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Calculus 3 : Cross Product

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Example Questions

Example Question #167 : Vectors And Vector Operations

Find the cross product between the vectors $\left \langle 1,5,2\right \rangle$ and $\left \langle 3,-2,-1\right \rangle$

Possible Answers:

$\left \langle 2,5,-2\right \rangle$

$\left \langle -1,-7,-17\right \rangle$

$\left \langle 2,-4,5\right \rangle$

$\left \langle -1,7,-17\right \rangle$

Correct answer:

$\left \langle -1,7,-17\right \rangle$

Explanation:

To find the cross product between vectors $a=\left \langle x_1,y_1,z_1\right \rangle$ and $b=\left \langle x_2,y_2,z_2\right \rangle$ , you find the determinant of the 3x3 matrix $\begin{bmatrix} i& j&k \\ x_1&y_1 &z_1 \\ x_2&y_2 &z_2 \end{bmatrix}$ . The determinant in this case is $i(-5+4)-j(-1-6)+k(-2-15)=\left \langle -1,7,-17\right \rangle$

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Example Question #168 : Vectors And Vector Operations

Two vectors u and v and their cross product have the following magnitudes:

$|\textbf{u}|=2\sqrt{2}\:,\: \: |\textbf{v}|=4, \:\: |\textbf{u}\times \textbf{v}|=8$

What is the angle between the two vectors?

Possible Answers:

$90\degree$

$30\degree$

$45\degree$

$20\degree$

$60\degree$

Correct answer:

$45\degree$

Explanation:

The magnitude of the cross product of two vectors u and v can be defined by the cross product and the angle theta between them as follows:

$|\textbf{u}\times\textbf{v}|=|\textbf{u}||\textbf{v}|\sin\theta$

Solving for theta and substituting the give quantities, we obtain:

$\theta=\sin^{-1}\left ( \frac{|\mathbf{u\times v|}}{|\mathbf{u}||\mathbf{v}|}\right )= \sin^{-1}\left ( \frac{8}{2\sqrt{2}\cdot 4}\right )=\sin^{-1}\left ( \frac{1}{\sqrt{2}}\right ) =45\degree$

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Example Question #169 : Vectors And Vector Operations

Find the cross product of the two vectors, written in vector form:

$\left \langle z, 3y, x^2\right \rangle \times \left \langle 9, 0, 0\right \rangle$

Possible Answers:

$\left \langle 0, 9x^2, 27y\right \rangle$

$\left \langle 0, 9x^2, -27y\right \rangle$

$\left \langle 0, 0, 0\right \rangle$

$\left \langle 9x^2,0, -27y\right \rangle$

Correct answer:

$\left \langle 0, 9x^2, -27y\right \rangle$

Explanation:

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

$\begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ z& 3y&x^2 \\ 9&0 &0 \end{vmatrix}$

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:

$9x^2\hat{j}-(27y\hat{k})=\left \langle 0, 9x^2, -27y\right \rangle$

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Example Question #170 : Vectors And Vector Operations

Find the cross product between the vectors $\left \langle 4,5,1\right \rangle$ and $\left \langle 2,0,1\right \rangle$

Possible Answers:

$\left \langle 5,-2,10\right \rangle$

$\left \langle 5,2,10\right \rangle$

$\left \langle 1,-2,5\right \rangle$

$\left \langle 3,-5,1\right \rangle$

Correct answer:

$\left \langle 5,-2,10\right \rangle$

Explanation:

To find the cross product between the vectors, we find the determinant of the 3x3 matrix $\begin{bmatrix} i& j& k\\ a&b &c \\ x&y &z \end{bmatrix}$ , where one vector is $\left \langle a,b,c\right \rangle$ and the other is $\left \langle x,y,z\right \rangle$ .

Using the formula for the determinant as determinant=i(bz-yc)-j(az-xc)+k(ay-xb) . we get:

$i(5-0)-j(4-2)+k(0-10)=\left \langle 5,-2,-10\right \rangle$

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Example Question #171 : Vectors And Vector Operations

Find the cross product of the two vectors, given in standard form:

$\left \langle x, z, z^2\right \rangle, \left \langle 2, 2, 2\right \rangle$

Possible Answers:

$(2z-2z^2)\hat{i}+(2z^2-2x)\hat{j}+(2x-2z)\hat{k}$

$(2z+2z^2)\hat{i}+(2z^2+2x)\hat{j}+(2x+2z)\hat{k}$

$2z\hat{i}+2z^2\hat{j}+2x\hat{k}$

Correct answer:

$(2z-2z^2)\hat{i}+(2z^2-2x)\hat{j}+(2x-2z)\hat{k}$

Explanation:

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

$\begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ x&z &z^2 \\ 2&2 &2 \end{vmatrix}$

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

$2z\hat{i}+2x\hat{k}+2z^2\hat{j}-(2z\hat{k}+2z^2\hat{i}+2x\hat{j})=(2z-2z^2)\hat{i}+(2z^2-2x)\hat{j}+(2x-2z)\hat{k}$

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Example Question #172 : Vectors And Vector Operations

Find the cross product of the two vectors, in vector form:

$\left \langle 2e^x, x, y\right \rangle, \left \langle 3, 4, 5\right \rangle$

Possible Answers:

