Answer: $60°$

Compute the dot product: $\mathbf{a} \cdot \mathbf{b} = (1)(0) + (0)(1) + (1)(1) = 1$. Compute the magnitudes: $|\mathbf{a}| = \sqrt{1+0+1} = \sqrt{2}$ and $|\mathbf{b}| = \sqrt{0+1+1} = \sqrt{2}$. Find the angle: $\cos\theta = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$, so $\theta = 60°$. Why distractors fail: Option A ($30°$) has $\cos 30° = \frac{\sqrt{3}}{2} \neq \frac{1}{2}$. Option B ($45°$) has $\cos 45° = \frac{\sqrt{2}}{2}$. Option D ($90°$) requires $\cos\theta = 0$.

Answer: The area of the parallelogram spanned by $\mathbf{a}$ and $\mathbf{b}$

Geometric meaning of the cross product magnitude: $|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta$, which equals the area of the parallelogram formed by $\mathbf{a}$ and $\mathbf{b}$. Why distractors fail: Option A involves cosine, which relates to the dot product. Option B describes a volume (which involves a triple product with three vectors). Option D is the scalar projection $\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|}$, related to the dot product.

Answer: The angle between the vectors is obtuse (greater than $90°$).

Interpret the sign of the dot product: Since magnitudes $|\mathbf{a}|$ and $|\mathbf{b}|$ are always positive, a negative dot product means $\cos\theta < 0$, which occurs when $90° < \theta \leq 180°$. Why distractors fail: Option A (perpendicular) gives a dot product of exactly zero, not negative. Option C is impossible since magnitudes are always non-negative. Option D (same direction) gives a positive dot product.

Answer: The claim is incorrect because $\mathbf{b} = -3\mathbf{a}$ — the vectors are antiparallel, not orthogonal.

Check whether b is a scalar multiple of a: $-3\mathbf{a} = -3\langle 1, -2, 2 \rangle = \langle -3, 6, -6 \rangle = \mathbf{b}$. So $\mathbf{b} = -3\mathbf{a}$, meaning the vectors are antiparallel (parallel but pointing in opposite directions). Compute the dot product to confirm: $\mathbf{a} \cdot \mathbf{b} = (1)(-3) + (-2)(6) + (2)(-6) = -3 - 12 - 12 = -27 \neq 0$. Since the dot product is nonzero, the vectors are not orthogonal. Why distractors fail: Option A is wrong because the dot product is $-27$, not $0$. Option B gives the wrong dot product value ($-21$ instead of $-27$). Option D: while it is true that the cross product is $\mathbf{0}$ (since the vectors are parallel), this does not support the claim of orthogonality — a zero cross product indicates parallelism, the opposite of orthogonality.

Answer: Because area depends on $\sin\theta$ (the perpendicular component), which is captured by the cross product magnitude, not the dot product.

Relate area to perpendicular height: The area of a parallelogram is base $\times$ height $= |\mathbf{a}| \cdot |\mathbf{b}|\sin\theta$. The cross product magnitude equals $|\mathbf{a}||\mathbf{b}|\sin\theta$, making it the correct tool. Why distractors fail: Option A is incorrect; the dot product does not give a triangle area. Option C is wrong; the dot product is always defined for any two vectors of the same dimension. Option D confuses the cross product (a vector) with its magnitude (a scalar).

Answer: A scalar value

Recall the nature of the dot product: The dot product $\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3$ sums the products of corresponding components, yielding a single number (scalar). Why distractors fail: Option A describes the cross product, not the dot product. Options C and D describe vectors, but the dot product always returns a scalar, not a vector.

Answer: Compute $\mathbf{a} \cdot \mathbf{b}$ for the angle (via $\cos\theta$), and $|\mathbf{a} \times \mathbf{b}|$ for the area.

Match each product to the right application: The dot product gives $\cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$, which is used to find the angle. The cross product magnitude $|\mathbf{a}\times\mathbf{b}|$ gives the parallelogram area. Why distractors fail: Option A reverses the roles. Option C is incorrect because the direction of the cross product indicates the normal to the plane, not the angle. Option D: the dot product alone cannot give the area.

