Question

If the volume of parallelopiped whose coterminous edges are $\overline{a}=3\overline{i}-\overline{j}+4\overline{k},\overline{b}=2\overline{i}+3\overline{j}-\overline{k}$ and $\overline{c}=-5\overline{i}+2\overline{j}+3\overline{k}$ is three times the volume of parallelopiped whose coterminous edges are $\overline{p}=\overline{i}+\overline{j}+3\overline{k},\overline{q}=\overline{i}-2\overline{j}+\lambda \overline{k}$ and $\overline{r}=2\overline{i}+3\overline{j}$ then the value of $\lambda$ is

• A. $\frac{15}{21}$
• B. $\frac{-47}{3}$
• C. $\frac{-37}{15}$
• D. $\frac{-15}{4}$

Answer 1

The correct option is B $\frac{-47}{3}$

The volume of a parallelopiped, whose coterminus edges are given, is given by $\overrightarrow{a}.(\overrightarrow{b}\times \overrightarrow{c})$

First volume $=\left|\begin{array}{ccc}3& -1& 4\\ 2& 3& -1\\ -5& 2& 3\end{array}\right|$

$=3(9+2)+1(6-5)+4(4+15)$

$=33+1+76=110$

Second volume $=\left|\begin{array}{ccc}1& 1& 3\\ 1& -2& \lambda \\ 2& 3& 0\end{array}\right|$

$=1(0-3\lambda )-1(0-2\lambda )+3(3+4)$

$=-\lambda +21$

This volume also has to be $\frac{110}{3}$ and so, $\lambda =-\frac{47}{3}$