Cross Product - Multivariable Calculus

Let $\vec{a}$ and $\vec{b}$ be vectors in $\mathbb{R}^3$ . The parallelogram $P_1$ spanned by $\vec{a}$ and $\vec{b}$ has area $A$ . A new parallelogram $P_2$ is spanned by the vectors $\vec{u} = 2\vec{a} + \vec{b}$ and $\vec{v} = \vec{a} - 3\vec{b}$ . What is the area of $P_2$ in terms of $A$ ?
Answer: $7A$ When you encounter problems about areas of parallelograms formed by different vectors, the key insight is that area scales according to how linear transformations affect the original vectors. The area of a parallelogram spanned by two vectors equals the absolute value of their cross product. For the original parallelogram $P_1$ , we have area $A = |\vec{a} \times \vec{b}|$ . To find the area of $P_2$ spanned by $\vec{u} = 2\vec{a} + \vec{b}$ and $\vec{v} = \vec{a} - 3\vec{b}$ , we need to compute $|\vec{u} \times \vec{v}|$ . Using the distributive property of cross products: $\vec{u} \times \vec{v} = (2\vec{a} + \vec{b}) \times (\vec{a} - 3\vec{b})$ $= 2\vec{a} \times \vec{a} - 6\vec{a} \times \vec{b} + \vec{b} \times \vec{a} - 3\vec{b} \times \vec{b}$ Since $\vec{a} \times \vec{a} = \vec{0}$ and $\vec{b} \times \vec{b} = \vec{0}$ , and $\vec{b} \times \vec{a} = -\vec{a} \times \vec{b}$ : $\vec{u} \times \vec{v} = -6\vec{a} \times \vec{b} - \vec{a} \times \vec{b} = -7(\vec{a} \times \vec{b})$ Therefore, the area of $P_2$ is $|-7(\vec{a} \times \vec{b})| = 7|\vec{a} \times \vec{b}| = 7A$ . Choice A ( $5A$ ) likely comes from incorrectly adding coefficients. Choice B ( $6A$ ) might result from forgetting the $\vec{b} \times \vec{a}$ term. Choice C is wrong because the cross product properties give us a definitive relationship regardless of the specific vectors. Study tip: Remember that cross products distribute over addition, and the determinant of the transformation matrix gives you the area scaling factor directly.
Let $\vec{u} = \langle 1, 2, 2 \rangle$ and $\vec{v} = \langle 3, 0, 4 \rangle$ . Let $A$ be the area of the parallelogram spanned by $\vec{u}$ and $\vec{v}$ . Let $A_{xy}$ be the area of the projection of this parallelogram onto the $xy$ -plane. What is the value of $A_{xy}$ ?
Answer: 6 The area of the projection of the parallelogram onto the $xy$ -plane is given by the magnitude of the $\vec{k}$ -component of the cross product $\vec{u} \times \vec{v}$ . First, compute the cross product: $\vec{u} \times \vec{v} = \det \begin{pmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 2 & 2 \\ 3 & 0 & 4 \end{pmatrix} = \vec{i}(8-0) - \vec{j}(4-6) + \vec{k}(0-6) = \langle 8, 2, -6 \rangle$ The area of the projection onto the $xy$ -plane is the absolute value of the $z$ -component, so $A_{xy} = |-6| = 6$ . Alternatively, one can project the vectors onto the $xy$ -plane first: $\vec{u}_{xy} = \langle 1, 2, 0 \rangle$ and $\vec{v}_{xy} = \langle 3, 0, 0 \rangle$ . Then $A_{xy} = \|\vec{u}_{xy} \times \vec{v}_{xy}\| = \|\langle 0, 0, -6 \rangle\| = 6$ . Distractor D, $\sqrt{104}$ , is the full area $A$ . Distractor B, 8, is the area of the projection onto the $yz$ -plane. Distractor C, 11, is the dot product $\vec{u} \cdot \vec{v}$ .
Three points $P(1,0,0)$ , $Q(0,1,0)$ , and $R(0,0,1)$ define a triangle. Let $\vec{v} = \vec{PQ} \times \vec{PR}$ . Which of the following statements correctly describes the relationship between the vector $\vec{v}$ and the area of triangle PQR?
