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 $(-8,5,-9)$

$L:\frac{x-1}{3}=\frac{y-2}{-1}=\frac{z-3}{4}=r$
Any point on the line $L$ be $B(3r+1,2-r,3+4r)$

$\overrightarrow{AB}=3r,-r,4r+6$

Hence, $\overrightarrow{AB}$ is parallel to $x+y-z=1$
$\Rightarrow 3r-r-4r-6=0$
$\Rightarrow r=-3$
$\therefore B$ is $(-8,5,-9)$