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Calculus 3 : Cross Product

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Example Questions

Example Question #191 : Vectors And Vector Operations

Find the cross product of the two vectors.

u=<-3,-3,6>

v=<-7,-5,-1>

Possible Answers:

<33,-45,-6>

<-7,-3,-1>

<-3,-5,6>

<15,-10,8>

Correct answer:

<33,-45,-6>

Explanation:

To find the cross product we solve for the determinant of the matrix

$\begin{vmatrix} i &j &k \\ -3&-3 &6 \\ -7& -5& -1\end{vmatrix}$

$=det(\begin{vmatrix} -3&6 \\ -5& -1\end{vmatrix})i-det(\begin{vmatrix} -3&6 \\ -7& -1\end{vmatrix})j+det(\begin{vmatrix} -3&-3 \\ -7& -5\end{vmatrix})k$

=[(-3)(-1)-6(-5)]i-[(-3)(-1)-6(-7)]j+[(-3)(-5)-(-3)(-7)]k

=[3+30]i-[3+42]j+[15-21]k

=33i-45j-6k

As such, the cross product is

<33,-45,-6>

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Example Question #191 : Vectors And Vector Operations

Find the angle between the vectors, given $\vec{A}=\left \langle 1, 0, 2\right \rangle, \vec{B}=\left \langle -1, 1, 3\right \rangle$ , $\left | AXB\right | = \sqrt{30}$

Possible Answers:

$\sin(\sqrt{\frac{6}{11}})$

$\sin^{-1}(\sqrt{\frac{11}{6}})$

$\sin^{-1}(\sqrt{\frac{6}{11}})$

$\cos^{-1}(\sqrt{\frac{6}{11}})$

Correct answer:

$\sin^{-1}(\sqrt{\frac{6}{11}})$

Explanation:

To find the angle between the vectors (denoted by theta), we must use the following fact:

$\left | \vec{A}\right |\left | \vec{B}\right |\sin(\theta)=\left | \vec{A}X\vec{B}\right |$

We must find the magnitude of vectors A and B, by taking the square root of the sum of the squares of each component:

$\left | \vec{A}\right |= \sqrt{1^2+0^2+2^2}= \sqrt{5}$ , $\left | \vec{B}\right |= \sqrt{(-1)^2+1^2+3^2}=\sqrt{11}$

Plugging this into our formula, we get

$\sin(\theta)=\frac{\sqrt{30}}{\sqrt{55}}$

$\theta=\sin^{-1}(\sqrt{\frac{6}{11}})$

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Example Question #191 : Vectors And Vector Operations

Find the cross product of the two vectors:

$\left \langle 4, 0, 1\right \rangle, \left \langle 2, 7, 3\right \rangle$

Possible Answers:

$\left \langle -7, 28, -10\right \rangle$

$\left \langle 7, 10, 28\right \rangle$

$\left \langle 7, 14, 28\right \rangle$

$\left \langle -7, -10, 28\right \rangle$

Correct answer:

$\left \langle -7, -10, 28\right \rangle$

Explanation:

First, we can write the determinant in order to take the cross product of the two vectors:

$\begin{vmatrix} \hat{i}& \hat{j}&\hat{k} \\ 4&0 &1 \\ 2&7 &3 \end{vmatrix}$

where i, j, and k are the unit vectors corresponding to the x, y, and z direction respectively.

Next, we take the cross product. One can do this by multiplying across from the top left to the lower right, and continuing downward, and then subtracting the terms multiplied from top right to the bottom left:

$0\hat{i}+28\hat{k}+2\hat{j}-(0\hat{k}+7\hat{i}+12\hat{j})=-7\hat{i}-10\hat{j}+28\hat{k} =\left \langle -7, -10, 28\right \rangle$

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Example Question #191 : Vectors And Vector Operations

Find the cross product of the vectors $\left \langle 2,4,-6\right \rangle$ and $\left \langle 6,-1,1\right \rangle$

Possible Answers:

$\left \langle 10,38,22\right \rangle$

$\left \langle 10,-37,22\right \rangle$

$\left \langle -2,-38,22\right \rangle$

$\left \langle 19,-18,25\right \rangle$

Correct answer:

