Vector Operations - Statics and Dynamics

1. At a point, $F_1 = 3i + 2j$ N, $F_2 = -i + 4j$ N. Find $F_3$ for equilibrium.
   Answer: $-2i - 6j$ N Add the two known forces: 3i+2j plus -i+4j gives 2i+6j. For equilibrium the third force must cancel that total, so F3 is the negative: -2i - 6j. The tempting mistake is to give 2i + 6j, which is just the resultant of F1 and F2, not the balancing force.
2. Vectors $\mathbf{P} = 4\hat{i} + p\hat{j}$ and $\mathbf{Q} = q\hat{i} - 3\hat{j}$ are given, where $p$ and $q$ are nonzero scalars. It is known that $\mathbf{P}$ and $\mathbf{Q}$ are anti-parallel (parallel but in opposite directions). Which relationship between $p$ and $q$ must hold, and what is $\mathbf{P} \times \mathbf{Q}$?
   Answer: Anti-parallelism requires $\frac{4}{q}=\frac{p}{-3}$, i.e., $pq=-12$; under this condition $\mathbf{P}\times\mathbf{Q}=\mathbf{0}$ because anti-parallel vectors have zero cross product. When two vectors are parallel or anti-parallel, their directions are collinear — one is a scalar multiple of the other. For anti-parallel vectors specifically, that scalar is negative, meaning $\mathbf{P} = k\mathbf{Q}$ where $k < 0$. This is the condition to apply, not perpendicularity. Setting $\mathbf{P} = k\mathbf{Q}$ gives $4\hat{i} + p\hat{j} = k(q\hat{i} - 3\hat{j})$, so $4 = kq$ and $p = -3k$. Eliminating $k$: $k = 4/q$, so $p = -3(4/q)$, giving $pq = -12$. This is also written as $\frac{4}{q} = \frac{p}{-3}$ — matching the ratio condition in answer B. Now, for any two parallel (or anti-parallel) vectors, the angle between them is $180°$, and $\sin 180° = 0$, so $\mathbf{P} \times \mathbf{Q} = \mathbf{0}$. Answer B is correct. Answer A confuses anti-parallelism with perpendicularity — $\mathbf{P} \cdot \mathbf{Q} = 0$ is the condition for orthogonal vectors, not collinear ones. Answer C correctly derives $pq = -12$ and correctly states the cross product is zero, but then makes a serious error by claiming anti-parallel vectors are "a special case of perpendicularity" and that $\mathbf{P} \cdot \mathbf{Q} = 0$. Anti-parallel vectors actually have $\mathbf{P} \cdot \mathbf{Q} = -|\mathbf{P}||\mathbf{Q}|$, the most negative possible dot product. Answer D has the right scalar condition but wrong trigonometry — $\sin 180° = 0$, not $1$ (that would be $\sin 90°$). A useful memory anchor: cross product uses sine, dot product uses cosine. Parallel/anti-parallel → $\sin\theta = 0$ → zero cross product. Perpendicular → $\cos\theta = 0$ → zero dot product. Keep these straight and this question type becomes straightforward.
3. Two unit vectors $\hat{u}$ and $\hat{v}$ satisfy $\hat{u} \cdot \hat{v} = 0.6$. A student needs the magnitude of $\hat{u} \times \hat{v}$. The student reasons: *'Since $\hat{u} \cdot \hat{v} = |\hat{u}||\hat{v}|\cos\theta = \cos\theta = 0.6$, then $\sin\theta = \sqrt{1-0.6^2} = 0.8$, so $|\hat{u}\times\hat{v}| = 0.8$.'* Which statement about this reasoning is most accurate?
