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Calculus 3 : Cross Product

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Example Questions

Example Question #51 : Cross Product

Find the cross product of the two vectors

u=<4,4,0>

v=<0,0,5>

Possible Answers:

<4,4,5>

<20,-20,0>

<0,13,0>

<0,0,0>

Correct answer:

<20,-20,0>

Explanation:

The cross product of the two vectors

a=<a_1,a_2,a_3>

b=<b_1,b_2,b_3>

is defined as the determinant of the matrix

$\begin{vmatrix} i&j &k \\ a_1&a_2 &a_3 \\ b_1& b_2&b_3 \end{vmatrix}$

For the vectors in the problem we solve the determinant of the matrix

$\begin{vmatrix} i&j &k \\4&4 &0 \\ 0& 0&5 \end{vmatrix}$

which is

$det(\begin{vmatrix} 4&0 \\ 0&5 \end{vmatrix})i-det(\begin{vmatrix} 4&0 \\ 0&5 \end{vmatrix})j +det(\begin{vmatrix} 4&4 \\ 0&0 \end{vmatrix})k$

=(4*5-0*0)i-(4*5-0*0)j+(4*0-0*0)k

=20i-20j+0k

=<20,-20,0>

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Example Question #52 : Cross Product

Find the cross product of the two vectors:

$\left \langle -2, 4, 5\right \rangle, \left \langle 1, 2, 3\right \rangle$

Possible Answers:

$\left \langle 2, -1, -8\right \rangle$

$\left \langle 2, 11, -8\right \rangle$

$\left \langle 16, 11, 0\right \rangle$

$\left \langle 1, 2, 3\right \rangle$

Correct answer:

$\left \langle 2, 11, -8\right \rangle$

Explanation:

To find the cross product of two vectors, we must write the determinant of the vectors:

$\vec{a} X \vec{b} = \begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ -2& 4&5 \\ 1& 2 &3 \end{vmatrix}$

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:

$12\hat{i}-4\hat{k}+5\hat{j}-(4\hat{k}+10{i}-6\hat{j})=2\hat{i}+11\hat{j}-8\hat{k}$

The vector is written in unit vector notation. We simply take the coefficients of our unit vectors and correspond them to x, y, and z:

$\left \langle 2, 11, -8\right \rangle$

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Example Question #53 : Cross Product

Determine the cross product (in vector notation) of the vectors

$\left \langle 4, 5, 2 \right \rangle$ and $\left \langle 2, 0, 1 \right \rangle$

Possible Answers:

$\left \langle 5, 8, 10 \right \rangle$

$\left \langle 5, 0, -10 \right \rangle$

$\left \langle 0, 5, 10 \right \rangle$

Correct answer:

$\left \langle 5, 0, -10 \right \rangle$

Explanation:

We must first write the determinant in order to take the cross product of the two vectors:

$\begin{vmatrix} \hat{i} & \hat{j} &\hat{k} \\4 &5 &2 \\2 &0 & 1 \end{vmatrix}$

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

$5\hat{i}+0\hat{k}+4\hat{j}-(10\hat{k}+0\hat{i}+4\hat{j})=5\hat{i}-10\hat{k}$

Writing this in vector notation, we get

$\left \langle 5, 0, -10 \right \rangle$

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Example Question #54 : Cross Product

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

$\left \langle 5, 1, 3\right \rangle, \left \langle x, y^2, z\right \rangle$

Possible Answers:

$z\hat{i}+(3x-5z)\hat{j}+(5y^2-x)\hat{k}$

$(z-3y^2)\hat{i}+(3x-5z)\hat{j}+(5y^2-x)\hat{k}$

$(z-3y^2)\hat{i}+(3x-5z)\hat{j}+(y^2+x)\hat{k}$

$(z+3y^2)\hat{i}+(3x+5z)\hat{j}+(5y^2+x)\hat{k}$

Correct answer:

$(z-3y^2)\hat{i}+(3x-5z)\hat{j}+(5y^2-x)\hat{k}$

Explanation:

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

$\begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ 5&1 &3 \\ x& y^2&z \end{vmatrix}$

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

$z\hat{i}+5y^2\hat{k}+3x\hat{j}-(x\hat{k}+3y^2\hat{i}+5z\hat{j})$

which simplified becomes

$(z-3y^2)\hat{i}+(3x-5z)\hat{j}+(5y^2-x)\hat{k}$

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Example Question #161 : Vectors And Vector Operations

Find the cross product of

$\left \langle x, \cos(z), y^2\right \rangle, \left \langle 3, 4, 6\right \rangle$

written in vector form

Possible Answers:

$\left \langle 6\cos(z)-4y^2, 3y^2-6x, 4x-3\cos(z)\right \rangle$

$\left \langle 6\cos(z)+4y^2, 3y^2+6x, 4x+3\cos(z)\right \rangle$

$\left \langle 6\cos(z), 3y^2-6x, 4x-3\cos(z)\right \rangle$

$\left \langle -6\cos(z)+4y^2, -3y^2+6x, -4x+3\cos(z)\right \rangle$

Correct answer:

$\left \langle 6\cos(z)-4y^2, 3y^2-6x, 4x-3\cos(z)\right \rangle$

Explanation:

We first write the determinant in order to take the cross product of the two vectors:

$\begin{vmatrix} \hat{i}& \hat{j}& \hat{k}\\ x&\cos(z) &y^2 \\ 3&4 & 6 \end{vmatrix}$

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

$6\cos(z)\hat{i}+4x\hat{k}+3y^2\hat{j}-(3\cos(z)\hat{k}+4y^2\hat{i}+6x\hat{j})=(6\cos(z)-4y^2)\hat{i}+(3y^2-6x)\hat{j}+(4x-3\cos(z))\hat{k}$

which written as a vector becomes

$\left \langle 6\cos(z)-4y^2, 3y^2-6x, 4x-3\cos(z)\right \rangle$

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Example Question #162 : Vectors And Vector Operations

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

$\left \langle 3y, 0, 5z\right \rangle, \left \langle 4, 1, 2\right \rangle$

Possible Answers:

$-5z\hat{i}+(20z-6y)\hat{j}+3y\hat{k}$

-5z+20z-3y

$-5z\hat{i}+20z\hat{j}-3y\hat{k}$

$5z\hat{i}+(20z+6y)\hat{j}+3y\hat{k}$

Correct answer:

$-5z\hat{i}+(20z-6y)\hat{j}+3y\hat{k}$

Explanation:

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

$\begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ 3y&0 &5z \\ 4& 1& 2 \end{vmatrix}$

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

$3y\hat{k}+20z\hat{j}-(5z\hat{i}+6y\hat{j})=-5z\hat{i}+(20z-6y)\hat{j}+3y\hat{k}$

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Example Question #163 : Vectors And Vector Operations

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

$\left \langle xy, 3z, z\right \rangle X \left \langle 5, 0, 1\right \rangle$

Possible Answers:

$\left \langle 3z, 5z+xy, 15z\right \rangle$

$\left \langle 3z, 5z+xy, -15z\right \rangle$

$\left \langle 3z, 5z+xyz, -15z\right \rangle$

$\left \langle 3z, 5z-xy, -15z\right \rangle$

Correct answer:

$\left \langle 3z, 5z-xy, -15z\right \rangle$

Explanation:

We first must write the determinant in order to take the cross product of the two vectors:

$\begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ xy& 3z& z\\ 5& 0& 1 \end{vmatrix}$

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

$3z\hat{i}+5z\hat{j}-(15z\hat{k}+xy\hat{j})=3z\hat{i}+(5z-xy)\hat{j}-15z\hat{k}=\left \langle 3z, 5z-xy, -15z\right \rangle$

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Example Question #164 : Vectors And Vector Operations

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

$\left \langle x^2, y^2, z\right \rangle, \left \langle 0, 1, 4\right \rangle$

Possible Answers:

$(4y^2-z)\hat{i}-4x\hat{j}$

$(4y^2-z)\hat{i}+x^2\hat{k}$

$(4y^2+z)\hat{i}+4x\hat{j}+x^2\hat{k}$

$(4y^2-z)\hat{i}-4x\hat{j}+x^2\hat{k}$

Correct answer:

$(4y^2-z)\hat{i}-4x\hat{j}+x^2\hat{k}$

Explanation:

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

$\begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ x^2&y^2 &z \\ 0&1 &4 \end{vmatrix}$

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

$4y^2\hat{i}+x^2\hat{k}-(z\hat{i}+4x^2\hat{j})=(4y-2)\hat{i}-4x^2\hat{j}+x^2\hat{k}$

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Example Question #165 : Vectors And Vector Operations

Find the cross product of the following vectors, given in vector form:

$\left \langle 2, 4, 4\right \rangle,\left \langle x, y, z\right \rangle$

Possible Answers:

$\left \langle 4z-4y, 4x-4, 2y-4x\right \rangle$

$\left \langle 4z+4y, 4x+4, 2y+4x\right \rangle$

$\left \langle 2z-2y, 2x-2, y-2x\right \rangle$

4z-2y-4

Correct answer:

$\left \langle 4z-4y, 4x-4, 2y-4x\right \rangle$

Explanation:

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

$\begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ 2&4 &4 \\ x& y& z \end{vmatrix}$

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

$4z\hat{i}+2y\hat{k}+4x\hat{j}-(4x\hat{k}+4y\hat{i}+4\hat{j})=(4z-4y)\hat{i}+(4x-4)\hat{j}+(2y-4x)\hat{k}=\left \langle 4z-4y, 4x-4, 2y-4x\right \rangle$

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Example Question #166 : Vectors And Vector Operations

Find the cross product between the vectors $\left \langle 3,-1,-2\right \rangle$ and $\left \langle 4,-2,0\right \rangle$

Possible Answers:

$\left \langle 1,5,4\right \rangle$

$\left \langle 4,8,-2\right \rangle$

$\left \langle 4,6,2\right \rangle$

Correct answer:

$\left \langle 4,8,-2\right \rangle$

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 $i(0+4)-j(0-8)+k(-6+4)=\left \langle 4,8,-2\right \rangle$

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Calculus 3 : Cross Product

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #51 : Cross Product

Example Question #52 : Cross Product

Example Question #53 : Cross Product

Example Question #54 : Cross Product

Example Question #161 : Vectors And Vector Operations

Example Question #162 : Vectors And Vector Operations

Example Question #163 : Vectors And Vector Operations

Example Question #164 : Vectors And Vector Operations

Example Question #165 : Vectors And Vector Operations

Example Question #166 : Vectors And Vector Operations

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