Answer: $\vec{v}$ is orthogonal to the plane of the triangle, and its magnitude is twice the area of the triangle. When you encounter cross products in multivariable calculus, remember that they create vectors with two key properties: direction (always perpendicular to both original vectors) and magnitude (related to the area of the parallelogram formed by those vectors). Let's find $\vec{PQ} = Q - P = (0,1,0) - (1,0,0) = (-1,1,0)$ and $\vec{PR} = R - P = (0,0,1) - (1,0,0) = (-1,0,1)$ . Computing the cross product: $\vec{v} = \vec{PQ} \times \vec{PR} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{vmatrix} = (1,1,1)$ The cross product is always orthogonal to the plane containing the original vectors, so $\vec{v}$ is perpendicular to triangle PQR's plane. The magnitude is $|\vec{v}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$ . For area, recall that $|\vec{PQ} \times \vec{PR}|$ equals the area of the parallelogram formed by these vectors. Since a triangle has half the area of the parallelogram formed by its sides, the triangle's area is $\frac{\sqrt{3}}{2}$ . Therefore, $|\vec{v}| = 2 \times \text{(triangle area)}$ . Answer B correctly identifies both properties. Answer A incorrectly claims the vector is parallel to the plane (cross products are never parallel to the plane of their component vectors). Answers C and D incorrectly state the magnitude equals the triangle's area, missing that cross product magnitude gives the parallelogram area. Remember: cross products always point perpendicular to the original plane, and their magnitude equals twice the triangle area.
Consider the basis vectors $\vec{u} = \langle 1, 1, 0 \rangle$ , $\vec{v} = \langle 0, 1, 1 \rangle$ , and $\vec{w} = \langle 1, 0, 1 \rangle$ . This basis forms a parallelepiped with the origin. How is this basis oriented, and what is the volume of the parallelepiped?
Answer: Right-handed, volume 2 The orientation and volume are determined by the scalar triple product $(\vec{u} \times \vec{v}) \cdot \vec{w}$ . A positive result indicates a right-handed system, while a negative result indicates a left-handed system. The absolute value of the result is the volume of the parallelepiped. The scalar triple product can be computed as the determinant of the matrix formed by the vectors: $V = \det \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{pmatrix} = 1(1\cdot1 - 1\cdot0) - 1(0\cdot1 - 1\cdot1) + 0 = 1(1) - 1(-1) = 1 + 1 = 2.$ Since the result is positive ( $+2$ ), the basis is right-handed. The volume of the parallelepiped is the absolute value of the determinant, which is $|2| = 2$ .
Let $\mathbf{u}$ and $\mathbf{v}$ be non-parallel vectors such that the area of the parallelogram they span is $12$ . The angle between $\mathbf{u}$ and $\mathbf{v}$ is $\pi/4$ . What is the area of the parallelogram formed by the vectors $3\mathbf{u}$ and $\mathbf{v} - \mathrm{proj}_{\mathbf{u}}(\mathbf{v})$ ?
Answer: $36$ The area of the new parallelogram is given by the magnitude of the cross product of the two new vectors: $|(3\mathbf{u}) \times (\mathbf{v} - \mathrm{proj}_{\mathbf{u}}(\mathbf{v}))|$ . Using the distributive property of the cross product: $|(3\mathbf{u}) \times \mathbf{v} - (3\mathbf{u}) \times \mathrm{proj}_{\mathbf{u}}(\mathbf{v})| = |3(\mathbf{u} \times \mathbf{v}) - 3(\mathbf{u} \times \mathrm{proj}_{\mathbf{u}}(\mathbf{v}))|$ . The projection of $\mathbf{v}$ onto $\mathbf{u}$ , denoted $\mathrm{proj}_{\mathbf{u}}(\mathbf{v})$ , is a vector that is parallel to $\mathbf{u}$ . The cross product of two parallel vectors is the zero vector, so $\mathbf{u} \times \mathrm{proj}_{\mathbf{u}}(\mathbf{v}) = \mathbf{0}$ . The expression for the new area simplifies to $|3(\mathbf{u} \times \mathbf{v})| = 3|\mathbf{u} \times \mathbf{v}|$ . We are given that the area of the parallelogram spanned by $\mathbf{u}$ and $\mathbf{v}$ is $12$ . This means $|\mathbf{u} \times \mathbf{v}| = 12$ . Therefore, the new area is $3 \times 12 = 36$ . The information about the angle $\theta = \pi/4$ is extraneous and intended to distract. A student might try to use it to find magnitudes, but it is not necessary for the solution.