Answer: They are perpendicular (orthogonal).

Recall the orthogonality condition: Since $\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}||\mathbf{v}|\cos\theta$ and both vectors are nonzero, $\mathbf{u} \cdot \mathbf{v} = 0$ implies $\cos\theta = 0$, so $\theta = 90°$. Why distractors fail: Option A (parallel) would require $\cos\theta = \pm 1$. Option B (opposite directions) gives $\cos\theta = -1$. Option D (equal magnitudes) is unrelated to the dot product being zero.

Answer: The claim is incorrect because scalars multiply: $(2\mathbf{a}) \times (3\mathbf{b}) = 6(\mathbf{a} \times \mathbf{b})$.

Apply scalar factoring rules: The cross product is bilinear, so $(c\mathbf{a}) \times (d\mathbf{b}) = cd(\mathbf{a} \times \mathbf{b})$. Here, $2 \times 3 = 6$, not $2 + 3 = 5$. Why distractors fail: Option A incorrectly claims scalars add. Option C is wrong; you can factor scalars out—they multiply. Option D incorrectly validates the claim.

Answer: $3\sqrt{3}$

Find the angle: $\cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|} = \frac{3}{6} = \frac{1}{2}$, so $\theta = 60°$. Compute the cross product magnitude: $|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin 60° = 2 \cdot 3 \cdot \frac{\sqrt{3}}{2} = 3\sqrt{3}$. Why distractors fail: Option A ($3$) equals the dot product, not the cross product magnitude. Option C ($6$) uses $\sin 90° = 1$ instead of $\sin 60°$, giving $|\mathbf{a}||\mathbf{b}|$ directly. Option D ($9$) may arise from squaring $3$ or other arithmetic errors.

Answer: $\cos\theta = \dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$

Recall the angle formula: From $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta$, solving for $\cos\theta$ gives $\cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$. Why distractors fail: Option A uses the cross product magnitude, which gives $\sin\theta$, not $\cos\theta$. Option C swaps cosine for sine. Option D, while algebraically derivable, is not the standard formula used for finding the angle.

Answer: $6$

Apply the geometric dot product formula: $\mathbf{p} \cdot \mathbf{q} = |\mathbf{p}||\mathbf{q}|\cos 60° = (3)(4)\left(\frac{1}{2}\right) = 6$. Why distractors fail: Option A ($12$) forgets to multiply by $\cos 60°$. Option B ($6\sqrt{3}$) uses $\sin 60°$ instead of $\cos 60°$. Option D ($12\sqrt{3}$) applies $\sin 60°$ to the product without halving.

Answer: $\langle 1, -1, 0 \rangle$ and $\langle 1, 1, 0 \rangle$

Check dot products: Option A: $3 + 4 + 3 = 10 \neq 0$. Option B: $1 + (-1) + 0 = 0$. Option C: $2 + 0 + 1 = 3 \neq 0$. Option D: $2 + 2 + 2 = 6 \neq 0$. Conclusion: Only Option B has a dot product of zero, confirming orthogonality.

Answer: Compute the dot product and check if it equals zero.

Apply the orthogonality test: Two vectors are perpendicular if and only if $\mathbf{a} \cdot \mathbf{b} = 0$. This is the simplest and most direct test. Why distractors fail: Option A: the cross product magnitude equals $|\mathbf{a}||\mathbf{b}|\sin\theta$, which is $|\mathbf{a}||\mathbf{b}|$ when perpendicular—not necessarily $1$. Option C: a zero cross product means the vectors are parallel, not perpendicular. Option D: the sum being zero means the vectors are opposites, not perpendicular.

