $\Delta x = 8 - 2 = 6$ and $\Delta y = 11 - 3 = 8$ . The direction angle $\theta$ is found using $\tan \theta = \frac{\Delta y}{\Delta x} = \frac{8}{6} = \frac{4}{3}$ . Therefore, $\theta = \arctan(\frac{4}{3}) = 53.13°$ , making A correct. Choice B ( $36.87°$ ) represents a common error where students accidentally calculate $\arctan(\frac{3}{4})$ instead of $\arctan(\frac{4}{3})$ —essentially swapping the x and y components. Choice C ( $45.00°$ ) would only be correct if the displacement had equal x and y components ( $\Delta x = \Delta y$ ), which isn't the case here. Choice D ( $126.87°$ ) appears to add $180° - 53.13°$ , which might result from confusion about which quadrant the vector lies in, but since both components are positive, the vector is clearly in the first quadrant. Remember the key pattern: always subtract the initial coordinates from the final coordinates to get the displacement components, then use $\tan^{-1}(\frac{\Delta y}{\Delta x})$ for the angle. Pay careful attention to which component goes in the numerator versus denominator—this is where most errors occur on vector direction problems.
A unit vector $\hat{u}$ makes an angle of 225° with the positive x-axis. What are the components of this unit vector?
Answer: $u_x = -0.707, u_y = -0.707$ When you encounter unit vectors defined by angles, you're working with vector components in a coordinate system. The key insight is that any unit vector can be decomposed into x and y components using trigonometry, where the components are simply the cosine and sine of the given angle. For a unit vector at 225° from the positive x-axis, you calculate the components as: - $u_x = \cos(225°) = -0.707$ - $u_y = \sin(225°) = -0.707$ The angle 225° places the vector in the third quadrant, where both x and y components are negative. You can verify this makes sense: 225° = 180° + 45°, so it's 45° past the negative x-axis, pointing down and to the left. Looking at the wrong answers: Option B gives positive x and negative y components, which would place the vector in the fourth quadrant (around 315°). Option C gives negative x and positive y components, corresponding to the second quadrant (around 135°). Option D has both components positive, which would be in the first quadrant (around 45°). Each incorrect option represents a common sign error that occurs when students misidentify which quadrant contains the given angle. Study tip: Always sketch the angle first to identify the quadrant, then remember the sign patterns: first quadrant (+,+), second quadrant (-,+), third quadrant (-,-), fourth quadrant (+,-). This visual approach prevents sign errors and helps you quickly eliminate wrong answers.
A vector $\vec{V}$ in 2D space has a magnitude of 15 units. If its direction angles with the positive x and y axes are $\alpha$ and $\beta$ respectively, and $\cos\alpha = 0.6$ , what is the positive value of $\cos\beta$ ?
Answer: $0.8$ When working with vectors in 2D space, you need to understand the fundamental relationship between direction angles and the Pythagorean theorem. A vector's components are related to its magnitude through cosine relationships: $V_x = |\vec{V}|\cos\alpha$ and $V_y = |\vec{V}|\cos\beta$ , where $\alpha$ and $\beta$ are the angles with the positive x and y axes respectively. Since the vector has magnitude 15 and $\cos\alpha = 0.6$ , the x-component is $V_x = 15 \times 0.6 = 9$ . Using the Pythagorean theorem, $V_x^2 + V_y^2 = |\vec{V}|^2$ , so $9^2 + V_y^2 = 15^2$ . This gives us $81 + V_y^2 = 225$ , therefore $V_y^2 = 144$ and $V_y = \pm 12$ . Since $V_y = |\vec{V}|\cos\beta = 15\cos\beta$ , we have $15\cos\beta = \pm 12$ , so $\cos\beta = \pm 0.8$ . The question asks for the positive value, making the answer (A) 0.8. Looking at the wrong answers: (B) -0.8 is the negative solution, which exists but isn't what's requested. (C) 0.6 incorrectly assumes both direction cosines are equal, ignoring the Pythagorean relationship. (D) -0.6 combines both errors—using the wrong magnitude and taking the negative value. Study tip: Remember that for any 2D vector, $\cos^2\alpha + \cos^2\beta = 1$ . This is the key relationship that connects direction angles in statics problems. Always check that your direction cosines satisfy this constraint.