Let $\vec{u}, \vec{v}, \vec{w}$ be three vectors in $\mathbb{R}^3$ such that $\vec{u} \times \vec{v} = \vec{u} \times \vec{w}$ . Which of the following statements must be true?
Answer: The vector $\vec{v} - \vec{w}$ is parallel to $\vec{u}$ . The given equation is $\vec{u} \times \vec{v} = \vec{u} \times \vec{w}$ . We can rearrange this to $\vec{u} \times \vec{v} - \vec{u} \times \vec{w} = \vec{0}$ . Using the distributive property of the cross product, this becomes $\vec{u} \times (\vec{v} - \vec{w}) = \vec{0}$ . The cross product of two vectors is the zero vector if and only if the two vectors are parallel. Thus, the vector $\vec{u}$ must be parallel to the vector $(\vec{v} - \vec{w})$ . This means $\vec{v} - \vec{w} = k\vec{u}$ for some scalar $k$ . Choice A is not necessarily true; for example, if $\vec{u}$ is parallel to $\vec{v}-\vec{w}$ , the equation holds even if $\vec{v} \ne \vec{w}$ . Choice B describes the condition for a dot product being zero. Choice C is incomplete; it misses the case where $\vec{u}$ and $\vec{v}-\vec{w}$ are non-zero but parallel.
Let $\mathbf{u}$ and $\mathbf{v}$ be vectors in $\mathbb{R}^3$ that span a parallelogram of area $A$ . What is the area of the parallelogram spanned by the vectors $2\mathbf{u} + \mathbf{v}$ and $\mathbf{u} - 3\mathbf{v}$ ?
Answer: $7A$ The area of the parallelogram spanned by two vectors is the magnitude of their cross product. The original area is $A = |\mathbf{u} \times \mathbf{v}|$ . The new area is the magnitude of the cross product of the new vectors: $|(2\mathbf{u} + \mathbf{v}) \times (\mathbf{u} - 3\mathbf{v})|$ . Using the distributive and anti-commutative properties of the cross product: $(2\mathbf{u} + \mathbf{v}) \times (\mathbf{u} - 3\mathbf{v}) = (2\mathbf{u} \times \mathbf{u}) - (2\mathbf{u} \times 3\mathbf{v}) + (\mathbf{v} \times \mathbf{u}) - (\mathbf{v} \times 3\mathbf{v})$ $= 2(\mathbf{u} \times \mathbf{u}) - 6(\mathbf{u} \times \mathbf{v}) + (\mathbf{v} \times \mathbf{u}) - 3(\mathbf{v} \times \mathbf{v})$ . Since the cross product of any vector with itself is the zero vector (i.e., $\mathbf{u} \times \mathbf{u} = \mathbf{0}$ and $\mathbf{v} \times \mathbf{v} = \mathbf{0}$ ), and $\mathbf{v} \times \mathbf{u} = -(\mathbf{u} \times \mathbf{v})$ , the expression simplifies to: $\mathbf{0} - 6(\mathbf{u} \times \mathbf{v}) - (\mathbf{u} \times \mathbf{v}) - \mathbf{0} = -7(\mathbf{u} \times \mathbf{v})$ . The magnitude of this new vector is $|-7(\mathbf{u} \times \mathbf{v})| = |-7| |\mathbf{u} \times \mathbf{v}| = 7A$ . Distractor A results from a sign error, calculating $|-6+1|=5$ instead of $|-6-1|=7$ . Distractor B arises from ignoring the $\mathbf{v} \times \mathbf{u}$ term. Distractor D would be chosen if one incorrectly believes this linear transformation of vectors does not change the area.
