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Calculus 3 : Vectors and Vector Operations

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Example Questions

Example Question #361 : Calculus 3

For what angle(s) is the cross product $\vec{A} \times \vec{B}=0$ ?

Possible Answers:

$30^{\circ}$

$270^{\circ}$

$90^{\circ}$

$0 ^{\circ}, 180 ^{\circ}, 360 ^{\circ}$

$45 ^{\circ}$

Correct answer:

$0 ^{\circ}, 180 ^{\circ}, 360 ^{\circ}$

Explanation:

We have the following equation that relates the cross product of two vectors $\vec{A} \times \vec{B}$ to the relative angle between them $\theta$ , written as

$\frac{\vec{A} \times \vec{B}}{\left| \vec{A} \right| \left| \vec{B} \right|}= \sin \theta$ .

From this, we can see that the numerator, or cross product, will be whenever $\sin \theta=0$ . This will be true for all even multiples of $\frac{\pi}{2}$ . Therefore, we find that the cross product of two vectors will be for $\theta = 0(0^{\circ}), \pi (180^{\circ}), 2\pi (360^{\circ})$ .

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Example Question #111 : Vectors And Vector Operations

Evaluate $<1,0> \times <0,8>$

Possible Answers:

<0,0,0>

<0,0>

<0>

None of the other answers

<0,8>

Correct answer:

None of the other answers

Explanation:

It is not possible to take the cross product of -component vectors. The definition of the cross product states that the two vectors must each have components. So the above problem is impossible.

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Example Question #7 : Cross Product

Compute $<1,0,4> \times <1,1,1>$ .

Possible Answers:

<3,4,-1>

<-3,-4,1>

<0,0,0>

<4,-3,-1>

<-4,3,1>

Correct answer:

<-4,3,1>

Explanation:

To evaluate the cross product, we use the determinant formula

$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1& a_2 & a_3 \\ b_1& b_2 & b_3 \end{vmatrix}$

So we have

$<1,0,4> \times <1,1,1>$

$= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1& 0 & 4 \\ 1& 1 & 1 \end{vmatrix}$

$= \begin{vmatrix} 0 & 4 \\ 1 & 1 \end{vmatrix} \mathbf{i} -\begin{vmatrix} 1 & 4 \\ 1& 1 \end{vmatrix} \mathbf{j} + \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} \mathbf{k}$ . (Use cofactor expansion along the top row. This is typically done when taking any cross products)

$= -4\mathbf{i}-(-3)\mathbf{j}+1\mathbf{k}$

= <-4,3,1>

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Example Question #2 : Cross Product

Evaluate $<1,1,0> \times <8,0,8>$ .

Possible Answers:

None of the other answers

<-8,8,8>

<8,-8,-8>

<-1,-1,1>

<1,1,-1>

Correct answer:

<8,-8,-8>

Explanation:

To evaluate the cross product, we use the determinant formula

$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1& a_2 & a_3 \\ b_1& b_2 & b_3 \end{vmatrix}$

So we have

$<1,1,0> \times <8,0,8>$

$= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1& 1 & 0 \\ 8& 0 & 8 \end{vmatrix}$

$= \begin{vmatrix} 1 & 0 \\ 0 & 8 \end{vmatrix} \mathbf{i} -\begin{vmatrix} 1 & 0 \\ 8& 8 \end{vmatrix} \mathbf{j} + \begin{vmatrix} 1 & 1 \\ 8 & 0 \end{vmatrix} \mathbf{k}$ . (Use cofactor expansion along the top row. This is typically done when taking any cross products)

$= 8\mathbf{i}-(8)\mathbf{j}+(-8)\mathbf{k}$

= <8,-8,-8>

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Example Question #9 : Cross Product

Find the cross product of the two vectors.

u=<4,3,7>

v=<5,7,10>

Possible Answers:

<9,10,17>

<-19,-5,13>

<20,21,70>

<-1,-4,-3>

Correct answer:

<-19,-5,13>

Explanation:

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

$\begin{vmatrix} i&j &k \\ 4&3 &7 \\ 5&7 &10 \end{vmatrix}$

The determinant equals

$=det(\begin{vmatrix} 3&7 \\ 7&10 \end{vmatrix})i-det(\begin{vmatrix} 4&7 \\ 5&10 \end{vmatrix})j+det(\begin{vmatrix} 4&3 \\ 5&7 \end{vmatrix})k$

=(3*10-7*7)i-(4*10-5*7)j+(4*7-5*3)k

=-19i-5j+13k

=<-19,-5,13>

As the cross-product.

