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Calculus 3 : Cross Product

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Example Questions

Example Question #1 : Cross Product

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

Find $u\times v$ .

Possible Answers:

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

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

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

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

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

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 #1 : Cross Product

Let u=(1,0,1) , and v=(0,0,1) .

Find $u\times v$ .

Possible Answers:

$u\times v=(0,0,-1)$

$u\times v=(-1,0,0)$

$u\times v=(0,0,0)$

$u\times v=(0,-1,0)$

$u\times v=(0,1,0)$

Correct answer:

$u\times v=(0,-1,0)$

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.

$u\times v=(0(1)-1(0),-1(1)+1(0), 1(0)-0(0))$

$u\times v=(0-0,-1+0, 0-0)$

$u\times v=(0,-1,0)$

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Example Question #3 : Cross Product

True or False: The cross product can only be taken of two 3-dimensional vectors.

Possible Answers:

False

True

Correct answer:

True

Explanation:

This is true. The cross product is defined this way. The dot product however can be taken for two vectors of dimension n (provided that both vectors are the same dimension).

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Example Question #4 : Cross Product

Which of the following choices is true?

Possible Answers:

$\begin{matrix}\vec{A}\cdot \vec{B}=-\vec{B}\cdot \vec{A}, & & \vec{A} \times \vec{B}= \vec{B}\times \vec{A} \end{matrix}$

$\begin{matrix}\vec{A}\cdot \vec{B}=\vec{B}\cdot \vec{A}, & & \vec{A} \times \vec{B}= - \vec{B}\times \vec{A} \end{matrix}$

$\begin{matrix}\vec{A}\cdot \vec{B}=\vec{B}\cdot \vec{A}, & & \vec{A} \times \vec{B}= \vec{B}\times \vec{A} \end{matrix}$

$\begin{matrix}\vec{A}\cdot \vec{B}=-\vec{B}\cdot \vec{A}, & & \vec{A} \times \vec{B}= - \vec{B}\times \vec{A} \end{matrix}$

Correct answer:

$\begin{matrix}\vec{A}\cdot \vec{B}=\vec{B}\cdot \vec{A}, & & \vec{A} \times \vec{B}= - \vec{B}\times \vec{A} \end{matrix}$

Explanation:

By definition, the order of the dot product of two vectors does not matter, as the final output is a scalar. However, the cross product of two vectors will change signs depending on the order that they are crossed. Therefore

$\begin{matrix}\vec{A}\cdot \vec{B}=\vec{B}\cdot \vec{A}, & & \vec{A} \times \vec{B}= - \vec{B}\times \vec{A} \end{matrix}$ .

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Example Question #2 : Cross Product

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

Possible Answers:

$45 ^{\circ}$

$270^{\circ}$

$90^{\circ}$

$30^{\circ}$

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

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 #361 : Calculus 3

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

Possible Answers:

<0,0,0>

<0,8>

<0>

<0,0>

None of the other answers

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:

<-1,-1,1>

None of the other answers

<1,1,-1>

<8,-8,-8>

<-8,8,8>

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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Calculus 3 : Cross Product

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1 : Cross Product

Example Question #1 : Cross Product

Example Question #3 : Cross Product

Example Question #4 : Cross Product

Example Question #2 : Cross Product

Example Question #361 : Calculus 3

Example Question #7 : Cross Product

Example Question #2 : Cross Product

Example Question #9 : Cross Product

Example Question #1 : Cross Product

All Calculus 3 Resources

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