Answer: $\langle 7, -3, -5 \rangle$

Set up the determinant: $\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & 1 \\ 1 & -1 & 2 \end{vmatrix}$. Compute each component: $\mathbf{i}(3 \cdot 2 - 1 \cdot (-1)) - \mathbf{j}(2 \cdot 2 - 1 \cdot 1) + \mathbf{k}(2 \cdot (-1) - 3 \cdot 1) = \mathbf{i}(7) - \mathbf{j}(3) + \mathbf{k}(-5) = \langle 7, -3, -5 \rangle$. Why distractors fail: Option B is $-(\mathbf{a} \times \mathbf{b})$, i.e., the result of computing $\mathbf{b} \times \mathbf{a}$. Option C has a sign error in the $\mathbf{j}$-component (forgetting the negative in the cofactor expansion). Option D scrambles the components.

Answer: $\sqrt{46}$

Compute the cross product: $\mathbf{u} \times \mathbf{v} = \langle (2)(3)-(0)(1),\; (0)(0)-(1)(3),\; (1)(1)-(2)(0) \rangle = \langle 6, -3, 1 \rangle$. Take the magnitude: $|\mathbf{u} \times \mathbf{v}| = \sqrt{36 + 9 + 1} = \sqrt{46}$. Why distractors fail: Option B ($7$) could come from summing the absolute values of the cross product components ($6+3+1=10$... actually that's $10$, so $7$ may come from a different arithmetic error). Option C ($\sqrt{58}$) results from an error in the cross product computation. Option D ($10$) sums absolute values of the cross product components.

Answer: The cross product is anticommutative: $\mathbf{a} \times \mathbf{b} = -(\mathbf{b} \times \mathbf{a})$.

Recall the anticommutativity property: By the right-hand rule and the determinant definition, swapping the two vectors reverses the sign: $\mathbf{a} \times \mathbf{b} = -(\mathbf{b} \times \mathbf{a})$. Why distractors fail: Option A is false; the cross product is specifically defined for 3D vectors. Option C is wrong because the cross product returns a vector, not a scalar. Option D is misleading; when vectors are parallel, $\mathbf{a} \times \mathbf{b} = \mathbf{0}$, so both orders give $\mathbf{0}$, but this doesn't mean commutativity holds in general.

Answer: Compute $\mathbf{a} \times \mathbf{b}$.

Key property of the cross product: By definition, $\mathbf{a} \times \mathbf{b}$ is perpendicular to both $\mathbf{a}$ and $\mathbf{b}$. Why distractors fail: Option A gives a scalar, not a vector. Option B gives a vector in the plane of $\mathbf{a}$ and $\mathbf{b}$, generally not perpendicular to either. Option D gives a scalar measuring how much $\mathbf{a}$ points in the direction of $\mathbf{b}$.

Answer: They reversed the sign on the $\mathbf{k}$-component or computed $\mathbf{b} \times \mathbf{a}$ instead of $\mathbf{a} \times \mathbf{b}$.

Compute the correct cross product: $\mathbf{i} \times \mathbf{j} = \mathbf{k} = \langle 0, 0, 1 \rangle$. The student got $\langle 0, 0, -1 \rangle = -\mathbf{k}$. Identify the error: Getting $-\mathbf{k}$ instead of $\mathbf{k}$ is consistent with computing $\mathbf{j} \times \mathbf{i}$ (reversing the order) or making a sign error in the $\mathbf{k}$-component of the determinant. Why distractors fail: Option A: the dot product would give a scalar ($0$), not a vector. Option B: $|\mathbf{i} \times \mathbf{j}| = 1$, not $0$. Option D: confusing basis vectors would change which component is nonzero, not just the sign.

Answer: They are parallel (one is a scalar multiple of the other).

Interpret a zero cross product: $|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta = 0$. Since both vectors are nonzero, $\sin\theta = 0$, meaning $\theta = 0°$ or $180°$. In either case the vectors are parallel. Why distractors fail: Option A: perpendicular vectors have $\sin\theta = 1$, giving a nonzero cross product. Option C: magnitudes are unrelated to the cross product being zero. Option D: parallel vectors have $\mathbf{a}\cdot\mathbf{b} = \pm|\mathbf{a}||\mathbf{b}| \neq 0$.

