Question

The line $\frac{X}{k}=\frac{Y}{2}=\frac{Z}{-12}$ makes an isosceles triangle with the planes 2x+y+3z-1 =0 and x+2y-3z-1 =0, then value of k is

• A. 3
• B. -2
• C. 5
• D. 0

Answer 1

The correct option is B -2

The equation of the line is $\frac{x}{k}=\frac{y}{2}=\frac{z}{-12}$

The equation of the planes $2x+y+3z-1=0$ and $x+2y-3z-1=0$

As the given one makes an isosceles triangle with the given planes, it means line is parallel to any one of the bisector of these planes. Now the equation of the bisector of these planes can be written as,

$\frac{2x+y+3z-1}{\sqrt{2^2+1^2+3^2}}=\pm \frac{x+2y-3z-1}{\sqrt{1^2+2^2+3^2}}$

$\therefore \frac{2x+y+3z-1}{\sqrt{14}}=\pm \frac{x+2y-3z-1}{\sqrt{14}}$

$\therefore 2x+y+3x-1=x+2y-3z-1$

or, $x-y+6z=0$ -----eq.1

$\therefore 2x+y+3x-1=-x-2y+3z+1$

or, $3x+3y-2=0$ -----eq.2

The line matches with the equation 1 and the co-efficient of x and y is $1$ & $-1$

Therefore the value of k is -2 (ans)