Q. 24 - Calculus

Question

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

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

Step-by-Step Solution

Answer

The value of $v\times w=-6i+22j+24k$ and $w\times v=6i-22j-24k$.

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 $v\times w \mathrm{and} w\times v$

Step 2. Firstly finding the value of v × w

The value of vectors $v=(4,0,1), \mathrm{and} w=(-2,6,5)$

The cross product of  $v\times w=\det \left[\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ 4& 0& 1\\ -2& 6& 5\end{array}\right]$

Step 3. Now solving the matrix.

$$\begin{gathered} v\times w=\det \left[\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ 4& 0& 1\\ -2& 6& 5\end{array}\right] \\ v\times w=\left(\right(0\left)\right(5)-(1\left)\right(6\left)\right)i+\left(\right(4\left)\right(5)-(1\left)\right(-2\left)\right)j+\left(\right(4\left)\right(6)-(0\left)\right(-2\left)\right)k \\ v\times w=(0-6)i+(20+2)j+(24+0)k \\ v\times w=-6i+22j+24k \end{gathered}$$

Step 4. Firstly finding the value of w × v

The value of vectors $v=(4,0,1), \mathrm{and} w=(-2,6,5)$

The cross product of $v\times w=\det \left[\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ 4& 0& 1\\ -2& 6& 5\end{array}\right]$

Step 5. Now solving the matrix.

$$\begin{gathered} w\times v=\det \left[\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ -2& 6& 5\\ 4& 0& 1\end{array}\right] \\ w\times v=\left(\right(6\left)\right(1)-(5\left)\right(0\left)\right)i+\left(\right(-2\left)\right(1)-(5\left)\right(4\left)\right)j+\left(\right(-2\left)\right(0)-(6\left)\right(4\left)\right)k \\ w\times v=(6-0)i+(-2-20)j+(0-24)k \\ w\times v=6i-22j-24k \end{gathered}$$