Answer: $30°$

Use the cross product magnitude formula: $|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta$, so $20 = 5 \cdot 8 \cdot \sin\theta = 40\sin\theta$. Solve for the angle: $\sin\theta = \frac{20}{40} = \frac{1}{2}$, so $\theta = 30°$. Why distractors fail: Option B ($45°$) gives $\sin 45° = \frac{\sqrt{2}}{2}$. Option C ($60°$) gives $\sin 60° = \frac{\sqrt{3}}{2}$. Option D ($90°$) gives $\sin 90° = 1$, which would require $|\mathbf{a} \times \mathbf{b}| = 40$.

Answer: $0$

Apply the key property of the cross product: The cross product $\mathbf{a} \times \mathbf{b}$ is always perpendicular to both $\mathbf{a}$ and $\mathbf{b}$. Therefore $(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{a} = 0$. Verification by computation: $\mathbf{a} \times \mathbf{b} = \langle (2)(0)-(3)(1),\; (3)(-2)-(1)(0),\; (1)(1)-(2)(-2)\rangle = \langle -3, -6, 5 \rangle$. Then $\langle -3,-6,5\rangle \cdot \langle 1,2,3\rangle = -3 -12 + 15 = 0$. Why distractors fail: Options A, B, and D are all nonzero, contradicting the fundamental property that the cross product is perpendicular to both input vectors.

Answer: $\mathbf{b} = \mathbf{0}$.

Analyze both conditions: $\mathbf{a} \times \mathbf{b} = \mathbf{0}$ (with $\mathbf{a} \neq \mathbf{0}$) implies either $\mathbf{b} = \mathbf{0}$ or $\mathbf{b}$ is parallel to $\mathbf{a}$. If $\mathbf{b}$ is parallel to $\mathbf{a}$ and nonzero, then $\mathbf{a} \cdot \mathbf{b} = \pm|\mathbf{a}||\mathbf{b}| \neq 0$, contradicting the first condition. Conclusion: The only possibility satisfying both conditions simultaneously is $\mathbf{b} = \mathbf{0}$. Why distractors fail: Option A: no information about magnitude. Option B: a nonzero perpendicular vector would give a nonzero cross product. Option D: a nonzero parallel vector would give a nonzero dot product.

Answer: $\frac{\sqrt{49}}{2} = \frac{7}{2}$

Form edge vectors: $\overrightarrow{PQ} = \langle -1, 2, 0 \rangle$ and $\overrightarrow{PR} = \langle -1, 0, 3 \rangle$. Compute the cross product: $\overrightarrow{PQ} \times \overrightarrow{PR} = \langle (2)(3)-(0)(0),\; (0)(-1)-((-1)(3)),\; ((-1)(0)-(2)(-1)) \rangle = \langle 6, 3, 2 \rangle$. Find the triangle area: $|\overrightarrow{PQ} \times \overrightarrow{PR}| = \sqrt{36+9+4} = \sqrt{49} = 7$. The triangle area is half the parallelogram area: $\frac{7}{2}$. Why distractors fail: Option B ($7$) gives the parallelogram area, not the triangle area. Options C and D use $\sqrt{14}$, which may come from errors in the cross product computation.

Answer: $-5$

Apply the dot product formula: $\mathbf{a} \cdot \mathbf{b} = (4)(-1) + (-2)(3) + (1)(5) = -4 - 6 + 5 = -5$. Why distractors fail: Option B ($5$) ignores the signs. Option C ($-15$) may result from multiplying magnitudes or errors. Option D ($15$) takes absolute values of all products.

Answer: Compute $\mathbf{a} \times \mathbf{b}$, then divide the result by its magnitude.

