Statics and Dynamics Quiz: Vector Operations

What this quiz covers

This quiz focuses on Vector Operations, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics and Dynamics.

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.

Question 1

1. The reasoning is flawed because $\sin\theta$ could be $-0.8$ if $\theta > 180°$, making the magnitude ambiguous without knowing the quadrant; the correct answer requires additional information about the orientation of $\hat{u}$ relative to $\hat{v}$.
2. 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 $0.64+0.36=1.00$. (correct answer)
3. The reasoning is flawed because the cross-product magnitude formula $|\hat{u}\times\hat{v}|=\sin\theta$ holds only in 2D; in 3D the result depends on the component of each vector perpendicular to the plane and cannot be determined from the dot product alone.
4. The reasoning is partially correct but incomplete; $\sin\theta = 0.8$ only if $0 \leq \theta \leq 90°$. Since $\cos\theta = 0.6 > 0$, $\theta$ is indeed in the first quadrant, so $\sin\theta = +0.8$. However, $|\hat{u}\times\hat{v}| = |\sin\theta|$, not $\sin\theta$, so the magnitude is 0.8 only after taking the absolute value, which the student omitted explicitly.

Question 2

1. Anti-parallelism requires $\mathbf{P}\cdot\mathbf{Q}=0$, so $4q-3p=0$, giving $q=3p/4$; under this condition $|\mathbf{P}\times\mathbf{Q}|=|\mathbf{P}||\mathbf{Q}|\sin90°=|\mathbf{P}||\mathbf{Q}|$.
2. 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. (correct answer)
3. Anti-parallelism requires $pq=-12$; under this condition $|\mathbf{P}\times\mathbf{Q}|=|\mathbf{P}||\mathbf{Q}|\sin180°=0$, but the dot product $\mathbf{P}\cdot\mathbf{Q}$ is also zero since anti-parallel vectors are a special case of perpendicularity.
4. Anti-parallelism requires $\frac{4}{q}=\frac{p}{-3}$, i.e., $pq=-12$; under this condition $|\mathbf{P}\times\mathbf{Q}|=|\mathbf{P}||\mathbf{Q}|$ since anti-parallel vectors subtend $180°$ and $\sin180°=1$.

Question 3

1. The particle is speeding up because $\mathbf{v}\cdot\mathbf{a} = (3)(-4)+(4)(3) = 0$, meaning the acceleration is entirely centripetal, which increases speed; the radius of curvature is $\rho = |\mathbf{v}|^2/|\mathbf{a}| = 25/5 = 5$ m.
2. 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, so $\rho = 125/25 = 5$ m.
3. The particle is slowing down because $\mathbf{v}\cdot\mathbf{a} = 0$ means the acceleration opposes the velocity; the radius of curvature is $\rho = |\mathbf{v}|^3/|\mathbf{v}\times\mathbf{a}| = 125/25 = 5$ m.
4. 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. (correct answer)

Question 4

1. -10 J (correct answer)
2. 10 J
3. -18 J
4. 8 J

Question 5

1. $2i - 3j$ N
2. $10i + j$ N
3. $2i + 3j$ N (correct answer)
4. $10i - j$ N

Question 6

1. $2i + 6j$ N
2. $-2i - 6j$ N (correct answer)
3. $-2i + 6j$ N
4. $2i - 6j$ N

Question 7

1. $5i - j - 3k$ N·m
2. $-5i - j + 3k$ N·m
3. $5i + j - 3k$ N·m (correct answer)
4. $5i + j + 3k$ N·m

Question 8

1. $-6i + 9j - 18k$ N
2. $42i - 63j + 126k$ N
3. $6i + 9j - 18k$ N
4. $6i - 9j + 18k$ N (correct answer)

Question 9

1. The claim is correct; the dot product magnitude gives the parallelogram area when both vectors originate from the same vertex.
2. 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. (correct answer)
3. The claim is incorrect in reasoning but accidentally correct numerically; dividing and multiplying by $|\mathbf{A}||\mathbf{B}|$ recovers $|\mathbf{A}\cdot\mathbf{B}|=8$, which happens to equal the true area in this special case.
4. The claim is partially correct; the dot product gives area only when $\mathbf{A}$ and $\mathbf{B}$ are perpendicular. Since $\mathbf{A}\cdot\mathbf{B}=8\neq0$, they are not perpendicular, and the correct area is $|\mathbf{A}\times\mathbf{B}| = \sqrt{230} \approx 15.2$ square units.

