Question

Consider a plane $x+y-z=1$ and the point $A(1,2,-3)$. A line $L$ has the equation $x=1+3r$, $y=2-r$, $z=3+4r$

The coordinate of a point $B$ of line $L$, such that $AB$ is parallel to the plane, is

• A. $(10,-1,15)$
• B. $(-5,4,-5)$
• C. $(4,1,7)$
• D. $(-8,5,-9)$

Answer 1

The correct option is D $(-8,5,-9)$

Let $\overrightarrow{OB}=(1+3r)\hat{i}+(2-r)\hat{j}+(3+4r)\hat{k}$
$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=(1+3r)\hat{i}+(2-r)\hat{j}+(3+4r)\hat{k}-\hat{i}-2\hat{j}+3\hat{k}=3r\hat{i}-r\hat{j}+(6+4r)\hat{k}$
Since, $\overrightarrow{AB}$ is parallel to $x+y-z=1$
Therefore, $\overrightarrow{AB}.(\hat{i}+\hat{j}-\hat{k})=0$
$\Rightarrow (3r\hat{i}-r\hat{j}+(6+4r)\hat{k}).(\hat{i}+\hat{j}-\hat{k})=0$
$\Rightarrow 3r-r-6-4r=0$
$\Rightarrow r=-3$
Therefore, $\overrightarrow{OB}=-8i+5j-9k$
Ans: D