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UPSC 2026 Maths Optional Paper 1 Q2c — Step-by-Step Solution 15 marks · Section A
Shortest distance between two skew lines · Analytic Geometry · asked 5× in 14 yrs · Read the full method →
Question
Find the points on the lines x − 3 1 = y − 5 − 2 = z − 7 1 \dfrac{x-3}{1} = \dfrac{y-5}{-2} = \dfrac{z-7}{1} 1 x − 3 = − 2 y − 5 = 1 z − 7 x + 1 7 = y + 1 − 6 = z + 1 1 \dfrac{x+1}{7} = \dfrac{y+1}{-6} = \dfrac{z+1}{1} 7 x + 1 = − 6 y + 1 = 1 z + 1
Technique
Parametrise a general point on each line, impose that the joining vector P Q ⃗ \vec{PQ} P Q both direction vectors (P Q ⃗ ⋅ d ⃗ 1 = 0 \vec{PQ}\cdot\vec d_1=0 P Q ⋅ d 1 = 0 P Q ⃗ ⋅ d ⃗ 2 = 0 \vec{PQ}\cdot\vec d_2=0 P Q ⋅ d 2 = 0 2 × 2 2\times2 2 × 2 ∣ ( a ⃗ 2 − a ⃗ 1 ) ⋅ ( d ⃗ 1 × d ⃗ 2 ) ∣ ∣ d ⃗ 1 × d ⃗ 2 ∣ \dfrac{|(\vec a_2-\vec a_1)\cdot(\vec d_1\times\vec d_2)|}{|\vec d_1\times\vec d_2|} ∣ d 1 × d 2 ∣ ∣ ( a 2 − a 1 ) ⋅ ( d 1 × d 2 ) ∣ d ⃗ 1 × d ⃗ 2 \vec d_1\times\vec d_2 d 1 × d 2
Solution
Step 1 — Read off points and directions.
L 1 : x − 3 1 = y − 5 − 2 = z − 7 1 = t , a ⃗ 1 = ( 3 , 5 , 7 ) , d ⃗ 1 = ( 1 , − 2 , 1 ) , L_1:\ \frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}=t,\qquad \vec a_1=(3,5,7),\quad \vec d_1=(1,-2,1), L 1 : 1 x − 3 = − 2 y − 5 = 1 z − 7 = t , a 1 = ( 3 , 5 , 7 ) , d 1 = ( 1 , − 2 , 1 ) , L 2 : x + 1 7 = y + 1 − 6 = z + 1 1 = s , a ⃗ 2 = ( − 1 , − 1 , − 1 ) , d ⃗ 2 = ( 7 , − 6 , 1 ) . L_2:\ \frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}=s,\qquad \vec a_2=(-1,-1,-1),\quad \vec d_2=(7,-6,1). L 2 : 7 x + 1 = − 6 y + 1 = 1 z + 1 = s , a 2 = ( − 1 , − 1 , − 1 ) , d 2 = ( 7 , − 6 , 1 ) . d ⃗ 1 \vec d_1 d 1 d ⃗ 2 \vec d_2 d 2 1 / 7 ≠ − 2 / − 6 1/7\neq -2/-6 1/7 = − 2/ − 6
Step 2 — General points and the joining vector.
P = ( 3 + t , 5 − 2 t , 7 + t ) ∈ L 1 , Q = ( − 1 + 7 s , − 1 − 6 s , − 1 + s ) ∈ L 2 , P=(3+t,\;5-2t,\;7+t)\in L_1,\qquad Q=(-1+7s,\;-1-6s,\;-1+s)\in L_2, P = ( 3 + t , 5 − 2 t , 7 + t ) ∈ L 1 , Q = ( − 1 + 7 s , − 1 − 6 s , − 1 + s ) ∈ L 2 , P Q ⃗ = Q − P = ( − 4 + 7 s − t , − 6 − 6 s + 2 t , − 8 + s − t ) . \vec{PQ}=Q-P=\big(-4+7s-t,\;\; -6-6s+2t,\;\; -8+s-t\big). P Q = Q − P = ( − 4 + 7 s − t , − 6 − 6 s + 2 t , − 8 + s − t ) . Step 3 — Impose perpendicularity to both lines.
