UPSC 2019 Maths Optional Paper 1 Q1e — Step-by-Step Solution

10 marks · Section A

Straight lines in 3D · Analytic Geometry · asked 5× in 13 yrs · Read the full method →

Question

Show that the lines $\dfrac{x+1}{-3}=\dfrac{y-3}{2}=\dfrac{z+2}{1}$ and $\dfrac{x}{1}=\dfrac{y-7}{-3}=\dfrac{z+7}{2}$ intersect. Find the coordinates of the point of intersection and the equation of the plane containing them.

Technique

Parametrize, solve two of the three equations for $(t,s)$ , verify the third for consistency (intersection test); plane normal via cross product of the two directions, anchored at the meeting point.

Solution

Step 1 — Parametrize both lines

L_1:\ (x,y,z)=(-1,3,-2)+t(-3,2,1),\qquad L_2:\ (x,y,z)=(0,7,-7)+s(1,-3,2).

Direction vectors $\mathbf d_1=(-3,2,1)$ , $\mathbf d_2=(1,-3,2)$ ; points $P_1=(-1,3,-2)$ , $P_2=(0,7,-7)$ .

Step 2 — Test for intersection (coplanarity / solvability)

Set the coordinates equal:

-1-3t=s,\qquad 3+2t=7-3s,\qquad -2+t=-7+2s.

From the first, $s=-1-3t$ . Substitute into the third: $-2+t=-7+2(-1-3t)=-9-6t\Rightarrow 7t=-7\Rightarrow t=-1$ , hence $s=-1-3(-1)=2$ . Consistency check in the second equation: LHS $3+2(-1)=1$ ; RHS $7-3(2)=1$ . ✓ All three equations are satisfied, so the lines intersect.

(Equivalently, the lines are coplanar: the scalar triple product

[\,\mathbf{P_2-P_1},\ \mathbf d_1,\ \mathbf d_2\,]=\begin{vmatrix}1&4&-5\\-3&2&1\\1&-3&2\end{vmatrix}=1(4+3)-4(-6-1)-5(9-2)=7+28-35=0,

and $\mathbf d_1,\mathbf d_2$ are not parallel, so they meet in a unique point.)

Step 3 — Point of intersection

Use $t=-1$ in $L_1$ :

(x,y,z)=(-1-3(-1),\ 3+2(-1),\ -2+(-1))=(2,1,-3).

Check in $L_2$ with $s=2$ : $(0+2,\ 7-6,\ -7+4)=(2,1,-3)$ . ✓

\boxed{\,\text{Point of intersection } (2,\,1,\,-3).\,}

Step 4 — Plane containing the two lines

The plane’s normal is $\mathbf n=\mathbf d_1\times\mathbf d_2$ :

\mathbf n=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\-3&2&1\\1&-3&2\end{vmatrix}=\mathbf i(2\cdot2-1\cdot(-3))-\mathbf j((-3)(2)-1\cdot1)+\mathbf k((-3)(-3)-2\cdot1)=(7,7,7).

Take $\mathbf n=(1,1,1)$ . The plane passes through $(2,1,-3)$ :

1(x-2)+1(y-1)+1(z+3)=0\ \Rightarrow\ x+y+z=0.

Answer

\boxed{\,x+y+z=0.\,}