   Answer: The reasoning is correct and the result $|\hat{u}\times\hat{v}| = 0.8$ is valid; the identity $\sin^2\theta + \cos^2\theta = 1$ applied to unit vectors guarantees that $|\hat{u}\times\hat{v}|^2 + (\hat{u}\cdot\hat{v})^2 = 1$, which is satisfied here since $0.64+0.36=1.00$. When you see a question combining dot products and cross products of unit vectors, think about the fundamental Pythagorean identity that connects them. For any two unit vectors, $|\hat{u} \times \hat{v}|^2 + (\hat{u} \cdot \hat{v})^2 = \sin^2\theta + \cos^2\theta = 1$. This identity is your anchor. The student's reasoning in B is completely sound. Since $\hat{u} \cdot \hat{v} = \cos\theta = 0.6$, we get $\cos^2\theta = 0.36$, so $\sin^2\theta = 0.64$, giving $|\hat{u} \times \hat{v}| = |\sin\theta| = 0.8$. Crucially, the cross-product magnitude is always non-negative by definition, so we take the positive root — and since $\cos\theta = 0.6 > 0$, the angle $\theta$ lies in the first quadrant, confirming $\sin\theta = 0.8$ directly. The verification $0.64 + 0.36 = 1.00$ closes the argument cleanly. A is wrong because the angle between two vectors is always defined on $[0°, 180°]$ — you never encounter negative sines in this context. There's no quadrant ambiguity to resolve. C is wrong because the formula $|\hat{u} \times \hat{v}| = |\hat{u}||\hat{v}|\sin\theta$ is the standard 3D definition, not a 2D restriction. Perpendicular components don't add any hidden complexity here. D is a clever distractor — it raises a valid point about absolute value, but since magnitude is inherently non-negative and $\theta \in [0°, 180°]$ ensures $\sin\theta \geq 0$, the distinction is moot. The student's answer is fully correct as stated, making D's critique a non-issue. Study tip: Memorize $|\hat{u} \times \hat{v}|^2 + (\hat{u} \cdot \hat{v})^2 = 1$ for unit vectors — it lets you convert between dot and cross product information instantly.
4. Forces $\mathbf{F}_1 = 6\hat{i} - 2\hat{j} + 3\hat{k}$ N and $\mathbf{F}_2 = -6\hat{i} + 2\hat{j} - 3\hat{k}$ N act on a particle. A student is asked to find the resultant force and its magnitude. The student correctly computes $\mathbf{F}_R = \mathbf{0}$ but then states: *'Because $\mathbf{F}_R = \mathbf{0}$, the two forces are perpendicular to each other.'* Which response most precisely identifies the error?
   Answer: The student's conclusion is incorrect. $\mathbf{F}_R = \mathbf{0}$ implies the forces are equal and opposite (anti-parallel), not perpendicular. Perpendicular forces satisfy $\mathbf{F}_1 \cdot \mathbf{F}_2 = 0$, but here $\mathbf{F}_1 \cdot \mathbf{F}_2 = -36-4-9 = -49 \neq 0$, confirming the forces are anti-parallel with $\mathbf{F}_2 = -\mathbf{F}_1$. When two forces act on a particle, a zero resultant tells you about balance, not orientation. The key insight is that $\mathbf{F}_R = \mathbf{0}$ means $\mathbf{F}_1 + \mathbf{F}_2 = \mathbf{0}$, which directly implies $\mathbf{F}_2 = -\mathbf{F}_1$. That algebraic relationship has a precise geometric meaning: the vectors point in exactly opposite directions, making them anti-parallel with $\theta = 180°$. This is confirmed by the dot product: $\mathbf{F}_1 \cdot \mathbf{F}_2 = (6)(-6)+(-2)(2)+(3)(-3) = -36-4-9 = -49$ N². Using the formula $\mathbf{F}_1 \cdot \mathbf{F}_2 = |\mathbf{F}_1||\mathbf{F}_2|\cos\theta = (7)(7)\cos180° = -49$ N², everything is consistent. Answer A captures this precisely and completely — the student's error is a specific conceptual mix-up, and A names it correctly while providing the mathematical evidence. Answer B is wrong because it introduces a false premise: equal magnitudes do not imply anything about perpendicularity, and stating the forces are "not in equilibrium" contradicts the correctly computed $\mathbf{F}_R = \mathbf{0}$. Answer C is wrong because it overstates ambiguity. Once $\mathbf{F}_R = \mathbf{0}$, you know exactly that $\mathbf{F}_2 = -\mathbf{F}_1$ — the angle is unambiguously $180°$, not "any angle." Answer D is wrong because, while it correctly identifies anti-parallelism, it introduces an irrelevant and misleading claim about coplanarity and indeterminate direction, which muddies an otherwise correct conclusion. Study tip: Always match your conclusion to what the math actually encodes. Zero resultant → anti-parallel. Zero dot product → perpendicular. Don't conflate equilibrium conditions with geometric angle conditions.
