Question

If the line $\frac{x-1}{2}=\frac{y-3}{a}=\frac{z+1}{3}$ lies in the plane $bx+2y+3z–4=0$, then

• A. $a=\frac{11}{2},b=1$
• B. $a=-\frac{5}{2},b=-7$
• C. $a=-\frac{11}{2},b=1$
• D. $a=1,b=-\frac{11}{2}$

Answer 1

The correct option is C

$a=-\frac{11}{2},b=1$

$\because$ line lies in plane.

$\Rightarrow (1,3,-1)$ also lies in plane

$\Rightarrow b+6-3-4=0$

$\Rightarrow b=1$

Also, normal of the plane will be perpendicular to the line.

$\Rightarrow 2b+2a+9=0$

$\Rightarrow 2+2a+9=0$

$\Rightarrow a=-\frac{11}{2}$