Discussion Forum

Let $a=2 \hat{i}+\hat{j}-3\hat{k}$ and $b= \hat{i}+3\hat{j}+2\hat{k}$ . Then the volume of the parallelopiped having coterminous edges as a,b and c, where c is the vector perpendicular to the plane of a, b and |c|=2 is 

Answer:

Explanation:

We have, 

$a=2 \hat{i}+\hat{j}-3\hat{k}$

and $b= \hat{i}+3\hat{j}+2\hat{k}$ 

Clearly , c is  parallel to $a \times b$ 

Here,  $a\times b=\begin{bmatrix}\hat{i} & \hat{j}&\hat{k} \\2 & 1&-3\\1&3&2 \end{bmatrix}=\hat{i}(11)-\hat{j}(7)+\hat{k}(5)$

        $11 \hat{i}-7 \hat{j}+5\hat{k}=d$    (say)

Now, $\hat{d}=\frac{a\times b}{|a \times b|}=\frac{11\hat{i}-7\hat{j}+5\hat{k}}{\sqrt{195}}$

Thus,   $c=|c|\hat{d}=\frac{2}{\sqrt{195}}(11 \hat{i}-7\hat{j}+5\hat{k})$

Hence, volume of the parallelopiped= [a b c]

= $\begin{bmatrix}2 & 1&-3 \\1 & 3&2\\\frac{22}{\sqrt{195}}&\frac{-14}{\sqrt{195}}&\frac{10}{\sqrt{195}} \end{bmatrix}$

    =$\frac{2}{\sqrt{195}}\begin{bmatrix}2 & 1&-3 \\1 & 3&2\\11&-7&5 \end{bmatrix}$

   = $\frac{2}{\sqrt{195}}[2(29)-1(-17)-3(-40)]$

    =$\frac{2}{\sqrt{195}}[58+17+120]$

=  $\frac{2}{\sqrt{195}}\times195$

    =$2\sqrt{195}$