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Example Question #1 : Cross Product

Find the cross product of the two vectors.

u=<1,0,0>

v=<0,1,0>

Possible Answers:

<0,0,1>

<1,-1,0>

<1,1,0>

<0,0,0>

Correct answer:

<0,0,1>

Explanation:

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

$\begin{vmatrix} i&j &k \\ 1&0 &0 \\ 0&1 &0 \end{vmatrix}$

The determinant equals

$=det(\begin{vmatrix} 0&0 \\ 1&0 \end{vmatrix})i-det(\begin{vmatrix} 1&0 \\ 0&0 \end{vmatrix})j+det(\begin{vmatrix} 1&0 \\ 0&1 \end{vmatrix})k$

=(0*0-1*0)i-(1*0-0*0)j+(1*1-0*0)k

=0i-0j+1k

=<0,0,1>

As the cross-product.

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Example Question #11 : Cross Product

Determine the cross product $\overrightarrow{a}\times\overrightarrow{b}$ , if $\overrightarrow{a}=2\widehat{i}+4\widehat{k}$ and $\overrightarrow{b}=\widehat{i}+3\widehat{j}-2\widehat{k}$

Possible Answers:

$-12\widehat{i}+8\widehat{j}+6\widehat{k}$

$-12\widehat{i}-8\widehat{j}+6\widehat{k}$

$-12\widehat{i}+8\widehat{j}-6\widehat{k}$

$12\widehat{i}+8\widehat{j}+6\widehat{k}$

$8\widehat{i}-12\widehat{j}+6\widehat{k}$

Correct answer:

$-12\widehat{i}+8\widehat{j}+6\widehat{k}$

Explanation:

The cross product of two vectors is a new vector. In order to find the cross product, it is useful to know the following relationships between the directional vectors:

$\widehat{i}\times\widehat{j}=\widehat{k};\widehat{j}\times\widehat{i}=-\widehat{k}$

$\widehat{j}\times\widehat{k}=\widehat{i};\widehat{k}\times\widehat{j}=-\widehat{i}$

$\widehat{k}\times\widehat{i}=\widehat{j};\widehat{i}\times\widehat{k}=-\widehat{j}$

$\widehat{i}\times\widehat{i}=\widehat{j}\times\widehat{j}=\widehat{k}\times\widehat{k}=0$

Note how the sign changes when terms are reordered; order matters!

Scalar values (the numerical coefficients) multiply through, e.g:

$r\widehat{i}\times s\widehat{j}=(rs)\widehat{k}$

With these principles in mind, we can calculate the cross product of our vectors $\overrightarrow{a}=2\widehat{i}+4\widehat{k}$ and $\overrightarrow{b}=\widehat{i}+3\widehat{j}-2\widehat{k}$

$\overrightarrow{a}\times \overrightarrow{b}=(-12)\widehat{i}+(4+4)\widehat{j}+(6)\widehat{k}=-12\widehat{i}+8\widehat{j}+6\widehat{k}$

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Example Question #11 : Cross Product

Determine the cross product $\overrightarrow{a}\times\overrightarrow{b}$ , if $\overrightarrow{a}=5\widehat{i}-5\widehat{j}$ and $\overrightarrow{b}=\widehat{i}+\widehat{j}$

Possible Answers:

$10\widehat{k}$

$15\widehat{k}$

$-20\widehat{k}$

$25\widehat{k}$

Correct answer:

$10\widehat{k}$

Explanation:

The cross product of two vectors is a new vector. In order to find the cross product, it is useful to know the following relationships between the directional vectors:

$\widehat{i}\times\widehat{j}=\widehat{k};\widehat{j}\times\widehat{i}=-\widehat{k}$

$\widehat{j}\times\widehat{k}=\widehat{i};\widehat{k}\times\widehat{j}=-\widehat{i}$

$\widehat{k}\times\widehat{i}=\widehat{j};\widehat{i}\times\widehat{k}=-\widehat{j}$

$\widehat{i}\times\widehat{i}=\widehat{j}\times\widehat{j}=\widehat{k}\times\widehat{k}=0$

Note how the sign changes when terms are reordered; order matters!

