← 2021 Paper 1

UPSC 2021 Maths Optional Paper 1 Q4c — Step-by-Step Solution

15 marks · Section A

Straight lines in 3D · Analytic Geometry · asked 5× in 14 yrs · Read the full method →

Question

Find the plane containing the lines $L_1:\dfrac{x+1}{3}=\dfrac{y+3}{5}=\dfrac{z+5}{7}$ and $L_2:\dfrac{x-2}{1}=\dfrac{y-4}{3}=\dfrac{z-6}{5}$. Also find the point of intersection.

Technique

Coplanarity test via determinant; plane normal = $\vec d_1\times\vec d_2$; parametrise both lines to find intersection.

Solution

Step 1 — Identify points and directions

$L_1$: point $P_1=(-1,-3,-5)$, direction $\vec d_1=(3,5,7)$.

$L_2$: point $P_2=(2,4,6)$, direction $\vec d_2=(1,3,5)$.

Step 2 — Check coplanarity

Lines are coplanar iff $[\vec{P_1 P_2},\vec d_1,\vec d_2]=0$.

$\vec{P_1 P_2}=P_2-P_1=(3,7,11)$.

$\det\begin{pmatrix}3 & 7 & 11\\ 3 & 5 & 7\\ 1 & 3 & 5\end{pmatrix}=?$

Expand along row 1: $3(5\cdot 5-7\cdot 3)-7(3\cdot 5-7\cdot 1)+11(3\cdot 3-5\cdot 1)$ $=3(25-21)-7(15-7)+11(9-5)$ $=3(4)-7(8)+11(4)=12-56+44=0$ ✓.

Lines are coplanar.

Step 3 — Plane equation

Normal to plane: $\vec n=\vec d_1\times\vec d_2$.

$\vec d_1\times\vec d_2=\det\begin{pmatrix}\hat\imath & \hat\jmath & \hat k\\ 3 & 5 & 7\\ 1 & 3 & 5\end{pmatrix}$

$\hat\imath:5\cdot 5-7\cdot 3=25-21=4$. $\hat\jmath:-(3\cdot 5-7\cdot 1)=-(15-7)=-8$. $\hat k:3\cdot 3-5\cdot 1=9-5=4$.

$\vec n=(4,-8,4)$, simplify by dividing by 4: $\vec n=(1,-2,1)$.

Plane through $P_1=(-1,-3,-5)$ with normal $(1,-2,1)$: $(x+1)-2(y+3)+(z+5)=0$, $x-2y+z+1-6+5=0$, $x-2y+z=0$.

Step 4 — Point of intersection of $L_1$ and $L_2$

Parameterise: $L_1$: $(-1+3t,-3+5t,-5+7t)$. $L_2$: $(2+s,4+3s,6+5s)$.

Set equal: $-1+3t=2+s\Rightarrow 3t-s=3.\qquad(i)$ $-3+5t=4+3s\Rightarrow 5t-3s=7.\qquad(ii)$ $-5+7t=6+5s\Rightarrow 7t-5s=11.\qquad(iii)$

From $(i)$: $s=3t-3$. Sub into $(ii)$: $5t-3(3t-3)=7\Rightarrow 5t-9t+9=7\Rightarrow -4t=-2\Rightarrow t=1/2$.

Then $s=3/2-3=-3/2$.

Check $(iii)$: $7(1/2)-5(-3/2)=7/2+15/2=22/2=11$ ✓.

Point: from $L_1$ at $t=1/2$: $(-1+3/2,-3+5/2,-5+7/2)=(1/2,-1/2,-3/2)$.

Verify from $L_2$ at $s=-3/2$: $(2-3/2,4-9/2,6-15/2)=(1/2,-1/2,-3/2)$ ✓.

Step 5 — Verify point on plane

$x-2y+z=1/2-2(-1/2)+(-3/2)=1/2+1-3/2=0$ ✓.

Answer