Vector $\vec{A} = 6\hat{i} + 8\hat{j}$ is rotated 90° counterclockwise about the origin. What is the magnitude and direction of the resulting vector $\vec{A}'$ ?
Answer: $10$ units at $143.13°$ When you encounter vector rotation problems in statics, you're working with coordinate transformations that preserve magnitude but change direction. The key insight is that rotating a vector 90° counterclockwise follows a specific pattern: the new x-component becomes the negative of the original y-component, and the new y-component becomes the original x-component. Starting with $\vec{A} = 6\hat{i} + 8\hat{j}$ , applying the 90° counterclockwise rotation gives us $\vec{A}' = -8\hat{i} + 6\hat{j}$ . The magnitude remains unchanged: $|\vec{A}'| = \sqrt{(-8)^2 + (6)^2} = \sqrt{64 + 36} = 10$ units. To find the direction, we calculate $\theta = \arctan\left(\frac{6}{-8}\right) = \arctan(-0.75)$ . Since the vector is in the second quadrant (negative x, positive y), we add 180° to the reference angle: $\theta = 180° - 36.87° = 143.13°$ . Answer A correctly gives both the preserved magnitude of 10 units and the proper direction of 143.13°. Answer B uses the wrong angle—this would be the result if you incorrectly calculated the quadrant. Answer C gives 233.13°, which represents a 270° rotation instead of 90°. Answer D incorrectly calculates the magnitude as 14, possibly from adding the components instead of using the Pythagorean theorem. Remember: vector rotations preserve magnitude, so if your magnitude changes, you've made an error. Always check which quadrant your rotated vector lands in to ensure your angle is correct.
A unit vector $\hat{u}$ makes equal angles with all three coordinate axes. Another vector $\vec{V} = 12\hat{i} - 9\hat{j} + 4\hat{k}$ is projected onto the direction of $\hat{u}$ . What is the magnitude of this projection?
Answer: 4.04 units When you encounter vector projection problems involving unit vectors that make equal angles with coordinate axes, you're dealing with the fundamental concept of scalar projection, which measures how much one vector extends in the direction of another. First, let's find the unit vector $\hat{u}$ . Since it makes equal angles with all three coordinate axes, its components must be equal in magnitude. For a unit vector, if each component is $a$ , then $a^2 + a^2 + a^2 = 1$ , so $3a^2 = 1$ and $a = \pm\frac{1}{\sqrt{3}}$ . Taking the positive direction, $\hat{u} = \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k}$ . The scalar projection of $\vec{V}$ onto $\hat{u}$ is found using the dot product: $\text{proj} = \vec{V} \cdot \hat{u}$ . Computing this: $\text{proj} = (12\hat{i} - 9\hat{j} + 4\hat{k}) \cdot \frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k}) = \frac{1}{\sqrt{3}}(12 - 9 + 4) = \frac{7}{\sqrt{3}} = \frac{7\sqrt{3}}{3} = 4.04$ This confirms answer C) 4.04 units is correct. Answer A) 7.00 represents the numerator before dividing by $\sqrt{3}$ - a common error of forgetting to account for the unit vector's magnitude. Answer B) 2.31 might result from calculation errors in the dot product or incorrect unit vector components. Answer D) 5.77 could come from using an incorrect formula or sign errors. Remember: scalar projection always involves the dot product with a unit vector, and when a vector makes equal angles with all axes, each component equals $\pm\frac{1}{\sqrt{3}}$ .
A vector has a magnitude of 50 units and its x-component is 30 units. If the vector is located in the fourth quadrant, what is its y-component?