Let $\vec{a}, \vec{b}, \vec{c}$ be three non-coplanar vectors in $\mathbb{R}^3$ . Define $\vec{u} = \vec{a} \times \vec{b}$ and $\vec{v} = \vec{a} \times \vec{c}$ . The area of the parallelogram spanned by $\vec{u}$ and $\vec{v}$ is given by which of the following?
Answer: $|(\vec{b} \times \vec{c}) \cdot \vec{a}| \|\vec{a}\|$ The area of the parallelogram spanned by $\vec{u}$ and $\vec{v}$ is $\|\vec{u} \times \vec{v}\|$ . We substitute the definitions of $\vec{u}$ and $\vec{v}$ : $\|(\vec{a} \times \vec{b}) \times (\vec{a} \times \vec{c})\|$ . We use the vector triple product identity $\vec{X} \times (\vec{Y} \times \vec{Z}) = (\vec{X} \cdot \vec{Z})\vec{Y} - (\vec{X} \cdot \vec{Y})\vec{Z}$ . Let $\vec{X} = \vec{a} \times \vec{b}$ , $\vec{Y} = \vec{a}$ , and $\vec{Z} = \vec{c}$ . This does not fit the form. Instead, use the identity for $(\vec{A} \times \vec{B}) \times (\vec{C} \times \vec{D}) = ((\vec{A} \times \vec{B}) \cdot \vec{D}) \vec{C} - ((\vec{A} \times \vec{B}) \cdot \vec{C}) \vec{D}$ . Here, $\vec{A}=\vec{C}=\vec{a}$ . A simpler identity is $(\vec{a} \times \vec{b}) \times \vec{c} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a}$ . Let $\vec{d} = \vec{a} \times \vec{c}$ . Then we have $(\vec{a} \times \vec{b}) \times \vec{d}$ . There is a specific identity for this case: $(\vec{a} \times \vec{b}) \times (\vec{a} \times \vec{c}) = ((\vec{a} \times \vec{b}) \cdot \vec{c}) \vec{a}$ . This is because $((\vec{a} \times \vec{b}) \cdot \vec{a})\vec{c} = 0$ . The term $(\vec{a} \times \vec{b}) \cdot \vec{c}$ is the scalar triple product, which represents the signed volume of the parallelepiped formed by $\vec{a}, \vec{b}, \vec{c}$ . The area is the magnitude: $\|((\vec{a} \times \vec{b}) \cdot \vec{c}) \vec{a}\| = |(\vec{a} \times \vec{b}) \cdot \vec{c}| \|\vec{a}\|$ . The scalar triple product is cyclic, so $|(\vec{a} \times \vec{b}) \cdot \vec{c}| = |(\vec{b} \times \vec{c}) \cdot \vec{a}|$ . Therefore, the area is $|(\vec{b} \times \vec{c}) \cdot \vec{a}| \|\vec{a}\|$ . Distractor D is false because if $\vec{u}$ and $\vec{v}$ were parallel, $\vec{a}, \vec{b}, \vec{c}$ would be coplanar.
Let $\mathbf{u}$ , $\mathbf{v}$ , and $\mathbf{w}$ be three non-zero vectors in $\mathbb{R}^3$ such that $\mathbf{u} + \mathbf{v} + \mathbf{w} = \mathbf{0}$ . Which of the following statements must be true?
