Question

Image of the point $P$ with position vector $7\hat{i}-\hat{j}+2\hat{k}$ in the line whose vector equation is $\overline{r}=(9\hat{i}+5\hat{j}+5\hat{k})+\lambda (\hat{i}+3\hat{j}+5\hat{k})$ has position vector

• A. $-9\hat{i}+5\hat{j}+2\hat{k}$
• B. $9\hat{i}+5\hat{j}+2\hat{k}$
• C. $9\hat{i}+5\hat{j}-2\hat{k}$
• D. $-9\hat{i}-5\hat{j}-2\hat{k}$

Answer 1

Answer: (C)

$9\hat{i}+5\hat{j}-2\hat{k}$

Let $(\alpha ,\beta ,\gamma )$ be the position vectors
$\therefore$ $\frac{\alpha +7}{2}\hat{i}+\frac{\beta -1}{2}\hat{j}+\frac{\gamma +2}{2}\hat{k}=(9\hat{i}+5\hat{j}+5\hat{k})+\lambda (\hat{i}+3\hat{j}+5\hat{k})$

Comparing, we get $\alpha =11+2\lambda$, $\beta =11+6\lambda$, $\gamma =8+10\lambda$

$P{P}^{\prime }$ and given line are $\perp$
$\Rightarrow$ $(\alpha -7)+3(\beta +1)+5(\gamma -2)=0$
$\therefore 11+2\lambda +3(11+6\lambda )+5(8+10\lambda )=14$
$\therefore 70+70\lambda =0$
$\therefore$ $\lambda =-1$
$\therefore$ $(\alpha ,\beta ,\gamma )\equiv (9,5,-2)$