← 2015 Paper 1

UPSC 2015 Maths Optional Paper 1 Q3c-i — Step-by-Step Solution

6 marks · Section A

Plane · Analytic Geometry · asked 5× in 13 yrs · Read the full method →

Question

Obtain the equation of the plane passing through the points $(2,3,1)$ and $(4,-5,3)$ parallel to $x$-axis.

Technique

“Parallel to $x$-axis” kills the $x$-coefficient; two points give two linear equations in three unknowns ($B,C,D$ up to scale).

Solution

Strategy. Plane parallel to $x$-axis ⇒ its normal is perpendicular to $\hat\imath=(1,0,0)$ ⇒ the $x$-coefficient in the plane equation is zero. So the plane has the form $By+Cz+D=0$.

Step 1 — Use the two points

Both points satisfy $By+Cz+D=0$:

• $(2,3,1)$: $3B+C+D=0\quad(1)$.
• $(4,-5,3)$: $-5B+3C+D=0\quad(2)$.

Step 2 — Solve for $B:C:D$

$(2)-(1)$: $-8B+2C=0\Rightarrow C=4B$.

From $(1)$: $3B+4B+D=0\Rightarrow D=-7B$.

Take $B=1$: $C=4$, $D=-7$.

Step 3 — Equation

Answer