Answer: $\mathbf{u} \times \mathbf{v} = \mathbf{v} \times \mathbf{w} = \mathbf{w} \times \mathbf{u}$ The condition $\mathbf{u} + \mathbf{v} + \mathbf{w} = \mathbf{0}$ implies that the three vectors form a closed triangle when placed head to tail. We can use vector algebra to prove the relationship. Start with the given equation: $\mathbf{u} + \mathbf{v} + \mathbf{w} = \mathbf{0}$ . Cross product with $\mathbf{v}$ from the left: $\mathbf{v} \times (\mathbf{u} + \mathbf{v} + \mathbf{w}) = \mathbf{v} \times \mathbf{0}$ $\mathbf{v} \times \mathbf{u} + \mathbf{v} \times \mathbf{v} + \mathbf{v} \times \mathbf{w} = \mathbf{0}$ Using properties $\mathbf{v} \times \mathbf{u} = -(\mathbf{u} \times \mathbf{v})$ and $\mathbf{v} \times \mathbf{v} = \mathbf{0}$ : $-(\mathbf{u} \times \mathbf{v}) + \mathbf{0} + \mathbf{v} \times \mathbf{w} = \mathbf{0}$ This gives $\mathbf{u} \times \mathbf{v} = \mathbf{v} \times \mathbf{w}$ . Now, start again and cross product with $\mathbf{w}$ from the left: $\mathbf{w} \times (\mathbf{u} + \mathbf{v} + \mathbf{w}) = \mathbf{w} \times \mathbf{0}$ $\mathbf{w} \times \mathbf{u} + \mathbf{w} \times \mathbf{v} + \mathbf{w} \times \mathbf{w} = \mathbf{0}$ Using properties $\mathbf{w} \times \mathbf{v} = -(\mathbf{v} \times \mathbf{w})$ and $\mathbf{w} \times \mathbf{w} = \mathbf{0}$ : $\mathbf{w} \times \mathbf{u} - (\mathbf{v} \times \mathbf{w}) + \mathbf{0} = \mathbf{0}$ This gives $\mathbf{v} \times \mathbf{w} = \mathbf{w} \times \mathbf{u}$ . Combining the two results, we have $\mathbf{u} \times \mathbf{v} = \mathbf{v} \times \mathbf{w} = \mathbf{w} \times \mathbf{u}$ . Distractor C would imply $3(\mathbf{u} \times \mathbf{v}) = \mathbf{0}$ , which is not generally true. Distractor D implies the vectors are collinear, which is a special case but not a necessary condition.
Let $\mathbf{u} = \langle 1, -1, 0 \rangle$ and $\mathbf{v} = \langle 1, 0, 1 \rangle$ . Find a unit vector $\mathbf{n}$ such that the ordered set of vectors $(\mathbf{u}, \mathbf{v}, \mathbf{n})$ forms a left-handed system.
Answer: $\frac{1}{\sqrt{3}} \langle 1, 1, -1 \rangle$ By the right-hand rule, the vector $\mathbf{w} = \mathbf{u} \times \mathbf{v}$ is orthogonal to both $\mathbf{u}$ and $\mathbf{v}$ , and the set $(\mathbf{u}, \mathbf{v}, \mathbf{w})$ forms a right-handed system. To form a left-handed system, we need a vector pointing in the opposite direction of $\mathbf{u} \times \mathbf{v}$ . First, calculate $\mathbf{u} \times \mathbf{v}$ : $\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 0 \\ 1 & 0 & 1 \end{vmatrix} = \mathbf{i}(-1 - 0) - \mathbf{j}(1 - 0) + \mathbf{k}(0 - (-1)) = \langle -1, -1, 1 \rangle.$ For a left-handed system $(\mathbf{u}, \mathbf{v}, \mathbf{n})$ , the vector $\mathbf{n}$ must be in the direction of $-(\mathbf{u} \times \mathbf{v}) = -\langle -1, -1, 1 \rangle = \langle 1, 1, -1 \rangle$ . To find the unit vector $\mathbf{n}$ , we normalize this vector. Its magnitude is $\sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3}$ . So, $\mathbf{n} = \frac{1}{\sqrt{3}} \langle 1, 1, -1 \rangle$ . Distractor A is the unit vector for a right-handed system. Distractor C is in the correct direction but is not a unit vector. Distractor D is a unit vector, but it is not orthogonal to $\mathbf{v}$ .
The area of the parallelogram spanned by vectors $\mathbf{u}$ and $\mathbf{v}$ is $4$ . If $\mathbf{w}$ is a vector with magnitude $|\mathbf{w}|=3$ , what is the maximum possible value of the scalar triple product $(\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w}$ ?
