Question

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

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