5. Resultant of $F_1$ and $F_2$ is $6i + 2j$ N. $F_1 = 4i - j$ N. Find $F_2$.
   Answer: $2i + 3j$ N Since the resultant is F1 + F2, subtract F1 from it: (6i + 2j) - (4i - j) = 2i + 3j N. Adding F1 and F2 instead of subtracting gives 10i + j N, but that ignores that F2 is the difference between the resultant and F1.
6. A 21 N force acts in the direction $2i - 3j + 6k$. What is the force vector?
   Answer: $6i - 9j + 18k$ N The direction vector has magnitude sqrt(4 + 9 + 36) = 7, so the unit vector is (2/7)i - (3/7)j + (6/7)k. Multiplying by 21 N gives 6i - 9j + 18k N. A common error is multiplying 21 by the raw components to get 42i - 63j + 126k N, but that skips normalizing by the direction's magnitude 7.
7. Compute the moment $r \times F$ about the origin for $r = 3j + k$ m and $F = i - 2j + k$ N.
   Answer: $5i + j - 3k$ N·m Compute r x F component by component: i = 3(1) - 1(-2) = 5, j = 1(1) - 0(1) = 1, k = 0(-2) - 3(1) = -3, giving 5i + j - 3k N·m. A common trap is reversing the cross product order or mishandling the j-term sign, which yields -5i - j + 3k.
8. Using vector operations on $\mathbf{v}$ and $\mathbf{a}$, what are the correct answers to both questions?
   Answer: The particle moves at constant speed because $\mathbf{v}\cdot\mathbf{a} = 0$; the radius of curvature is $\rho = |\mathbf{v}|^3/|\mathbf{v}\times\mathbf{a}|$. Computing $\mathbf{v}\times\mathbf{a} = [(3)(3)-(4)(-4)]\hat{k} = 25\hat{k}$ m²/s³, and $|\mathbf{v}|=5$ m/s, so $\rho = 5^3/25 = 125/25 = 5$ m. When analyzing curvilinear motion, your two key tools are the dot product and cross product of velocity and acceleration. Each tells you something different: the dot product reveals whether speed is changing, while the cross product helps you find the radius of curvature. The dot product $\mathbf{v}\cdot\mathbf{a}$ equals the rate of change of speed ($\dot{v}$). Here, $\mathbf{v}\cdot\mathbf{a} = (3)(-4)+(4)(3) = -12+12 = 0$, which means the tangential acceleration is zero — speed is neither increasing nor decreasing. The particle moves at constant speed, curving through space with all acceleration directed normal to the path. For the radius of curvature, recall that the normal acceleration satisfies $a_n = v^2/\rho$, which rearranges to $\rho = v^3/|\mathbf{v}\times\mathbf{a}|$ (since $|\mathbf{v}\times\mathbf{a}| = v\cdot a_n$). The cross product gives $\mathbf{v}\times\mathbf{a} = [(3)(3)-(4)(-4)]\hat{k} = 25\hat{k}$ m²/s³, and with $|\mathbf{v}|=5$ m/s, you get $\rho = 5^3/25 = 5$ m. That's answer D. A makes a conceptual error: a zero dot product means zero tangential acceleration, which means constant speed — not speeding up. It also uses the wrong formula $|\mathbf{v}|^2/|\mathbf{a}|$, which only works when acceleration is purely normal. B gets the physics right but never explicitly confirms $|\mathbf{v}|=5$ m/s before substituting — it's incomplete and loses a step that could cause arithmetic errors. C misinterprets $\mathbf{v}\cdot\mathbf{a}=0$; perpendicular vectors don't oppose each other, so "slowing down" is wrong. Your go-to rule: dot product → speeding up/slowing down; cross product magnitude → radius of curvature. Memorize both formulas together.
