Question

The equation of plane passing through (2, 1, 0) and line of intersection of planes $x-2y+3z=4$ and $x-y+z=3$ is

• A. $x+y-z+4=0$
• B. $2x+y+z+1=0$
• C. $x-z=2$
• D. $x+y+z+1=0$

Answer 1

The correct option is C $x-z=2$

Since the plane passes through the line of intersection of the two given planes, the points lying on both the planes also lie on the new plane.

Let $x=0$, we get $-2y+3z=4$ and $-y+z=3$ for both the given planes.

Solving these two equations simultaneously, we get

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

$\Rightarrow z=-2$ and $y=-5$

The point becomes $(0,-5,-2)$

Now, if we take $y$ to be $0$, we get

$x+3z=4$ and $x+z=3$

Solving them simultaneously, we get $x+3z-4-(x+z-3)=0$

$\Rightarrow 2z-1=0orz=\frac{1}{2}$ and thus $x=\frac{5}{2}$

The point becomes $(\frac{5}{2},0,\frac{1}{2})$

The vector joining two of those gives

$2\hat{i}+6\hat{j}+2\hat{k}$ and $\frac{1}{2}\hat{i}-\hat{j}+\frac{1}{2}$

Normal to the plane is given by their cross product, which gives

$(3+2)\hat{i}-(1-1)\hat{j}+(-2-3)\hat{k}=5\hat{i}-5\hat{k}$

Thus, the equation of the plane can be written as

$5x-5z+d=0$

Substituting $(2,1,0)$ we get

$10-0+d=0$ or $d=-10$

The equation of the plane is $x-z=2$.