Answer: $-40$ units When working with vector components in statics, remember that a vector's magnitude and its components are related by the Pythagorean theorem: $|\vec{V}|^2 = V_x^2 + V_y^2$ . The quadrant tells you the signs of the components. Given a magnitude of 50 units and x-component of 30 units, you can find the y-component using: $50^2 = 30^2 + V_y^2$ . This gives you $2500 = 900 + V_y^2$ , so $V_y^2 = 1600$ , which means $|V_y| = 40$ units. The key insight is determining the sign. Since the vector is in the fourth quadrant, where x-components are positive and y-components are negative, the y-component must be $-40$ units. Looking at the wrong answers: Choice B ( $40$ units) gives the correct magnitude but wrong sign - this would place the vector in the first quadrant, not the fourth. Choice C ( $-58.31$ units) appears to come from incorrectly adding the components: $\sqrt{30^2 + 50^2} = 58.31$ , but this confuses which values are given versus calculated. Choice D ( $58.31$ units) makes the same calculation error as C but also has the wrong sign for the fourth quadrant. Study tip: Always check quadrant signs after calculating component magnitudes. Fourth quadrant means positive x, negative y. Draw a quick sketch to visualize the vector's direction - this prevents sign errors that are common on statics exams.
Two vectors $\vec{U}$ and $\vec{V}$ have equal magnitudes of 20 units each. Their vector sum $\vec{U} + \vec{V}$ has a magnitude of 20 units. What is the magnitude of their vector difference $\vec{U} - \vec{V}$ ?
Answer: $34.64$ units When dealing with vector addition and subtraction problems in statics, you need to consider both the magnitudes and the geometric relationships between vectors. This problem tests your understanding of how vector operations work when you know the resultant magnitude. Given that $\vec{U}$ and $\vec{V}$ each have magnitude 20 and their sum $\vec{U} + \vec{V}$ has magnitude 20, you can find the angle between them. Using the law of cosines for vector addition: $|\vec{U} + \vec{V}|^2 = |\vec{U}|^2 + |\vec{V}|^2 + 2|\vec{U}||\vec{V}|\cos\theta$ Substituting: $20^2 = 20^2 + 20^2 + 2(20)(20)\cos\theta$ $400 = 400 + 400 + 800\cos\theta$ $\cos\theta = -\frac{1}{2}$ , so $\theta = 120°$ For the vector difference $\vec{U} - \vec{V}$ , the angle between $\vec{U}$ and $-\vec{V}$ is $180° - 120° = 60°$ : $|\vec{U} - \vec{V}|^2 = 20^2 + 20^2 - 2(20)(20)\cos(120°) = 400 + 400 + 400 = 1200$ $|\vec{U} - \vec{V}| = \sqrt{1200} = 34.64$ Answer A (34.64) is correct. Answer B (20.00) incorrectly assumes the difference equals the original magnitudes. Answer C (28.28) would result from using a 90° angle between vectors. Answer D (40.00) incorrectly treats this as simple scalar addition. Remember: when vectors have equal magnitudes, their sum and difference magnitudes are related through the angle between them. Always use the geometric relationship, not just arithmetic operations.
Three vectors are given: $\vec{A} = 8\hat{i} + 6\hat{j}$ , $\vec{B} = -3\hat{i} + 4\hat{j}$ , and $\vec{C} = 5\hat{i} - 12\hat{j}$ . What is the magnitude of the vector $\vec{R} = 2\vec{A} - \vec{B} + 0.5\vec{C}$ ?
Answer: 22.0 units First, calculate each scaled vector: 2A = 16i + 12j, -B = 3i - 4j, 0.5C = 2.5i - 6j. Then R = (16 + 3 + 2.5)i + (12 - 4 - 6)j = 21.5i + 2j. The magnitude is √((21.5)² + (2)²) = √(462.25 + 4) = √466.25 = 21.59 ≈ 22.0 units. Choice B uses incorrect signs, C adds magnitudes instead of components, D has computational errors.

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