Design the procedure: Step 1: $\mathbf{n} = \mathbf{a} \times \mathbf{b}$ gives a vector perpendicular to the plane of $\mathbf{a}$ and $\mathbf{b}$. Step 2: $\hat{\mathbf{n}} = \frac{\mathbf{n}}{|\mathbf{n}|}$ normalizes it to a unit vector. Why distractors fail: Option A produces a scalar (the cosine of the angle), not a vector. Option C: $\mathbf{a} + \mathbf{b}$ lies in the plane of $\mathbf{a}$ and $\mathbf{b}$, so it's not perpendicular to the plane. Option D is vague and does not generalize to 3D; rotating by $90°$ in 3D requires specifying an axis.

Answer: $16$

Apply the Lagrange identity: $|\mathbf{a} \times \mathbf{b}|^2 = |\mathbf{a}|^2|\mathbf{b}|^2 - (\mathbf{a}\cdot\mathbf{b})^2 = 16 \cdot 25 - 144 = 400 - 144 = 256$. Find the magnitude: $|\mathbf{a} \times \mathbf{b}| = \sqrt{256} = 16$. Why distractors fail: Option A ($8$) is half the correct answer. Option C ($20$) equals $|\mathbf{a}||\mathbf{b}|$ without subtracting the dot product squared. Option D ($\sqrt{256}$) equals $16$ but is not in simplified form; the question asks for the value, and $16$ is the simplified answer.

Answer: $-7$

Apply the dot product formula: The dot product is $\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 = (3)(1) + (-1)(4) + (2)(-3) = 3 - 4 - 6 = -7$. Why the correct answer works: Option A gives $-7$, which is the correct computation of the component-wise products summed together. Why distractors fail: Option B ($7$) results from taking absolute values of each product before summing. Option C ($-5$) comes from an arithmetic error such as computing $2 \times (-3) = -4$. Option D ($1$) might arise from forgetting the sign on the middle term.

Answer: $k = -2$

Set the dot product equal to zero: $\mathbf{a} \cdot \mathbf{b} = (2)(1) + (k)(-2) + (3)(k) = 2 - 2k + 3k = 2 + k = 0$. Solve for k: $k = -2$. Verify: With $k = -2$: $\mathbf{a} = \langle 2, -2, 3 \rangle$ and $\mathbf{b} = \langle 1, -2, -2 \rangle$. Dot product: $2 + 4 - 6 = 0$. ✓ Why distractors fail: Option A ($k=-3$): $2+(-3) = -1 \neq 0$. Option B ($k=2$): $2+2 = 4 \neq 0$. Option D ($k=6$): $2+6 = 8 \neq 0$.

Answer: No, because $\mathbf{a} \cdot (\mathbf{b} - \mathbf{c}) = 0$ only means $\mathbf{b} - \mathbf{c}$ is perpendicular to $\mathbf{a}$, not that $\mathbf{b} - \mathbf{c} = \mathbf{0}$.

Analyze the claim: $\mathbf{a} \cdot \mathbf{b} = \mathbf{a} \cdot \mathbf{c}$ implies $\mathbf{a} \cdot (\mathbf{b} - \mathbf{c}) = 0$. This means $\mathbf{b} - \mathbf{c}$ is orthogonal to $\mathbf{a}$, but it could be any nonzero vector perpendicular to $\mathbf{a}$. Why distractors fail: Option A: the dot product is not injective; many different vectors can give the same dot product with $\mathbf{a}$. Option B: you cannot 'divide by a vector.' Option D: the dot product is defined for any vectors, not just unit vectors.

Answer: The determinant of a $3 \times 3$ matrix with $\mathbf{i}, \mathbf{j}, \mathbf{k}$ in the first row and the components of $\mathbf{a}$ and $\mathbf{b}$ in the next two rows

Recall the cross product computation: The cross product is computed as $\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$, yielding a vector. Why distractors fail: Option A gives only the magnitude $|\mathbf{a} \times \mathbf{b}|$, not the cross product itself which is a vector. Option C is the dot product formula. Option D is incorrect because the cross product is perpendicular to the plane of $\mathbf{a}$ and $\mathbf{b}$.