Question

If $\overrightarrow{r}=3\overrightarrow{i}+2\overrightarrow{j}-5\overrightarrow{k}$, $\overrightarrow{a}=2\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k}$, $\overrightarrow{b}=\overrightarrow{i}+3\overrightarrow{j}-2\overrightarrow{k}$ and $\overrightarrow{c}=-2\overrightarrow{i}+\overrightarrow{j}-3\overrightarrow{k}$ such that $\overrightarrow{r}=\lambda \overrightarrow{a}+\mu \overrightarrow{b}+\nu \overrightarrow{c}$ then

• A. $\mu ,\frac{\lambda }{2},\nu$ are in $AP$
• B. $\lambda ,\mu ,\nu$ are in $AP$
• C. $\lambda ,\mu ,\nu$ are in $HP$
• D. $\mu ,\lambda ,\nu$ are in $GP$

Answer 1

Answer: (A)

$\mu ,\frac{\lambda }{2},\nu$ are in $AP$

Given : $\overrightarrow{r}=3\overrightarrow{i}+2\overrightarrow{j}-5\overrightarrow{k}$, $\overrightarrow{a}=2\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k}$, $\overrightarrow{b}=\overrightarrow{i}+3\overrightarrow{j}-2\overrightarrow{k}$ and $\overrightarrow{c}=-2\overrightarrow{i}+\overrightarrow{j}-3\overrightarrow{k}$

$\overrightarrow{r}=\lambda \overrightarrow{a}+\mu \overrightarrow{b}+\nu \overrightarrow{c}$

$\Rightarrow 3\overrightarrow{i}+2\overrightarrow{j}-5\overrightarrow{k}$ $=\lambda (2\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k})+$ $\mu (\overrightarrow{i}+3\overrightarrow{j}-2\overrightarrow{k})+$ $\nu (-2\overrightarrow{i}+\overrightarrow{j}-3\overrightarrow{k})$

By comparing coefficients, we get

$2\lambda +\mu -2\nu =3$

$-\lambda +3\mu +\nu =2$

$\lambda -2\mu -3\nu =-5$

Solving these using Cramer's rule, we have

$D=$$\left|\begin{array}{ccc}2& 1& -2\\ -1& 3& 1\\ 1& -2& -3\end{array}\right|$

$=-14-2+2=-14$

$D_{\lambda }=\left|\begin{array}{ccc}3& 1& -2\\ 2& 3& 1\\ -5& -2& -3\end{array}\right|$ $=-21+1-22=-42.$

$\Rightarrow \lambda =\frac{-42}{-14}=3$

Similarly we get $\mu =1,\nu =2$

Thus option $A$ is satisfied.