Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

The given planes are:

$x-y-z=4$ (1)

and

$x+y+2z=4$ (2)

Since the required plane passes through the line of intersection of planes (1) and (2), its equation is:

$(x-y-z-4)+\alpha (x+y+2z-4)=0$

This simplifies to:

$(1+\alpha )x+(\alpha -1)y+(2\alpha -1)z-(4\alpha +4)=0$ (3)

Given that plane (1) and plane (3) are perpendicular, their dot product of normal vectors is zero:

$$\Rightarrow 1(1+\alpha )-1(\alpha -1)-1(2\alpha -1)=0$$

$$\Rightarrow 1+\alpha -\alpha +1-2\alpha +1=0$$

$$\Rightarrow 3-2\alpha =0\Rightarrow \alpha =\frac{3}{2}$$

Substituting $\alpha =\frac{3}{2}$ into the equation for the required plane (3):

$(x-y-z-4)+\frac{3}{2}(x+y+2z-4)=0$

Multiplying by 2 to clear the fraction:

$2(x-y-z-4)+3(x+y+2z-4)=0$

$2x-2y-2z-8+3x+3y+6z-12=0$

$5x+y+4z-20=0$