Scalar values (the numerical coefficients) multiply through, e.g:

$r\widehat{i}\times s\widehat{j}=(rs)\widehat{k}$

With these principles in mind, we can calculate the cross product of our vectors $\overrightarrow{a}=5\widehat{i}-5\widehat{j}$ and $\overrightarrow{b}=\widehat{i}+\widehat{j}$

$\overrightarrow{a}\times \overrightarrow{b}=(0+5+5+0)\widehat{k}=10\widehat{k}$

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Example Question #12 : Cross Product

Determine the cross product $\overrightarrow{a}\times\overrightarrow{b}$ , if $\overrightarrow{a}=3\widehat{i}+9\widehat{j}$ and $\overrightarrow{b}=4\widehat{i}-16\widehat{j}$

Possible Answers:

$156\widehat{k}$

$-84\widehat{k}$

$-132\widehat{k}$

$-37\widehat{k}$

$91\widehat{k}$

Correct answer:

$-84\widehat{k}$

Explanation:

The cross product of two vectors is a new vector. In order to find the cross product, it is useful to know the following relationships between the directional vectors:

$\widehat{i}\times\widehat{j}=\widehat{k};\widehat{j}\times\widehat{i}=-\widehat{k}$

$\widehat{j}\times\widehat{k}=\widehat{i};\widehat{k}\times\widehat{j}=-\widehat{i}$

$\widehat{k}\times\widehat{i}=\widehat{j};\widehat{i}\times\widehat{k}=-\widehat{j}$

$\widehat{i}\times\widehat{i}=\widehat{j}\times\widehat{j}=\widehat{k}\times\widehat{k}=0$

Note how the sign changes when terms are reordered; order matters!

Scalar values (the numerical coefficients) multiply through, e.g:

$r\widehat{i}\times s\widehat{j}=(rs)\widehat{k}$

With these principles in mind, we can calculate the cross product of our vectors $\overrightarrow{a}=3\widehat{i}+9\widehat{j}$ and $\overrightarrow{b}=4\widehat{i}-16\widehat{j}$

$\overrightarrow{a}\times \overrightarrow{b}=(0-48-36+0)\widehat{k}=-84\widehat{k}$

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Example Question #11 : Cross Product

Determine the cross product $\overrightarrow{a}\times\overrightarrow{b}$ , if $\overrightarrow{a}=-10\widehat{j}$ and $\overrightarrow{b}=3\widehat{i}+6\widehat{j}$

Possible Answers:

$-30\widehat{k}$

$90\widehat{k}$

$45\widehat{k}$

$-90\widehat{k}$

$30\widehat{k}$

Correct answer:

$30\widehat{k}$

Explanation:

The cross product of two vectors is a new vector. In order to find the cross product, it is useful to know the following relationships between the directional vectors:

$\widehat{i}\times\widehat{j}=\widehat{k};\widehat{j}\times\widehat{i}=-\widehat{k}$

$\widehat{j}\times\widehat{k}=\widehat{i};\widehat{k}\times\widehat{j}=-\widehat{i}$

$\widehat{k}\times\widehat{i}=\widehat{j};\widehat{i}\times\widehat{k}=-\widehat{j}$

$\widehat{i}\times\widehat{i}=\widehat{j}\times\widehat{j}=\widehat{k}\times\widehat{k}=0$

Note how the sign changes when terms are reordered; order matters!

Scalar values (the numerical coefficients) multiply through, e.g:

$r\widehat{i}\times s\widehat{j}=(rs)\widehat{k}$

With these principles in mind, we can calculate the cross product of our vectors $\overrightarrow{a}=-10\widehat{j}$ and $\overrightarrow{b}=3\widehat{i}+6\widehat{j}$

$\overrightarrow{a}\times \overrightarrow{b}=(30+0)\widehat{k}=30\widehat{k}$

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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 #361 : Calculus 3

Example Question #111 : Vectors And Vector Operations

Example Question #7 : Cross Product

Example Question #2 : Cross Product

Example Question #9 : Cross Product

Example Question #1 : Cross Product

Example Question #11 : Cross Product

Example Question #11 : Cross Product

Example Question #12 : Cross Product

Example Question #11 : Cross Product

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