Find the perpendicular distance from the point $P (1, 2, 3)$ to the plane $2 x - y + 2 z = 7$ .

medium 3 marks Class 12 · Mathematics · Three Dimensional Geometry

Answer

\frac{1}{3}

The correct answer is $\frac{1}{3}$ — Thus $d = \frac{1}{3}$ .

Step-by-step solution

Use the distance formula $d = \frac{∣ A x _{1} + B y _{1} + C z _{1} + D ∣}{A ^{2} + B ^{2} + C ^{2}}$ for plane $A x + B y + C z + D = 0$ .
Rewrite the plane as $2 x - y + 2 z - 7 = 0$ , so $A = 2$ , $B = - 1$ , $C = 2$ , $D = - 7$ .
Plug the point coordinates: $∣2 (1) - 1 (2) + 2 (3) - 7∣ = ∣2 - 2 + 6 - 7∣ = ∣ - 1∣ = 1$ .
Compute the denominator: $2^{2} + (- 1)^{2} + 2^{2} = 4 + 1 + 4 = 9 = 3$ .
Thus $d = \frac{1}{3}$ .

Common mistake: Forgetting the absolute value in the numerator or using the wrong denominator.

Concept tested: Distance of point from a plane