UPSC Maths 2013 Paper 1 Q1d — Solution

10 marks · Section A

Question

Find the equation of the plane which passes through the points $(0,1,1)$ and $(2,0,-1)$ , and is parallel to the line joining the points $(-1,1,-2),(3,-2,4)$ . Find also the distance between the line and the plane.

Technique

Build the plane’s normal from two non-parallel direction vectors via cross product; standard “distance from point to plane” with any point on the parallel line.

Solution

Strategy. The plane contains the vector joining its two given points, and is parallel to the vector along the given line. The cross product of these two vectors is the plane’s normal.

Step 1 — Two direction vectors in (or parallel to) the plane

Let $P_1=(0,1,1),\;P_2=(2,0,-1)$ , and let $A=(-1,1,-2),\;B=(3,-2,4)$ be the points defining the parallel line.

$\overrightarrow{P_1P_2}=(2,-1,-2)$ — lies in the plane. $\overrightarrow{AB}=(4,-3,6)$ — parallel to the plane.

Step 2 — Normal via cross product

\overrightarrow{P_1P_2}\times\overrightarrow{AB}=\begin{vmatrix}\hat i & \hat j & \hat k\\ 2 & -1 & -2\\ 4 & -3 & 6\end{vmatrix}=\hat i(-6-6)-\hat j(12+8)+\hat k(-6+4)=(-12,-20,-2).

Scale by $-1/2$ to get a cleaner normal: $\vec n=(6,10,1)$ .

Step 3 — Plane equation

Using $P_1=(0,1,1)$ as a point on the plane:

6(x-0)+10(y-1)+1(z-1)=0

\Longrightarrow\;\boxed{\;6x+10y+z=11.\;}

Step 4 — Distance line-to-plane

Since the line is parallel to the plane (by construction — its direction is perpendicular to $\vec n$ , check $(4,-3,6)\cdot(6,10,1)=24-30+6=0$ ✓), distance from any point on the line to the plane equals distance line-to-plane. Use $A=(-1,1,-2)$ :

d=\frac{|6(-1)+10(1)+(-2)-11|}{\sqrt{6^{2}+10^{2}+1^{2}}}=\frac{|-9|}{\sqrt{137}}=\frac{9}{\sqrt{137}}=\frac{9\sqrt{137}}{137}.

Answer

\boxed{\;d=\dfrac{9}{\sqrt{137}}\;\bigl(=\tfrac{9\sqrt{137}}{137}\approx 0.7689\bigr).\;}