Q. 26 - Calculus

Question

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

$(u\times v)·w \mathrm{and} u·(v\times w)$

Step-by-Step Solution

Answer

The value of $(u\times v)·w=(-1,20,1) \mathrm{and} u·(v\times w)=(-4,23,21)$

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 $(u\times v)·w \mathrm{and} u·(v\times w)$

Step 2. Firstly finding the value of ( u × v ) · w

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

Now solving the cross product.

$$\begin{gathered} u\times v=\left(\right(1\left)\right(1)-(-3\left)\right(0\left)\right)i+\left(\right(2\left)\right(1)-(-3\left)\right(4\left)\right)j+\left(\right(2\left)\right(0)-(1\left)\right(4\left)\right)k \\ u\times v=(1-0)i+(2+12)j+(0-4)k \\ u\times v=1i+14j-4k \end{gathered}$$

Step 3. Now finding the value of ( u × v ) · w

Now the vectors of $u\times v \mathrm{is} (1,14,-4)$

$$\begin{gathered} (u\times v)·w=(1,14,-4)·(-2,6,5) \\ (u\times v)·w=\{1+(-2),14+6,(-4)+5\} \\ (u\times v)·w=(1-2,14+6,-4+5) \\ (u\times v)·w=(-1,20,1) \end{gathered}$$

Step 4. Firstly finding the value of v × w

The cross product of 

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

Now solving the cross product

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

Step 5. Now finding the value of u · ( v × w )

The vectors of $v\times w \mathrm{is} (-6,22,24)$

$$\begin{gathered} u·(v\times w)=(2,1,-3)·(-6,22,24) \\ u·(v\times w)=\{2+(-6),1+22,-3+24\} \\ u·(v\times w)=(-4,23,21) \end{gathered}$$