Question

A plane passes through (1, -2, 1) and is perpendicular to two planes 2x – 2y + z = 0 and x – y + 2z = 4. The distance of the plane from the point (1, 2, 2) is

Answer 1

Hint – In this particular type of question use the concept that the general equation of plane from the given point say (l, m ,n) is given as, (a(x – l) + b(y – m) + c(z – n) = 0), where, a, b, c are the direction ratios and use the concept if that if two planes are perpendicular then the sum of individual multiplication of the direction ratios is zero so use this concept to reach the solution of the question.

Complete step by step solution:
Let the direction ratio of the plane passing through the point (1, 2, 2) be a, b and c
So the general equation of the plane passing through the point (1, 2, 2) is
$\Rightarrow$ a(x – 1) + b(y + 2) + c(z – 2) = 0...................... (1)
Now this plane is perpendicular to the planes which are given as
2x – 2y + z = 0 and x – y + 2z = 4
So the direction ratios of the planes are (2, -2, 1) and (1, -1, 2).
Now as the given planes are perpendicular to the plane a(x – 1) + b(y + 2) + c(z – 2) = 0.
So the sum of the individual multiplication of the direction ratios is always zero.
Therefore,
2a – 2b + c = 0................ (2)
And
a – b + 2c = 0................... (3)
Now multiply by 2 in equation (3) and subtract equation (2) from it. We have,
$\Rightarrow 2\left(a-b+2c\right)-\left(2a-2b+c\right)=0$
Now simplify this we have,
$\Rightarrow 4c-c=0$
$\Rightarrow 3c=0$
$\Rightarrow c=0$
Now from equation (2) we have,
$\Rightarrow 2a-2b+0=0$
$\Rightarrow a=b$
Now substitute these values in equation (1) we have,
$\Rightarrow$ a(x – 1) + a(y + 2) + 0(z – 2) = 0
$\Rightarrow$ a(x – 1) + a(y + 2) = 0
$\Rightarrow$ (x – 1) + (y + 2) = 0
$\Rightarrow$ x + y + 1 = 0
Now we have to find out the distance of the point (1, 2, 2) from this plane.
As we know that the distance (D) formula from any point (p, q, r) on the plane ax + by + cz + d = 0 is given as,
$\Rightarrow D={\displaystyle \frac{\left|ap+bq+cr+d\right|}{\sqrt{a^2+b^2+c^2}}}$
So the distance of the point (1, 2, 2) on the plane x + y + 1 = 0 is given as
$\Rightarrow D={\displaystyle \frac{\left|1+2+0+1\right|}{\sqrt{1^2+1^2+0^2}}}$
Now simplify this we have,
$\Rightarrow D={\displaystyle \frac{4}{\sqrt{2}}}$
Now multiply and divide by square root of 2 we have,
$\Rightarrow D={\displaystyle \frac{4}{\sqrt{2}}}\times {\displaystyle \frac{\sqrt{2}}{\sqrt{2}}}=2\sqrt{2}$
So this is the required answer.
Hence option (D) is the correct answer.

Note – Whenever we face such types of questions the key concept we have to remember is that always recall the distance formula between the given point and plane which is stated above, then first find out the plane as above then apply the formula as above and simplify, we will get the required answer.