$\left \langle 5x+4y, 3y+10e^x, 8e^x+3x\right \rangle$

$\left \langle 6e^x,4x,5y \right \rangle$

6e^x+4x+5y

$\left \langle 5x-4y, 3y-10e^x, 8e^x-3x\right \rangle$

Correct answer:

$\left \langle 5x-4y, 3y-10e^x, 8e^x-3x\right \rangle$

Explanation:

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

$\begin{vmatrix}\hat{i}& \hat{j}&\hat{k} \\ 2e^x&x & y\\ 3&4 &5 \end{vmatrix}$

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

$5x\hat{i}+8e^x\hat{k}+3y\hat{j}-(3x\hat{k}+4y\hat{i}+10e^x\hat{j})=(5x-4y)\hat{i}+(3y-10e^x)\hat{j}+(8e^x-3x)\hat{k}=\left \langle 5x-4y, 3y-10e^x, 8e^x-3x\right \rangle$

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Example Question #173 : Vectors And Vector Operations

Find the cross product of the two vectors, in standard form:

$\left \langle a, b, c\right \rangle, \left \langle 10, 15, 4\right \rangle$

Possible Answers:

$4b\hat{i}+ 10c\hat{j}+10b\hat{k}$

$(4b+15c)\hat{i}+ (10c+4a)\hat{j}+(15a+10b)\hat{k}$

$-15c\hat{i}-4a\hat{j}-10b\hat{k}$

$(4b-15c)\hat{i}+ (10c-4a)\hat{j}+(15a-10b)\hat{k}$

Correct answer:

$(4b-15c)\hat{i}+ (10c-4a)\hat{j}+(15a-10b)\hat{k}$

Explanation:

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

$\begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ a&b &c \\ 10&15 &4 \end{vmatrix}$

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

$4b\hat{i}+15a\hat{k}+10c\hat{j}-(10b\hat{k}+15c\hat{i}+4a\hat{j})=(4b-15c)\hat{i}+(10c-4a)\hat{j}+(15a-10b)\hat{k}$

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Example Question #174 : Vectors And Vector Operations

Find the cross product of the two vectors, given in standard form:

$\left \langle a, b, c\right \rangle, \left \langle 2x, 3y, z\right \rangle$

Possible Answers:

$(bz)\hat{i}+(2xc)\hat{j}+(3ya)\hat{k}$

$(bz+3yc)\hat{i}+(2xc+az)\hat{j}+(3ya+2xb)\hat{k}$

$(bz-3yc)\hat{i}+(2xc-az)\hat{j}+(3ya-2xb)\hat{k}$

Correct answer:

$(bz-3yc)\hat{i}+(2xc-az)\hat{j}+(3ya-2xb)\hat{k}$

Explanation:

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

$\begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ a&b &c \\ 2x&3y &z \end{vmatrix}$

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

$(bz)\hat{i}+(3ya)\hat{k}+(2xc)\hat{j}-((2xb)\hat{k}+(3yc)\hat{i}+(az)\hat{j})=(bz-3yc)\hat{i}+(2xc-az)\hat{j}+(3ya-2xb)\hat{k}$

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Example Question #175 : Vectors And Vector Operations

Find the cross product between the two vectors, given in vector form:

$\left \langle 9, 5, 3\right \rangle, \left \langle 2, 0, 0\right \rangle$

Possible Answers:

$\left \langle 0, -10, 6\right \rangle$

$\left \langle 0, 6, -10\right \rangle$

$\left \langle -10, 6, 0\right \rangle$

$\left \langle 0, 6, 10\right \rangle$

Correct answer:

$\left \langle 0, 6, -10\right \rangle$

Explanation:

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

$\begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ 9&5 &3 \\ 2&0 &0 \end{vmatrix}$

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

$6\hat{j}-(10\hat{k})=6\hat{j}-10\hat{k}=\left \langle 0, 6, -10\right \rangle$

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Example Question #176 : Vectors And Vector Operations

Find the cross product of the two vectors

<0,5,3>

<10,-2,13>

Possible Answers:

<50,-6,0>

<0,-10,26>

<71,30,-50>

<0,26,30>

Correct answer:

<71,30,-50>

Explanation:

The cross product is defined as the determinant of the matrix

$\begin{vmatrix}i &j &k \\ 0&5 &3 \\10 &-2 &13 \end{vmatrix}$

Which is

$=det(\begin{vmatrix} 5& 3\\-2 &13 \end{vmatrix})i-det(\begin{vmatrix} 0&3 \\10 &13 \end{vmatrix})j+det(\begin{vmatrix} 0&5 \\10 &-2 \end{vmatrix})k$

=(5*13-(-2)*3)i-(0*13-10*3)j+(0*(-2)-10*5)k

=(65+6)i-(-30)j+(-50)k

=71i+30j-50k

Thus the cross product is

<71,30,-50>

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Calculus 3 : Cross Product

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #167 : Vectors And Vector Operations

Example Question #168 : Vectors And Vector Operations

Example Question #169 : Vectors And Vector Operations

Example Question #170 : Vectors And Vector Operations

Example Question #171 : Vectors And Vector Operations

Example Question #172 : Vectors And Vector Operations

Example Question #173 : Vectors And Vector Operations

Example Question #174 : Vectors And Vector Operations

Example Question #175 : Vectors And Vector Operations

Example Question #176 : Vectors And Vector Operations

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