Question

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

• A. $(-9,5,2)$
• B. $(9,5,-2)$
• C. $(9,-5,-2)$
• D. None of these

Answer 1

The correct option is B $(9,5,-2)$

We will convert $\overrightarrow{r}=9\hat{i}+5\hat{j}+5\hat{k}+\lambda (\hat{i}+3\hat{j}+5\hat{k})$ into cartesian form

$\overrightarrow{r}=(9+\lambda )\hat{i}+(5+3\lambda )\hat{j}+(5+5\lambda )\hat{k}$

Comparing with $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$

$\Rightarrow x-9=\lambda ,\frac{y-5}{3}=\lambda ,\frac{z-5}{5}=\lambda$

$\Rightarrow \frac{x-9}{1}=\frac{y-5}{3}=\frac{z-5}{5}$

So, this line passes through $(9,5,7)$ and has direction ratios $1,3,5$.

Now, let the image of point $P(7,-1,2)$ be $Q(\alpha ,\beta ,\gamma )$.

Now, the direction ratios of $PQ$ are $\alpha -7,\beta +1,\gamma -2$

Since, $PQ$ is perpendicular to the given line,

$1(\alpha -7)+3(\beta +1)+5(\gamma -2)=0$

$\Rightarrow \alpha +3\beta +5\gamma -14=0$ ....$\left(1\right)$

Now, the coordinates of mid-point of $PQ$ are $(\frac{\alpha +7}{2},\frac{\beta -1}{2},\frac{\gamma +2}{2})$

Since, this point lies on the given line

$\frac{\frac{\alpha +7}{2}-9}{1}=\frac{\frac{\beta -1}{2}-5}{3}=\frac{\frac{\gamma +2}{2}-5}{5}=\lambda \left(say\right)$

$\Rightarrow \alpha =2\lambda +11,\beta =6\lambda +11,\gamma =10\lambda +8$

Substituting in $\left(1\right)$, we get

$\lambda =-1$

Thus, the image is $(9,5,-2)$.

Hence, option 'B' is correct.