Browse Questions

# Find the equation of the plane that contains the point $(1,-1,2)$ and is $\perp$ to the planes $2x+3y-2z=5$ and $x+2y-3z=8$. Hence find the distance of the point $(-2,5,5)$ from the plane obtained above.

Toolbox:
• Distance of the point $(x_1,y_1,z_1)$ from the plane $ax+by+cz+d=0$ is given by $\bigg|\large\frac{ax_1+by_1+cz_1+d}{\sqrt {a^2+b^2+c^2}}\bigg|$
To form the equation of a plane we need a point on it and normal vector $(\overrightarrow n)$
Step1
Given that the required plane contains the point $(1,-1,2)$
Let the normal to the required plane be $\overrightarrow n$
Also it is given that the required plane is $\perp$ to the planes
$2x+3y-2z-5=0$..............(i) and
$x+2y-3z-8$..........(ii)
Normal to the plane is $\overrightarrow n_1=(2,3,-2)$
Normal to the plane (ii) is $\overrightarrow n_2=(1,2,-3)$
Since the required plane is $\perp$ to (i) and (ii),
$\overrightarrow n=\overrightarrow n_1\times\overrightarrow n_2$
$\overrightarrow n=\left |\begin {array} \hat i & \hat j & \hat k\\2 & 3 & -2\\1 & 2 & -3\end {array}\right|=\hat i(-9+4)-\hat j(-6+2)+\hat k(4-3)$
$i.e., \overrightarrow n=(-5,4,1)$
Step 2
$\therefore$ Equation of the required plane is
$-5x+4y+z+d=0$
Given that this plane contains the point $(1,-1,2)$
$\Rightarrow\:$ It satisfies the equation of the plane.
$\Rightarrow\:-5\times 1+4\times (-1)+2+d=0$
$\Rightarrow\:d=7$
$\Rightarrow\:$ Equation of the required plane is $5x-4y-z-7=0$
Step 3
We know that distance of the point $(x_1,y_1,z_1)$ from the plane $ax+by+cz+d=0$ is given by
$\bigg|\large\frac{ax_1+by_1+cz_1+d}{\sqrt {a^2+b^2+c^2}}\bigg|$
The distance of the point $(-2,5,5)$ from the plane $5x-4y-z-7=0$ is given by
$\bigg|\large\frac{5\times (-2)+(-4)\times 5+(-1)\times 5-7}{\sqrt {5^2+(-4)^2+(-1)^2}}\bigg|$
$=\bigg|\large\frac{-10-20-5-7}{\sqrt 42}\bigg|$
$=\large\frac{42}{\sqrt {42}}$$=\sqrt {42}$ units.