Cross Product - AP Calculus BC

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Question

Let , and .

Find $u\times v$ .

Answer

We are trying to find the cross product between and .

Recall the formula for cross product.

If , and , 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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Question

Let , and .

Find $u\times v$ .

Answer

We are trying to find the cross product between and .

Recall the formula for cross product.

If , and , 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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Question

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

Answer

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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Question

Which of the following choices is true?

Answer

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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Question

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

Answer

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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Question

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

Answer

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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Question

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

Answer

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)

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

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Question

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

Answer

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)

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

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Question

Find the cross product of the two vectors.

Answer

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

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

As the cross-product.

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Question

Find the cross product of the two vectors.

Answer

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

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

As the cross-product.

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Question

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

Answer

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 $\overrightarrow{a}=2\widehat{i}+\widehat{j}$ and $\overrightarrow{b}=\widehat{i}+3\widehat{j}$

$\overrightarrow{a}\times \overrightarrow{b}=(6-1)\widehat{k}=5\widehat{k}$

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Question

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

Answer

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 $\overrightarrow{a}=2\widehat{j}$ and $\overrightarrow{b}=3\widehat{i}+4\widehat{j}$

$\overrightarrow{a}\times \overrightarrow{b}=(-6+0)\widehat{k}=-6\widehat{k}$

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Question

Determine the cross product $\overrightarrow{a}\times\overrightarrow{b}$ , if $\overrightarrow{a}=3\widehat{i}-\widehat{j}$ and $\overrightarrow{b}=5\widehat{i}+\widehat{j}$

Answer

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 $\overrightarrow{a}=3\widehat{i}-\widehat{j}$ and $\overrightarrow{b}=5\widehat{i}+\widehat{j}$

$\overrightarrow{a}\times \overrightarrow{b}=(0+3+5+0)\widehat{k}=8\widehat{k}$

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Question

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

Answer

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 $\overrightarrow{a}=\widehat{i}+2\widehat{j}$ and $\overrightarrow{b}=2\widehat{i}+4\widehat{j}$

$\overrightarrow{a}\times \overrightarrow{b}=(0+4-4+0)\widehat{k}=0$

Note that the zero answer means that these two vectors are parallel!

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Question

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

Answer

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 $\overrightarrow{a}=\widehat{i}+9\widehat{j}$ and $\overrightarrow{b}=9\widehat{i}-\widehat{j}$

$\overrightarrow{a}\times \overrightarrow{b}=(0-1-81+0)\widehat{k}=-82\widehat{k}$

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Question

Determine the cross product $\overrightarrow{a}\times\overrightarrow{b}$ , if $\overrightarrow{a}=7\widehat{i}-\widehat{j}$ and $\overrightarrow{b}=-21\widehat{i}+3\widehat{j}$ .

Answer

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 $\overrightarrow{a}=7\widehat{i}-\widehat{j}$ and $\overrightarrow{b}=-21\widehat{i}+3\widehat{j}$

$\overrightarrow{a}\times \overrightarrow{b}=(0+21-21+0)\widehat{k}=0$

These zero results means that the two vectors are parallel.

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Question

Determine the cross product $\overrightarrow{a}\times\overrightarrow{b}$ , if $\overrightarrow{a}=3\widehat{i}+4\widehat{j}$ and $\overrightarrow{b}=-8\widehat{i}+6\widehat{j}$ .

Answer

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 $\overrightarrow{a}=3\widehat{i}+4\widehat{j}$ and $\overrightarrow{b}=-8\widehat{i}+6\widehat{j}$

$\overrightarrow{a}\times \overrightarrow{b}=(0+18+32+0)\widehat{k}=50\widehat{k}$

Take note that if you were to find the magnitude of each vector (5 and 10) respectively and found their product, it'd be the same as the absolute value of the cross product. This equivalence indicates that the two vectors are perpendicular!

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Question

Determine the cross product $\overrightarrow{a}\times\overrightarrow{b}$ , if $\overrightarrow{a}=2\widehat{i}+11\widehat{j}$ and $\overrightarrow{b}=-\widehat{i}-\widehat{j}$ .

Answer

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 $\overrightarrow{a}=2\widehat{i}+11\widehat{j}$ and $\overrightarrow{b}=-\widehat{i}-\widehat{j}$

$\overrightarrow{a}\times \overrightarrow{b}=(0-2+11+0)\widehat{k}=9\widehat{k}$

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Question

Determine the cross product $\overrightarrow{a}\times\overrightarrow{b}$ , if $\overrightarrow{a}=6\widehat{i}+2\widehat{j}$ and $\overrightarrow{b}=12\widehat{i}+5\widehat{j}$

Answer

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 $\overrightarrow{a}=6\widehat{i}+2\widehat{j}$ and $\overrightarrow{b}=12\widehat{i}+5\widehat{j}$

$\overrightarrow{a}\times \overrightarrow{b}=(0+30-24+0)\widehat{k}=6\widehat{k}$

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Question

Determine the cross product $\overrightarrow{a}\times\overrightarrow{b}$ , if $\overrightarrow{a}=12\widehat{i}+9\widehat{j}$ and $\overrightarrow{b}=3\widehat{i}-2\widehat{j}$ .

Answer

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 $\overrightarrow{a}=12\widehat{i}+9\widehat{j}$ and $\overrightarrow{b}=3\widehat{i}-2\widehat{j}$

$\overrightarrow{a}\times \overrightarrow{b}=(-24-27)\widehat{k}=-51\widehat{k}$

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