Answer: $12$ The scalar triple product $(\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w}$ can be interpreted geometrically using the formula for the dot product: $(\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w} = |\mathbf{u} \times \mathbf{v}| |\mathbf{w}| \cos(\theta)$ , where $\theta$ is the angle between the vector $\mathbf{u} \times \mathbf{v}$ and the vector $\mathbf{w}$ . We are given that the area of the parallelogram spanned by $\mathbf{u}$ and $\mathbf{v}$ is $4$ . This area is equal to the magnitude of the cross product, so $|\mathbf{u} \times \mathbf{v}| = 4$ . We are also given $|\mathbf{w}| = 3$ . Substituting these values, we get: $(\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w} = (4)(3) \cos(\theta) = 12 \cos(\theta)$ . To maximize this value, we must maximize $\cos(\theta)$ . The maximum value of $\cos(\theta)$ is $1$ , which occurs when $\theta = 0$ . This means $\mathbf{w}$ is parallel to and has the same direction as $\mathbf{u} \times \mathbf{v}$ . The maximum value is $12 \times 1 = 12$ . Distractor B ignores the magnitude of $\mathbf{w}$ . Distractor A represents a division instead of multiplication. Distractor C is the sum of the magnitudes, not their product.
A rigid body rotates with angular velocity $\vec{\omega} = \langle 1, -2, 2 \rangle$ about an axis through the origin. The linear velocity of a point $P$ with position vector $\vec{r}$ is $\vec{v} = \vec{\omega} \times \vec{r}$ . Which statement correctly describes the orientation of the circular path of point $P$ ?
Answer: The plane of the circle is normal to the angular velocity vector $\vec{\omega}$ . The angular velocity vector $\vec{\omega}$ defines the axis of rotation. Any point on the rigid body, such as $P$ , travels in a circular path. The plane containing this circle is perpendicular (normal) to the axis of rotation. Therefore, the plane of rotation for point $P$ is normal to the vector $\vec{\omega}$ . The linear velocity vector $\vec{v}$ is tangent to the circular path at point $P$ , so it lies within the plane of the circle, making choice B incorrect. The position vector $\vec{r}$ extends from the origin to the point $P$ on the circle, so it is generally not normal to the plane of rotation, making choice A incorrect. The plane of the circle cannot contain $\vec{\omega}$ as $\vec{\omega}$ is normal to it, making choice D incorrect.
Let $\vec{u}$ and $\vec{v}$ be unit vectors in $\mathbb{R}^3$ . The area of the parallelogram spanned by $\vec{u}$ and $\vec{v}$ is $\frac{1}{2}$ . What is the value of $|(\vec{u} \times \vec{v}) \cdot \vec{u}|$ ?
Answer: 0 The question asks for the value of $|(\vec{u} \times \vec{v}) \cdot \vec{u}|$ . The vector $\vec{w} = \vec{u} \times \vec{v}$ is, by definition of the cross product, orthogonal to both $\vec{u}$ and $\vec{v}$ . The dot product of two orthogonal vectors is zero. Therefore, $(\vec{u} \times \vec{v}) \cdot \vec{u} = 0$ . The information about the vectors being unit vectors and the area of the parallelogram being $\frac{1}{2}$ is extraneous information designed to distract. This information would be used to find the angle between the vectors ( $\|\vec{u} \times \vec{v}\| = \|\vec{u}\|\|\vec{v}\|\sin\theta = (1)(1)\sin\theta = 1/2$ , so $\theta = \pi/6$ or $5\pi/6$ ), but it is not needed to answer the question about the scalar triple product.
Let $\mathbf{u}$ , $\mathbf{v}$ , and $\mathbf{w}$ be non-zero vectors in $\mathbb{R}^3$ . If the vector triple product $(\mathbf{u} \times \mathbf{v}) \times \mathbf{w} = \mathbf{0}$ , which of the following statements must be true?