Question 10

1. $\mathbf{F}$ should be directed along $+\hat{k}$, giving $M_{k,\max} = |\mathbf{r}||\mathbf{F}| = 100\sqrt{0.10} \approx 31.6$ N·m, since the axial force acts directly about the bolt axis.
2. $\mathbf{F}$ should be directed along $\mathbf{r}$, yielding $M_{k,\max} = |\mathbf{r}||\mathbf{F}| = 100\sqrt{0.10} \approx 31.6$ N·m, since a force along the handle arm creates the largest moment arm.
3. $\mathbf{F}$ 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. (correct answer)
4. $\mathbf{F}$ should be directed along $+\hat{k}$, giving $M_{k,\max} = |\mathbf{r}\times F\hat{k}|= F|\mathbf{r}|= 100\sqrt{0.10}\approx31.6$ N·m, since the cross product of $\mathbf{r}$ (in the $xy$-plane) with $\hat{k}$ lies in the $xy$-plane and its magnitude equals $|\mathbf{r}|$.

Question 11

1. $M_\lambda = -6$ N·m, computed by projecting $\mathbf{r}$ onto $\hat{\lambda}$ first to get the moment arm, then multiplying by $|\mathbf{F}|$: $\hat{\lambda}\cdot\mathbf{r} = \frac{1}{3}(2-2+0)=0$ m, then $M_\lambda = 0 \times |\mathbf{F}| = 0$, contradicting the result, so the student instead uses only the $\hat{k}$ components: $(2/3)(-6)=-4$ N·m, adjusted by $r_y=2$ to give $-6$ N·m.
2. $M_\lambda = 0$ N·m, computed by noting that $\hat{\lambda}\cdot\mathbf{r} = \frac{1}{3}(2-2+0)=0$, which means $\mathbf{r}$ is perpendicular to the axis, so the force produces no scalar moment about the axis.
3. $M_\lambda = -22$ 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. (correct answer)
4. $M_\lambda = +22$ N·m, computed by reversing the cross-product order to $\mathbf{F}\times\mathbf{r}$ instead of $\mathbf{r}\times\mathbf{F}$, yielding $+12\hat{i}-6\hat{j}+18\hat{k}$ N·m, then dotting with $\hat{\lambda}$: $\frac{1}{3}(24+6+36)=+22$ N·m.

Question 12

1. 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 $\mathbf{F}_2 = -\mathbf{F}_1$. (correct answer)
2. The student's conclusion is incorrect. $\mathbf{F}_R = \mathbf{0}$ implies perpendicularity only if the magnitudes are equal, and since $|\mathbf{F}_1|=|\mathbf{F}_2|=7$ N, the student should confirm perpendicularity via $\mathbf{F}_1\cdot\mathbf{F}_2=0$; here $\mathbf{F}_1\cdot\mathbf{F}_2=-49\neq0$, so the forces are neither perpendicular nor in equilibrium.
3. The student's conclusion is incorrect. Zero resultant is a necessary but not sufficient condition for perpendicularity; the forces could be anti-parallel, perpendicular, or at any angle as long as they cancel, so additional information is needed before any conclusion about orientation can be drawn.
4. The student's conclusion is incorrect. $\mathbf{F}_R = \mathbf{0}$ means $\mathbf{F}_1 = -\mathbf{F}_2$, so the vectors are anti-parallel ($\theta=180°$). The dot product $\mathbf{F}_1\cdot\mathbf{F}_2=|\mathbf{F}_1||\mathbf{F}_2|\cos180°=-49$ N² confirms this; however, since $|\mathbf{F}_1\times\mathbf{F}_2|=0$ for anti-parallel vectors, the student could correctly conclude the forces are coplanar with any shared plane, making the direction indeterminate.