9. Vectors $\mathbf{A} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\mathbf{B} = -\hat{i} + 4\hat{j} + 2\hat{k}$ represent two edge vectors of a parallelogram. A student claims: *'The area of the parallelogram equals $|\mathbf{A} \cdot \mathbf{B}|$, which is $|{-2+12-2}| = 8$ square units.'* Which of the following best evaluates this claim?
   Answer: The claim is incorrect; the dot product yields a scalar related to projected length, not area. The correct area is $|\mathbf{A} \times \mathbf{B}| = \sqrt{10^2+(-3)^2+11^2} = \sqrt{230} \approx 15.2$ square units. Whenever you see a question about the area of a parallelogram formed by two vectors, your instinct should immediately go to the cross product, not the dot product. These two operations serve fundamentally different geometric purposes, and confusing them is one of the most common errors in vector mechanics. The area of a parallelogram with edge vectors A and B equals $|\mathbf{A} \times \mathbf{B}|$, because the cross product's magnitude equals $|\mathbf{A}||\mathbf{B}|\sin\theta$, where $\theta$ is the angle between the vectors — and $\frac{1}{2}|\mathbf{A}||\mathbf{B}|\sin\theta$ is the triangle area formula you already know. Computing the cross product here gives $\mathbf{A} \times \mathbf{B} = (3\cdot2 - (-1)\cdot4)\hat{i} - (2\cdot2 - (-1)(-1))\hat{j} + (2\cdot4 - 3\cdot(-1))\hat{k} = 10\hat{i} - 3\hat{j} + 11\hat{k}$, so the area is $\sqrt{100 + 9 + 121} = \sqrt{230} \approx 15.2$ square units. This confirms B is correct. A is wrong because the dot product never gives area — regardless of whether vectors share a vertex. C is wrong because 8 does not equal $\sqrt{230}$; there is no numerical accident here, just a false claim. D contains a real truth (dot product = 0 implies perpendicularity, and $\sin 90° = 1$ simplifies area), but it misrepresents the dot product as an area formula under any condition — it never is. Your study tip: tattoo this distinction in your memory — dot product → scalar projection (uses cosine); cross product → area (uses sine). When you see "area of parallelogram," always reach for $|\mathbf{A} \times \mathbf{B}|$.
10. If $|\mathbf{F}| = 100$ N is fixed, which orientation of $\mathbf{F}$ maximizes the $\hat{k}$ component of $\mathbf{M} = \mathbf{r}\times\mathbf{F}$, and what is that maximum value?
    Answer: $\mathbf{F}$ should be directed perpendicular to $\mathbf{r}$ within the $xy$-plane, yielding $M_{k,\max} = |\mathbf{r}||\mathbf{F}|\sin90° = 100\sqrt{0.09+0.01} = 100\sqrt{0.10} \approx 31.6$ N·m. When a question asks you to maximize a specific component of a torque, your first move should be to write out that component explicitly using the cross product formula, then treat it as an optimization problem. The $\hat{k}$ component of $\mathbf{M} = \mathbf{r} \times \mathbf{F}$ is $M_k = r_x F_y - r_y F_x$. Notice that $F_z$ vanishes entirely from this expression — the out-of-plane force component contributes nothing to the bolt's rotation. So to maximize $M_k$, you should concentrate all 100 N into the $xy$-plane, setting $F_z = 0$. Once you're working purely in 2D, $M_k = |\mathbf{r}||\mathbf{F}_{xy}|\sin\theta$, which is maximized when $\theta = 90°$ — meaning $\mathbf{F}$ is perpendicular to $\mathbf{r}$ within the $xy$-plane. This gives $M_{k,\max} = \sqrt{0.3^2 + 0.1^2} \times 100 = 100\sqrt{0.10} \approx 31.6$ N·m, confirming C. Answer A and D both direct the force along $+\hat{k}$. Since $F_z$ drops out of $M_k$ entirely, this wastes all 100 N on a component that produces zero turning moment about the bolt axis — a critical conceptual error. Answer B points $\mathbf{F}$ along $\mathbf{r}$, which makes $\sin\theta = 0$ and gives $M_k = 0$. A force directed straight down the wrench handle tightens nothing. Note that A, B, and D all state the same numerical result (≈ 31.6 N·m) but for wrong reasons — the exam is testing whether you verify the direction, not just the formula output. Your takeaway: always expand the cross product and identify which force components actually appear in the target moment expression before optimizing.