Answer: Either $\mathbf{u}$ and $\mathbf{v}$ are parallel, or $\mathbf{w}$ is orthogonal to both $\mathbf{u}$ and $\mathbf{v}$ . When you encounter vector triple products, the key is understanding what conditions make cross products equal to zero and applying the vector triple product identity. The vector triple product $(\mathbf{u} \times \mathbf{v}) \times \mathbf{w} = \mathbf{0}$ means that either $\mathbf{u} \times \mathbf{v} = \mathbf{0}$ or $(\mathbf{u} \times \mathbf{v}) \perp \mathbf{w}$ doesn't hold in the usual sense. Instead, we need $(\mathbf{u} \times \mathbf{v}) \times \mathbf{w} = \mathbf{0}$ , which occurs when $\mathbf{u} \times \mathbf{v}$ and $\mathbf{w}$ are parallel (including when one is zero). This happens in two cases: either $\mathbf{u} \times \mathbf{v} = \mathbf{0}$ (meaning $\mathbf{u}$ and $\mathbf{v}$ are parallel), or $\mathbf{w}$ is parallel to $\mathbf{u} \times \mathbf{v}$ . Since $\mathbf{u} \times \mathbf{v}$ is orthogonal to both $\mathbf{u}$ and $\mathbf{v}$ , the second case means $\mathbf{w}$ is orthogonal to both $\mathbf{u}$ and $\mathbf{v}$ . This confirms answer A. Answer B is incorrect because coplanarity isn't necessary—you could have $\mathbf{u}$ and $\mathbf{v}$ parallel with $\mathbf{w}$ pointing anywhere. Answer C is wrong since mutual orthogonality is too restrictive; for example, parallel $\mathbf{u}$ and $\mathbf{v}$ satisfy the condition without being orthogonal to each other. Answer D contradicts the given information that all vectors are non-zero. Remember: when analyzing vector triple products, consider both when the first cross product is zero and when the result is parallel to one of the vectors. The "either-or" structure in answer A is a strong hint that multiple scenarios can satisfy the condition.
A parallelogram lies in the plane $x - 2y + 2z = 5$ . The area of this parallelogram is 12. Let $\vec{n}$ be the vector representing the cross product of the two adjacent edge vectors that define the parallelogram. Which of the following could be $\vec{n}$ ?
Answer: $\langle 4, -8, 8 \rangle$ The cross product of the two vectors spanning the parallelogram, $\vec{n} = \vec{a} \times \vec{b}$ , results in a vector that is normal to the plane containing the parallelogram. The normal vector to the plane $x - 2y + 2z = 5$ is $\vec{N} = \langle 1, -2, 2 \rangle$ . Therefore, $\vec{n}$ must be parallel to $\vec{N}$ , i.e., $\vec{n} = k \langle 1, -2, 2 \rangle$ for some scalar $k$ . The magnitude of $\vec{n}$ is the area of the parallelogram, which is given as 12. So, $\|\vec{n}\| = 12$ . We have $\|k \langle 1, -2, 2 \rangle\| = |k| \sqrt{1^2 + (-2)^2 + 2^2} = |k|\sqrt{9} = 3|k|$ . Setting the magnitude equal to the area: $3|k| = 12$ , which gives $|k| = 4$ . Thus, $\vec{n}$ can be either $4\langle 1, -2, 2 \rangle = \langle 4, -8, 8 \rangle$ or $-4\langle 1, -2, 2 \rangle = \langle -4, 8, -8 \rangle$ . Choice A is one of these possibilities. Choice B is parallel but has the wrong magnitude (3). Choice C is parallel but has the wrong magnitude (36). Choice D is not parallel to the normal vector.
The vertices of a triangle in $\mathbb{R}^3$ are given by the points $P(1, 0, 1)$ , $Q(2, 3, 1)$ , and $R(0, 1, 4)$ . What is the area of triangle $PQR$ ?
