Question

A plane whose equation is $2x-y+3z+5=0$ is rotated through ${90}^0$ about its line of intersection with plane $5x-4y-2z+1=0.$ The equation of plane in its new position is

• A. $27x-24y-26z=13$
• B. $27x+24y+26z=13$
• C. $27x+24y+26z=15$
• D. $27x-24y+26z-13=0$

Answer 1

Answer: (A)

$27x-24y-26z=13$

The new equation is $2x-y+3z+5+\lambda (5x-4y-2z+1)=0$
$(2+5\lambda )x-(1+4\lambda )y+(3-2\lambda )z+5+\lambda =0$
It is perpendicular to the plane $2x-y+3z+5=0$
$\therefore 2(2+5\lambda )+1+4\lambda +3(3-2\lambda )=0$
$\therefore 7+4\lambda =0$
$\therefore \lambda =-\frac{7}{4}$
Substituting for $\lambda$, we get $27x-24y-26z=13$

Hence, option A is correct.