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Calculus 3 : Vectors and Vector Operations

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Example Questions

Example Question #191 : Vectors And Vector Operations

Find the cross product of the two vectors:

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

Possible Answers:

$\left \langle 16, 10, 14\right \rangle$

$\left \langle -14, 10, 14\right \rangle$

$\left \langle 16, -14, -6\right \rangle$

$\left \langle -16, 10, 14\right \rangle$

Correct answer:

$\left \langle -16, 10, 14\right \rangle$

Explanation:

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

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

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:

$-15\hat{i}+4\hat{k}+-2\hat{j}-(-10\hat{k}+\hat{i}-12\hat{j})= -16\hat{i}+10\hat{j}+14\hat{k}=\left \langle -16, 10, 14\right \rangle$

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Example Question #82 : Cross Product

Find the cross product between the vectors $\left \langle 3,7,1\right \rangle$ and $\left \langle 0,4,5\right \rangle$

Possible Answers:

$\left \langle 31,-14,12\right \rangle$

$\left \langle 31,15,12\right \rangle$

$\left \langle 31,-15,12\right \rangle$

$\left \langle 25,-15,10\right \rangle$

Correct answer:

$\left \langle 31,-15,12\right \rangle$

Explanation:

To find the cross product between 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 take 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 the vectors from the problem statement, we get

$determinant=\hat{i}(35-4)-\hat{j}(15-0)+\hat{k}(12-0)=\left \langle 31,-15,12\right \rangle$

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Example Question #193 : Vectors And Vector Operations

Find the cross product between the vectors $\left \langle 0,1,-7\right \rangle$ and $\left \langle -9,3,0\right \rangle$

Possible Answers:

$\left \langle -21,63,9\right \rangle$

$\left \langle -21,-63,9\right \rangle$

$\left \langle -21,-63,10\right \rangle$

$\left \langle -19,-63,9\right \rangle$

Correct answer:

$\left \langle -21,-63,9\right \rangle$

Explanation:

To find the cross product between 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 take 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 the vectors from the problem statement, we get

$determinant=\hat{i}(0-21)-\hat{j}(0+63)+\hat{k}(0+9)=\left \langle -21,63,9\right \rangle$

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Example Question #192 : Vectors And Vector Operations

Find the cross product between the vectors $\left \langle 2,7,3\right \rangle$ and $\left \langle 0,5,3\right \rangle$

Possible Answers:

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

$\left \langle 9-6,11\right \rangle$

$\left \langle 3,-6,10\right \rangle$

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

Correct answer:

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

Explanation:

To find the cross product between 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 take 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 the vectors from the problem statement, we get

$determinant=\hat{i}(21-15)-\hat{j}(6-0)+\hat{k}(10-0)=\left \langle 6,-6,10\right \rangle$

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Example Question #193 : Vectors And Vector Operations

Find the cross product of the two vectors.

u=<2,6,7>

v=<-2,10,5>

Possible Answers:

<-40,-24,32>

<-2,6,5>

<8,12,16>

<2,10,7>

Correct answer:

<-40,-24,32>

Explanation:

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

$\begin{vmatrix} i &j &k \\ 2&6 &7 \\ -2& 10& 5\end{vmatrix}$

$=det(\begin{vmatrix} 6&7 \\ 10& 5\end{vmatrix})i-det(\begin{vmatrix} 2&7 \\ -2& 5\end{vmatrix})j+det(\begin{vmatrix} 2&6 \\ -2& 10\end{vmatrix})k$

=[6(5)-7(10)]i-[2(5)-7(-2)]j+[2(10)-6(-2)]k

=[30-70]i-[10+14]j+[20+12]k

=-40i-24j+32k

As such, the cross product is

<-40,-24,32>

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Example Question #194 : Vectors And Vector Operations

Find the cross product of the two vectors.

u=<6,-3,-1>

v=<4,1,-1>

Possible Answers:

<6,1,-1>

<0,5,3>

<2,10,18>

<4,-3,-1>

Correct answer:

<2,10,18>

Explanation:

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

$\begin{vmatrix} i &j &k \\ 6&-3 &\-1 \\ 4& 1& -1\end{vmatrix}$

$=det(\begin{vmatrix} -3&1 \\ 1& -1\end{vmatrix})i-det(\begin{vmatrix} 6&1 \\ 4& -1\end{vmatrix})j+det(\begin{vmatrix} 6&-3 \\ 4& 1\end{vmatrix})k$

=[-3(-1)-1(1)]i-[6(-1)-4(1)]j+[6(1)-4(-3)]k

=[3-1]i-[-6-4]j+[6+12]k

=2i+10j+18k

As such, the cross product is

<2,10,18>

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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.

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

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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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Calculus 3 : Vectors and Vector Operations

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #191 : Vectors And Vector Operations

Example Question #82 : Cross Product

Example Question #193 : Vectors And Vector Operations

Example Question #192 : Vectors And Vector Operations

Example Question #193 : Vectors And Vector Operations

Example Question #194 : 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 #191 : Vectors And Vector Operations

All Calculus 3 Resources

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