Question

If $\overline{r}=3\overline{i}+2\overline{j}-5\overline{k},\overline{a}=2\overline{i}-\overline{j}+\overline{k}$, $\overline{b}=\overline{i}+3\overline{j}-2\overline{k}$ and $\overline{c}=-2\overline{i}+\overline{j}-3\overline{k}$ such that $\overline{r}=\lambda \overline{a}+\mu \overline{b}+\delta \overline{c}$ then $\mu ,\frac{\lambda }{2},\delta$ are in

• A. $AP$
• B. $GP$
• C. $HP$
• D. $AGP$

Answer 1

Answer: (A)

$AP$

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

$\left(3\hat{i}+2\hat{j}-5\hat{k}\right)=\lambda \left(2\hat{i}-\hat{j}+\hat{k}\right)+\mu \left(\hat{i}+3\hat{j}-2\hat{k}\right)+\delta \left(-2\hat{i}+\hat{j}-3\hat{k}\right)$

$\left(3\hat{i}+2\hat{j}-5\hat{k}\right)=\left(2\lambda +\mu +2\delta \right)\hat{i}+\left(-\lambda +3\mu +\delta \right)\hat{j}+\left(\lambda -2\mu -3\lambda \right)\hat{k}$

On comparing both sides

$\begin{array}{c}2\lambda +\mu -2\delta =3----\left(i\right)\\ -\lambda +3\mu +\delta =2----\left(ii\right)\\ \lambda -2\mu -3\delta =-5---\left(iii\right)\end{array}$

multiply (ii) by $2$ and add with (i)

$7\mu =7;\mu =1$

Adding (ii) and (iii)

$\mu -2\delta =-3$

$2\delta =4$

$\delta =2$

putting $\mu =1$ and $\delta =2$ in (ii)

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

$\lambda =3$

$2\times \left(\frac{\lambda }{2}\right)=2\times \frac{3}{2}=3$

$\mu +\delta =3$

Therefore, $2\times \left(\frac{\lambda }{2}\right)=\mu +\delta$

Hence, $\mu ,\frac{\lambda }{2},\delta$ are in A.P.