$\left \langle -2,-38,22\right \rangle$

Explanation:

To find the cross product of two vectors $a=\left \langle a_1,a_2,a_3\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ , you find the determinant of the 3x3 matrix $\begin{bmatrix} \hat{i}&\hat{j} &\hat{k} \\ a_1&a_2 &a_3 \\ b_1&b_2 & b_3 \end{bmatrix}$

$determinant=\hat{i}(a_2b_3-b_2a_3)-\hat{j}(a_1b_3-b_1a_3)+\hat{k}(a_1b_2-b_1a_2)$

Using this formula, we evaluate using the vectors from the problem statement:

$\hat{i}(4-6)-\hat{j}(2+36)+\hat{k}(-2+24)=\left \langle 10,-38,22\right \rangle$

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Example Question #201 : Vectors And Vector Operations

Find the cross product of the vectors $\left \langle 10,-2,1\right \rangle$ and $\left \langle 5,-9,0\right \rangle$

Possible Answers:

$\left \langle 9,-5,-80\right \rangle$

$\left \langle 9,8,-40\right \rangle$

$\left \langle 18,10,-81\right \rangle$

$\left \langle 9,5,-80\right \rangle$

Correct answer:

$\left \langle 9,5,-80\right \rangle$

Explanation:

$determinant=\hat{i}(a_2b_3-b_2a_3)-\hat{j}(a_1b_3-b_1a_3)+\hat{k}(a_1b_2-b_1a_2)$

Using this formula, we evaluate using the vectors from the problem statement:

$\hat{i}(0+9)-\hat{j}(0-5)+\hat{k}(-90+10)=\left \langle 9,5,-80\right \rangle$

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Example Question #93 : Cross Product

Find the cross product of the vectors $\left \langle 0,-1,3\right \rangle$ and $\left \langle 7,7,-2\right \rangle$

Possible Answers:

$\left \langle -19,-21,7\right \rangle$

$\left \langle -19,20,1\right \rangle$

$\left \langle -19,21,7\right \rangle$

$\left \langle -16,2,7\right \rangle$

Correct answer:

$\left \langle -19,21,7\right \rangle$

Explanation:

$determinant=\hat{i}(a_2b_3-b_2a_3)-\hat{j}(a_1b_3-b_1a_3)+\hat{k}(a_1b_2-b_1a_2)$

Using this formula, we evaluate using the vectors from the problem statement:

$\hat{i}(2-21)-\hat{j}(0-21)+\hat{k}(0+7)=\left \langle -19,21,7\right \rangle$

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Example Question #94 : Cross Product

Find the cross product of the vectors $\left \langle 9,-7,2\right \rangle$ and $\left \langle 3,-9,0\right \rangle$

Possible Answers:

$\left \langle 18,-6,-60\right \rangle$

$\left \langle 8,16,-20\right \rangle$

$\left \langle 18,6,-60\right \rangle$

$\left \langle 18,6,-10\right \rangle$

Correct answer:

$\left \langle 18,6,-60\right \rangle$

Explanation:

$determinant=\hat{i}(a_2b_3-b_2a_3)-\hat{j}(a_1b_3-b_1a_3)+\hat{k}(a_1b_2-b_1a_2)$

Using this formula, we evaluate using the vectors from the problem statement:

$\hat{i}(0+18)-\hat{j}(0-6)+\hat{k}(-81+21)=\left \langle 18,6,-60\right \rangle$

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Example Question #95 : Cross Product

Find the cross product of the vectors $\left \langle 10,3,13\right \rangle$ and $\left \langle 0,5,6\right \rangle$

Possible Answers:

$\left \langle -47,-60,53\right \rangle$

$\left \langle -45,-65,55\right \rangle$

$\left \langle -47,-60,50\right \rangle$

$\left \langle -47,-63,50\right \rangle$

Correct answer:

$\left \langle -47,-60,50\right \rangle$

Explanation:

To determine the cross product of two vectors $a=\left \langle a_1,a_2,a_3\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ , we find the determinant of the 3x3 matrix $m=\begin{bmatrix} \hat{i}& \hat{j} & \hat{k}\\ a_1&a_2 &a_3 \\ b_1& b_2& b_3 \end{bmatrix}$ , using the formula

