Question

Let the position vectors of the points $P,A$ and $B$ be $\overrightarrow{r},\hat{i}+\hat{j}+\hat{k}$ and $-\hat{i}+\hat{k}$. If $PA$ is perpendicular to $PB$ but $\overrightarrow{r}$ is not perpendicular to $\overrightarrow{r}-(\hat{j}+2\hat{k}),$ then $\overrightarrow{r}$ is

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

Answer 1

Answer: (D)

$\hat{j}+2\hat{k}$

$\overrightarrow{PA}\perp \overrightarrow{PB}\Rightarrow (\overrightarrow{a}-\overrightarrow{r}).(\overrightarrow{b}-\overrightarrow{r})=0$ or
$r^2-\overrightarrow{r}\cdot (\overrightarrow{a}+\overrightarrow{b})+\overrightarrow{a}\cdot \overrightarrow{b}=0................\left(1\right)$
But $\overrightarrow{a}\cdot \overrightarrow{b}=-1+0+1=0$
and $\overrightarrow{a}+\overrightarrow{b}=\hat{j}+2\hat{k}$
Hence from $\left(1\right)$, we have,
$r^2-\overrightarrow{r}\cdot (\hat{j}+2\hat{k})+0=0$
$\overrightarrow{r}\cdot [r-(\hat{j}+2\hat{k})]=0$
Now we know that $\overrightarrow{a}\cdot \overrightarrow{b}=0\Rightarrow$ either $a=0$ or $b=0$ or $\overrightarrow{a}$ and $\overrightarrow{b}$ are perpendicular.
By virtue of given condition we have $b=0$
$\therefore r-(\hat{j}+2k)=0\therefore \overrightarrow{r}=\hat{j}+2\hat{k}$