Answer: $\frac{\sqrt{106}}{2}$ The area of a triangle with vertices $P, Q, R$ is half the magnitude of the cross product of two vectors forming sides of the triangle, for example, $\mathbf{u} = \vec{PQ}$ and $\mathbf{v} = \vec{PR}$ . First, we find these vectors: $\mathbf{u} = Q - P = \langle 2-1, 3-0, 1-1 \rangle = \langle 1, 3, 0 \rangle$ . $\mathbf{v} = R - P = \langle 0-1, 1-0, 4-1 \rangle = \langle -1, 1, 3 \rangle$ . Next, we compute their cross product: $\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 3 & 0 \\ -1 & 1 & 3 \end{vmatrix} = \mathbf{i}(3 \cdot 3 - 0 \cdot 1) - \mathbf{j}(1 \cdot 3 - 0 \cdot (-1)) + \mathbf{k}(1 \cdot 1 - 3 \cdot (-1)) = \langle 9, -3, 4 \rangle.$ The magnitude of this vector is $|\mathbf{u} \times \mathbf{v}| = \sqrt{9^2 + (-3)^2 + 4^2} = \sqrt{81 + 9 + 16} = \sqrt{106}$ . This magnitude represents the area of the parallelogram spanned by $\mathbf{u}$ and $\mathbf{v}$ . The area of the triangle is half of this value, which is $\frac{\sqrt{106}}{2}$ . Distractor A is the area of the parallelogram, a common error from forgetting to divide by 2. Distractor C is the squared magnitude. Distractor D results from a miscalculation, such as using vectors $\vec{PQ}$ and $\vec{RQ} = \langle 2, 2, -3 \rangle$ , whose cross product magnitude is $\sqrt{78}$ .
A force $\mathbf{F} = \langle 10, 0, 0 \rangle$ is applied at the point $P(0, 5, 2)$ . What is the magnitude of the torque produced by this force about the point $Q(1, 5, 0)$ ?
Answer: $20$ When you encounter torque problems in multivariable calculus, you're dealing with the cross product of position and force vectors. Torque measures the rotational effect of a force about a specific point. To find the torque about point Q, you first need the position vector from Q to P where the force is applied: $\vec{r} = P - Q = \langle 0,5,2 \rangle - \langle 1,5,0 \rangle = \langle -1,0,2 \rangle$ The torque is given by $\vec{\tau} = \vec{r} \times \mathbf{F}$ . Computing this cross product: $\vec{\tau} = \langle -1,0,2 \rangle \times \langle 10,0,0 \rangle = \langle 0,20,0 \rangle$ The magnitude is $|\vec{\tau}| = \sqrt{0^2 + 20^2 + 0^2} = 20$ . Looking at the wrong answers: Choice A (10) likely comes from just using the magnitude of the force vector, ignoring the position component entirely. Choice B ( $\sqrt{2900}$ ) appears to result from incorrectly computing $|\vec{r}| \cdot |\mathbf{F}|$ instead of using the cross product—this gives $\sqrt{5} \cdot 10 = 10\sqrt{5} = \sqrt{2500}$ , though the exact source of 2900 may involve other computational errors. Choice C (50) might come from mistakenly using the full distance calculation or incorrectly applying the cross product formula. Remember: torque always requires the cross product of position and force vectors, not their dot product or simple multiplication. The cross product naturally accounts for the perpendicular component of force that creates rotation.
A tetrahedron has vertices at points $O$ , $A$ , $B$ , and $C$ . The area of the triangular face $ABC$ is 10. Let $\mathbf{n} = (\vec{AB}) \times (\vec{AC})$ . What is the magnitude of the vector $\mathbf{n}$ ?
Answer: $20$ The magnitude of the cross product of two vectors, $|\mathbf{a} \times \mathbf{b}|$ , represents the area of the parallelogram spanned by the vectors $\mathbf{a}$ and $\mathbf{b}$ . The vectors $\vec{AB}$ and $\vec{AC}$ are two side vectors of the triangular face $ABC$ . The area of the triangle $ABC$ is half the area of the parallelogram spanned by these vectors. So, Area $(ABC) = \frac{1}{2} |\vec{AB} \times \vec{AC}|$ . We are given that the area of triangle $ABC$ is $10$ , and the vector $\mathbf{n}$ is defined as $\mathbf{n} = \vec{AB} \times \vec{AC}$ . Substituting the known values into the formula: $10 = \frac{1}{2} |\mathbf{n}|$ . Solving for $|\mathbf{n}|$ , we find that the magnitude of $\mathbf{n}$ is $2 \times 10 = 20$ . Distractor B incorrectly equates the magnitude of the cross product with the triangle's area. Distractor A incorrectly divides the area by 2 instead of multiplying.

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