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Calculus 3 : Vectors and Vector Operations

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #181 : Vectors And Vector Operations

Find the cross product between the vectors $\left \langle 2,-5,7\right \rangle$ and $\left \langle 0,-3,1\right \rangle$

Possible Answers:

$\left \langle 16,-2,-6\right \rangle$

$\left \langle 16,-1,-6\right \rangle$

$\left \langle 16,2,-6\right \rangle$

$\left \langle 15,-2,-6\right \rangle$

Correct answer:

$\left \langle 16,-2,-6\right \rangle$

Explanation:

To find the cross product between two vectors $a=\left \langle a_1,a_2,a_3\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ , we find the determinant of the 3x3 matrix $\begin{bmatrix} i&j &k \\ a_1&a_2 &a_3 \\ b_1&b_2 &b_3 \end{bmatrix}$ which follows the formula determinant= i(a_2b_3-b_2a_3)-j(a_1b_3-b_1a_3)+k(a_1b_2-b_1a_2)

Applying to the vectors from the problem statement, we get

$i(-5+21)-j(2-0)+k(-6-0)=\left \langle 16,-2,-6\right \rangle$

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Example Question #431 : Calculus 3

Find the cross product between the vectors $\left \langle 1,-4,6\right \rangle$ and $\left \langle -1,0,5\right \rangle$

Possible Answers:

$\left \langle -20,-10,-6\right \rangle$

$\left \langle -20,11,-4\right \rangle$

$\left \langle -20,-11,-4\right \rangle$

$\left \langle -22,-11,-4\right \rangle$

Correct answer:

$\left \langle -20,-11,-4\right \rangle$

Explanation:

Applying to the vectors from the problem statement, we get

$i(-20-0)-j(5+6)+k(0-4)=\left \langle -20,-11,-4\right \rangle$

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Example Question #71 : Cross Product

Find the cross product between the two vectors:

$\left \langle 4, 9, 1\right \rangle, \left \langle 3, -1, -1\right \rangle$

Possible Answers:

$\left \langle -8, 7, -26\right \rangle$

$\left \langle -9, -4, 3\right \rangle$

$\left \langle -10, -1, 18\right \rangle$

$\left \langle 8, 7, 18\right \rangle$

Correct answer:

$\left \langle -8, 7, -26\right \rangle$

Explanation:

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

$\begin{vmatrix} \hat{i}& \hat{j}&\hat{k} \\ 4& 9 &1 \\ 3& -1&-1 \end{vmatrix}$

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:

$-9\hat{i}+-4\hat{k}+3\hat{j}-(22\hat{k}-\hat{i}-4\hat{j})= -8\hat{i}+7\hat{j}+18\hat{k}=\left \langle -8, 7, 18\right \rangle$

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Example Question #432 : Calculus 3

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

Possible Answers:

$\left \langle -16,10,-7\right \rangle$

$\left \langle -15,12,-6\right \rangle$

$\left \langle -16,-12,-7\right \rangle$

$\left \langle -16,12,-7\right \rangle$

Correct answer:

$\left \langle -16,12,-7\right \rangle$

Explanation:

To find the cross product between two vectors $a=\left \langle a_1,a_2,a_3\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ , we 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

$determinant=\hat{i}(-16-0)-\hat{j}(-12-0)+\hat{k}(-3-4)=\left \langle -16,12,-7\right \rangle$

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Example Question #433 : Calculus 3

Find the cross product between the vectors $\left \langle 0,1,7\right \rangle$ and $\left \langle 5,1,-2\right \rangle$

Possible Answers:

$\left \langle 9,35,-5\right \rangle$

$\left \langle 9,35,-3\right \rangle$

$\left \langle 8,25,-5\right \rangle$

$\left \langle 9,-35,-5\right \rangle$

Correct answer:

$\left \langle 9,35,-5\right \rangle$

Explanation:

To find the cross product between two vectors $a=\left \langle a_1,a_2,a_3\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ , we 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

$determinant=\hat{i}(2+7)-\hat{j}(0-35)+\hat{k}(0-5)=\left \langle 9,35,-5\right \rangle$

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Example Question #76 : Cross Product

Find the cross product between the vectors $\left \langle 0,8,4\right \rangle$ and $\left \langle 1,4,-5\right \rangle$

Possible Answers:

$\left \langle -61,-4,-8\right \rangle$

$\left \langle -51,4,-4\right \rangle$

$\left \langle -61,4,-8\right \rangle$

$\left \langle -60,5,-9\right \rangle$

Correct answer:

$\left \langle -61,4,-8\right \rangle$

Explanation:

To find the cross product between two vectors $a=\left \langle a_1,a_2,a_3\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ , we 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

$determinant=\hat{i}(-45-16)-\hat{j}(0-4)+\hat{k}(0-8)=\left \langle -61,4,-8\right \rangle$

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Example Question #81 : Cross Product

Find the cross product between the vectors $\left \langle 9,1,8\right \rangle$ and $\left \langle -1,0,5\right \rangle$

Possible Answers:

$\left \langle 7,-51,1\right \rangle$

$\left \langle 5,-50,1\right \rangle$

$\left \langle 5,53,1\right \rangle$

$\left \langle 5,-53,1\right \rangle$

Correct answer:

$\left \langle 5,-53,1\right \rangle$

Explanation:

To find the cross product between two vectors $a=\left \langle a_1,a_2,a_3\right \rangle$ and $b=\left \langle b_1,b_2,b_3\right \rangle$ , we 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

$determinant=\hat{i}(5-0)-\hat{j}(45+8)+\hat{k}(0+1)=\left \langle 5,-53,1\right \rangle$

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Example Question #432 : Calculus 3

Determine the dot product between the two vectors:

$\left \langle x, xy, z\right \rangle, \left \langle 3y, 3z, 5x\right \rangle$

Possible Answers:

45x^3y^2z^2

$\left \langle 3xy, 3xyz, 5xz\right \rangle$

6xy+5xz

3xy+3xyz+5xz

Correct answer:

3xy+3xyz+5xz

Explanation:

The dot product is given by the sum of the products of the corresponding components of each vector (for example, $\left \langle a,b\right \rangle\cdot \left \langle x,y\right \rangle= ax+by$ )

Our final answer is

3xy+3xyz+5xz

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Example Question #435 : Calculus 3

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

$4\hat{i}+7\hat{j}, \hat{i}+8\hat{j}+11\hat{k}$

Possible Answers:

$77\hat{i}+44\hat{j}+25\hat{k}$

$77\hat{i}-44\hat{j}+25\hat{k}$

$77\hat{i}+44\hat{j}+39\hat{k}$

$44\hat{i}-77\hat{j}+25\hat{k}$

Correct answer:

$77\hat{i}-44\hat{j}+25\hat{k}$

Explanation:

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

$\begin{vmatrix} \hat{i}& \hat{j}&\hat{k} \\ 4&7 &0 \\ 1&8&11 \end{vmatrix}$

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

$77\hat{i}+32\hat{k}+0\hat{j}-(7\hat{k}+0\hat{i}+44\hat{j})=77\hat{i}-44\hat{j}+25\hat{k}$

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Example Question #433 : Calculus 3

Find the cross product of the following two vectors:

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

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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 #181 : Vectors And Vector Operations

Example Question #431 : Calculus 3

Example Question #71 : Cross Product

Example Question #432 : Calculus 3

Example Question #433 : Calculus 3

Example Question #76 : Cross Product

Example Question #81 : Cross Product

Example Question #432 : Calculus 3

Example Question #435 : Calculus 3

Example Question #433 : Calculus 3

All Calculus 3 Resources

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