Find the perpendicular distance of the point $P (2, - 1, 4)$ from the plane $3 x - 4 y + 2 z = 12$ .

easy 2 marks Class 12 · Mathematics · Three Dimensional Geometry

Answer

\frac{6}{29}

The correct answer is $\frac{6}{29}$ — Step 5: The distance is $\frac{6}{29}$ units.

Step-by-step solution

Step 1: Write the plane equation in standard form $3 x - 4 y + 2 z - 12 = 0$ , so the coefficients are $A = 3$ , $B = - 4$ , $C = 2$ , $D = - 12$ .
Step 2: Use the point‑to‑plane distance formula $\frac{∣ A x _{0} + B y _{0} + C z _{0} + D ∣}{A ^{2} + B ^{2} + C ^{2}}$ .
Step 3: Substitute the point $P (2, - 1, 4)$ : numerator $= ∣3 (2) - 4 (- 1) + 2 (4) - 12∣ = ∣6 + 4 + 8 - 12∣ = ∣6∣ = 6$ .
Step 4: Compute the denominator $3^{2} + (- 4)^{2} + 2^{2} = 9 + 16 + 4 = 29$ .
Step 5: The distance is $\frac{6}{29}$ units.

Common mistake: Using

\frac{∣ A x _{0} + B y _{0} + C z _{0} ∣}{A ^{2} + B ^{2} + C ^{2}}

and forgetting the constant term

D

Concept tested: Distance from point to plane