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Calculus 3 : Vectors and Vector Operations

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Example Questions

Example Question #451 : Calculus 3

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

Possible Answers:

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

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

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

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

Correct answer:

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

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:

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

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Example Question #452 : Calculus 3

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

Possible Answers:

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

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

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

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

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 #453 : Calculus 3

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

Possible Answers:

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

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

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

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

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 #94 : 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:

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

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

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

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

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 #91 : 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:

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

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

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

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

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

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 #92 : Cross Product

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

Find $u\times v$ .

Possible Answers:

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

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

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

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

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

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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Example Question #454 : Calculus 3

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

Find $u\times v$ .

Possible Answers:

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

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

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

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

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

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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Example Question #1 : Vector Addition

Let $\vec{a}=(1,2,3)$ , $\vec{b}=(10,-10,3)$ , and $\vec{c}=(0,0,1)$ ,

find $\vec{a}+\vec{b}+\vec{c}$

Possible Answers:

(11, 7, -8)

(0,0,1)

(11, 0, 1)

(7, -8,11)

(11, -8, 7)

Correct answer:

(11, -8, 7)

Explanation:

In order to find $\vec{a}+\vec{b}+\vec{c}$ , we need to add like components. $\vec{a}+\vec{b}+\vec{c}=(1+10+0,2-10+0,3+3+1)=(11,-8,7)$

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Example Question #1 : Vector Addition

Given the vectors

$\small v=(2,3)$ and $\small w=(2,100)$ , compute $\small v+w$ .

Possible Answers:

$\small \small \small v+w=(0,100)$

$\small \small \small v+w=(104,5)$

$\small \small v+w=(103,4)$

$\small \small v+w=(4,103)$

Correct answer:

$\small \small v+w=(4,103)$

Explanation:

To add two vectors $\small v,w$ , we simply add their components:

$\small \small \small v+w=(2,3)+(2,100)=(2+2,3+100)=(4,103)$

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Example Question #1 : Vector Addition

Given the vectors

$\small \small v=(0,10)$ and $\small \small w=(-10,0)$ , compute $\small v+w$ .

Possible Answers:

$\small \small \small \small v+w=(0,0)$

$\small \small \small \small v+w=(-10,10)$

$\small \small \small \small v+w=(10,10)$

$\small \small \small \small \small v+w=(10,-10)$

Correct answer:

$\small \small \small \small v+w=(-10,10)$

Explanation:

To add two vectors $\small v,w$ , we simply add their components:

$\small \small \small \small v+w=(0,10)+(-10,0)=(0-10,10+0)=(-10,10)$

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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 #451 : Calculus 3

Example Question #452 : Calculus 3

Example Question #453 : Calculus 3

Example Question #94 : Cross Product

Example Question #91 : Cross Product

Example Question #92 : Cross Product

Example Question #454 : Calculus 3

Example Question #1 : Vector Addition

Example Question #1 : Vector Addition

Example Question #1 : Vector Addition

All Calculus 3 Resources

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