$determinant(m)=\hat{i}(a_2b_3-b_2a_3)-\hat{j}(a_1b_3-b_1a_3)+\hat{k}(a_1b_2-b_1a_2)$

Using the vectors from the problem statement, we get

$\hat{i}(18-65)-\hat{j}(60-0)+\hat{k}(50-0)=\left \langle -47,-60,50\right \rangle$

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Example Question #95 : Cross Product

Find the cross product $\mathbf a\times \mathbf b$ , where

$\mathbf a = \left \langle 2,4,0\right \rangle$ and $\mathbf b \left \langle 5,-1,2\right \rangle$ .

Possible Answers:

$9\mathbf i-4 \mathbf j-18 \mathbf k$

$8\mathbf i-4 \mathbf j+22 \mathbf k$

$10\mathbf i+4 \mathbf j+18 \mathbf k$

$8\mathbf i+4 \mathbf j-18 \mathbf k$

$8\mathbf i-4 \mathbf j-22 \mathbf k$

Correct answer:

$8\mathbf i-4 \mathbf j-22 \mathbf k$

Explanation:

Recall the definition of the cross product $\mathbf a\times \mathbf b$ of two vectors $\mathbf a = \left \langle a_1, a_2, a_3\right \rangle$ and $\mathbf b = \left \langle b_1, b_2, b_3\right \rangle$ , in terms of determinants:

$\mathbf a\times \mathbf b= \begin{vmatrix} \mathbf i~\ \mathbf j~\ \mathbf k\\a_1\ a_2\ a_3\\b_1\ b_2\ b_3 \end{vmatrix}$

where $\mathbf i$ , $\mathbf j$ , and $\mathbf k$ are the standard basis vectors pointing in the directions of the positive -, - and -axes, respectively. We can apply this definition to calculate the cross product of $\mathbf a = \left \langle 2,4,0\right \rangle$ and $\mathbf b \left \langle 5,-1,2\right \rangle$ , as follows:

$\mathbf a\times \mathbf b= \begin{vmatrix} \mathbf i~\ \mathbf j~\ \mathbf k\\2~\ 4\ ~0\\5 -1\ 2 \end{vmatrix}$

$= \begin{vmatrix} 4\ \0\\ -1 \2\end{vmatrix} \mathbf i + \begin{vmatrix} 2\ ~0\\ 5 ~\2\end{vmatrix} \mathbf j + \begin{vmatrix} 2\ ~4\\ 5 -1\end{vmatrix} \mathbf k$

$=(8-0) \mathbf i-(4-0) \mathbf j+(-2-20) \mathbf k$

$=8 \mathbf i -4 \mathbf j-22 \mathbf k$

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Example Question #451 : Calculus 3

Let u=(2,4,5) , and v=(-2,6,10) .

Find $u\times v$ .

Possible Answers:

$u\times v=(-10,30,-20)$

$u\times v=(10,-30,20)$

$u\times v=(0,-3,2)$

$u\times v=(10,30,20)$

$u\times v=(40,-20,12)$

Correct answer:

$u\times v=(10,-30,20)$

Explanation:

We are trying to find the cross product between and .

Recall the formula for cross product.

If u=(u_x,u_y,u_z) , and v=(v_x,v_y,v_z) , then

$u\times v=(u_yv_z-u_zv_y, -u_xv_z+u_zv_x, u_xv_y-u_yv_x)$ .

Now apply this to our situation.

$u\times v=(4(10)-5(6),-2(10)+5(-2), 2(6)-4(-2))$

$u\times v=(40-30,-20-10, 12+8)$

$u\times v=(10,-30,20)$

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Calculus 3 : Cross Product

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #191 : Vectors And Vector Operations

Example Question #191 : Vectors And Vector Operations

Example Question #191 : Vectors And Vector Operations

Example Question #191 : Vectors And Vector Operations

Example Question #201 : Vectors And Vector Operations

Example Question #93 : Cross Product

Example Question #94 : Cross Product

Example Question #95 : Cross Product

Example Question #95 : Cross Product

Example Question #451 : Calculus 3

All Calculus 3 Resources

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