11. What is the work done by $F = 4i + 3j$ N over $d = 2i - 6j$ m?
    Answer: -10 J Multiply matching components: 4 times 2 is 8, and 3 times -6 is -18. Add them: 8 + (-18) = -10 J. The tempting positive 10 J comes from ignoring the negative sign on the j component, but work is a dot product and that sign matters.
12. What is the scalar moment of $\mathbf{F}$ about the given axis, computed as $M_\lambda = \hat{\lambda}\cdot(\mathbf{r}\times\mathbf{F})$?
    Answer: $M_\lambda = -22$ N·m, computed by evaluating $\mathbf{r}\times\mathbf{F} = -12\hat{i}+6\hat{j}-18\hat{k}$ N·m and dotting with $\hat{\lambda}=\frac{1}{3}(2\hat{i}-\hat{j}+2\hat{k})$, giving $\frac{1}{3}(-24-6-36)=-22$ N·m. When a question asks for the scalar moment of a force about an axis, your go-to formula is the scalar triple product: $M_\lambda = \hat{\lambda} \cdot (\mathbf{r} \times \mathbf{F})$. The order of operations matters — cross product first, then dot product — and the cross product is not commutative. Start by computing $\mathbf{r} \times \mathbf{F}$ with $\mathbf{r} = \hat{i} + 2\hat{j} + 0\hat{k}$ and $\mathbf{F} = 9\hat{i} + 0\hat{j} - 6\hat{k}$: $\mathbf{r} \times \mathbf{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 0 \\ 9 & 0 & -6 \end{vmatrix} = (2(-6)-0(0))\hat{i} - (1(-6)-0(9))\hat{j} + (1(0)-2(9))\hat{k} = -12\hat{i} + 6\hat{j} - 18\hat{k}$ Now dot with $\hat{\lambda} = \frac{1}{3}(2\hat{i} - \hat{j} + 2\hat{k})$: $M_\lambda = \frac{1}{3}[(-12)(2) + (6)(-1) + (-18)(2)] = \frac{1}{3}(-24 - 6 - 36) = \frac{-66}{3} = -22 \text{ N}\cdot\text{m}$ This confirms C is correct. Choice B commits a critical conceptual error: $\hat{\lambda} \cdot \mathbf{r} = 0$ simply means $\mathbf{r}$ is perpendicular to the axis direction, which has no bearing on whether the moment is zero — you must dot $\hat{\lambda}$ with the cross product, not with $\mathbf{r}$ alone. Choice A uses an invalid patchwork method with no physical basis. Choice D reverses the cross-product order to $\mathbf{F} \times \mathbf{r}$, which flips all signs and gives $+22$ N·m — the magnitude is right but the sign is wrong. Study tip: Always respect the formula order $\hat{\lambda} \cdot (\mathbf{r} \times \mathbf{F})$. A sign error from reversing the cross product is one of the most common traps on statics exams, and the dot product of $\hat{\lambda}$ with $\mathbf{r}$ alone is meaningless for computing a moment about an axis.

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