Question

The equation of the plane passing through $(2,-3,1)$ and is normal to the line joining the points $(3,4,-1)$ and $(2,-1,5)$ is given by

• A. $x+5y-6z+19=0$
• B. $x-5y+6z-19=0$
• C. $x+5y+6z+19=0$
• D. $x-5y-6z-19=0$

Answer 1

The correct option is A $x+5y-6z+19=0$

$A(3,4,-1)$ and $B(2,-1,5)$

$\overrightarrow{AB}=-\hat{i}-5\hat{j}+6\hat{k}$

$\overrightarrow{AB}$ is normal to the required plane $\Rightarrow$ Directions of normal $(-1,-5,6)$

$\therefore$ Equation , $-x-5y+6z=k$

Point $(2,-3,1)$ passes through the plane,

$\therefore -2+15+6=k\Rightarrow k=19\therefore -x-5y+6z=19x+5y-6z+19=0$