Question

In Exercises 22–29 compute the indicated quantities when $u=(2,1,-3),v=(4,0,1),\mathrm{and}w=(-2,6,5)$

$u\times v\mathrm{and}v\times u$

Short answer

The value of $u\times v=1i+14j-4k$andlocalid="1649397999598" $v\times u=-1i-14j+4k$

Step-by-step solution

Step 1. Given Information 

In Exercises 22–29 compute the indicated quantities when$u=(2,1,-3),v=(4,0,1),\mathrm{and}w=(-2,6,5)$

We have to find the value of$u\times v\mathrm{and}v\times u$

Step 2. Firstly finding the value of u×v

The value of vectors $u=(2,1,-3),v=(4,0,1)$

The cross product of $u\times v$

localid="1649398046712" $u\times v=\det \left[\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ 2& 1& -3\\ 4& 0& 1\end{array}\right]$

Step 3. Now solving the matrix.

$$\begin{gathered} u\times v=\det \left[\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ 2& 1& -3\\ 4& 0& 1\end{array}\right] \\ u\times v=\left(\right(1\left)\right(1)-(-3\left)\right(0\left)\right)i+\left(\right(2\left)\right(1)-(-3\left)\right(4\left)\right)j+\left(\right(2\left)\right(0)-(1\left)\right(4\left)\right)k \\ u\times v=(1+0)i+(2+12)j+(0-4)k \\ u\times v=1i+14j-4k \end{gathered}$$

Step 4. Now finding the value of v×u

The value of vectors$u=(2,1,-3),v=(4,0,1)$

The cross product of$v\times u=\det \left[\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ 4& 0& 1\\ 2& 1& -3\end{array}\right]$

Step 5. Now solving the matrix. 

$$\begin{gathered} v\times u=\det \left[\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ 4& 0& 1\\ 2& 1& -3\end{array}\right] \\ v\times u=\left(\right(0\left)\right(-3)-(1\left)\right(1\left)\right)i+\left(\right(4\left)\right(-3)-(1\left)\right(2\left)\right)j+\left(\right(4\left)\right(1)-(0\left)\right(2\left)\right)k \\ v\times u=(0-1)i+(-12-2)j+(4-0)k \\ v\times u=-1i-14j+4k \end{gathered}$$