P Q PQ P Q P Q ⃗ ⊥ d ⃗ 1 \vec{PQ}\perp\vec d_1 P Q ⊥ d 1 P Q ⃗ ⊥ d ⃗ 2 \vec{PQ}\perp\vec d_2 P Q ⊥ d 2
P Q ⃗ ⋅ d ⃗ 1 = 0 \vec{PQ}\cdot\vec d_1=0 P Q ⋅ d 1 = 0
( − 4 + 7 s − t ) ( 1 ) + ( − 6 − 6 s + 2 t ) ( − 2 ) + ( − 8 + s − t ) ( 1 ) = 0 (-4+7s-t)(1)+(-6-6s+2t)(-2)+(-8+s-t)(1)=0 ( − 4 + 7 s − t ) ( 1 ) + ( − 6 − 6 s + 2 t ) ( − 2 ) + ( − 8 + s − t ) ( 1 ) = 0 ( − 4 + 12 − 8 ) + ( 7 + 12 + 1 ) s + ( − 1 − 4 − 1 ) t = 0 ⟹ 20 s − 6 t = 0 ⟹ 10 s − 3 t = 0. (i) (-4+12-8)+(7+12+1)s+(-1-4-1)t=0\ \Longrightarrow\ 20s-6t=0\ \Longrightarrow\ 10s-3t=0. \tag{i} ( − 4 + 12 − 8 ) + ( 7 + 12 + 1 ) s + ( − 1 − 4 − 1 ) t = 0 ⟹ 20 s − 6 t = 0 ⟹ 10 s − 3 t = 0. ( i ) P Q ⃗ ⋅ d ⃗ 2 = 0 \vec{PQ}\cdot\vec d_2=0 P Q ⋅ d 2 = 0
( − 4 + 7 s − t ) ( 7 ) + ( − 6 − 6 s + 2 t ) ( − 6 ) + ( − 8 + s − t ) ( 1 ) = 0 (-4+7s-t)(7)+(-6-6s+2t)(-6)+(-8+s-t)(1)=0 ( − 4 + 7 s − t ) ( 7 ) + ( − 6 − 6 s + 2 t ) ( − 6 ) + ( − 8 + s − t ) ( 1 ) = 0 ( − 28 + 36 − 8 ) + ( 49 + 36 + 1 ) s + ( − 7 − 12 − 1 ) t = 0 ⟹ 86 s − 20 t = 0 ⟹ 43 s − 10 t = 0. (ii) (-28+36-8)+(49+36+1)s+(-7-12-1)t=0\ \Longrightarrow\ 86s-20t=0\ \Longrightarrow\ 43s-10t=0. \tag{ii} ( − 28 + 36 − 8 ) + ( 49 + 36 + 1 ) s + ( − 7 − 12 − 1 ) t = 0 ⟹ 86 s − 20 t = 0 ⟹ 43 s − 10 t = 0. ( ii ) Step 4 — Solve the system.
From (i), t = 10 s 3 t=\dfrac{10s}{3} t = 3 10 s
43 s − 10 ⋅ 10 s 3 = 0 ⟹ 129 s − 100 s 3 = 0 ⟹ 29 s = 0 ⟹ s = 0 , 43s-10\cdot\frac{10s}{3}=0\ \Longrightarrow\ \frac{129s-100s}{3}=0\ \Longrightarrow\ 29s=0\ \Longrightarrow\ s=0, 43 s − 10 ⋅ 3 10 s = 0 ⟹ 3 129 s − 100 s = 0 ⟹ 29 s = 0 ⟹ s = 0 , and hence t = 0 t=0 t = 0 ∣ 10 − 3 43 − 10 ∣ = − 100 + 129 = 29 ≠ 0 \begin{vmatrix}10&-3\\43&-10\end{vmatrix}=-100+129=29\neq0 10 43 − 3 − 10 = − 100 + 129 = 29 = 0
Therefore the nearest points are the base points of the two given lines themselves:
P = ( 3 , 5 , 7 ) , Q = ( − 1 , − 1 , − 1 ) . P=(3,\,5,\,7),\qquad Q=(-1,\,-1,\,-1). P = ( 3 , 5 , 7 ) , Q = ( − 1 , − 1 , − 1 ) . Step 5 — Shortest distance.
P Q ⃗ = Q − P = ( − 4 , − 6 , − 8 ) , \vec{PQ}=Q-P=(-4,-6,-8), P Q = Q − P = ( − 4 , − 6 , − 8 ) , S D = ∣ P Q ⃗ ∣ = 16 + 36 + 64 = 116 = 2 29 ≈ 10.77. SD=|\vec{PQ}|=\sqrt{16+36+64}=\sqrt{116}=2\sqrt{29}\approx 10.77. S D = ∣ P Q ∣ = 16 + 36 + 64 = 116 = 2 29 ≈ 10.77. Perpendicularity check (worth writing down):
P Q ⃗ ⋅ d ⃗ 1 = − 4 + 12 − 8 = 0 , P Q ⃗ ⋅ d ⃗ 2 = − 28 + 36 − 8 = 0. ✓ \vec{PQ}\cdot\vec d_1=-4+12-8=0,\qquad \vec{PQ}\cdot\vec d_2=-28+36-8=0.\ \checkmark P Q ⋅ d 1 = − 4 + 12 − 8 = 0 , P Q ⋅ d 2 = − 28 + 36 − 8 = 0. ✓ Step 6 — Cross-check by the scalar-triple-product formula.
d ⃗ 1 × d ⃗ 2 = ∣ i ^ j ^ k ^ 1 − 2 1 7 − 6 1 ∣ = i ^ ( − 2 + 6 ) − j ^ ( 1 − 7 ) + k ^ ( − 6 + 14 ) = ( 4 , 6 , 8 ) , \vec d_1\times\vec d_2=\begin{vmatrix}\hat i & \hat j & \hat k\\ 1 & -2 & 1\\ 7 & -6 & 1\end{vmatrix}
=\hat i(-2+6)-\hat j(1-7)+\hat k(-6+14)=(4,\,6,\,8), d 1 × d 2 = i ^ 1 7 j ^ − 2 − 6 k ^ 1 1 = i ^ ( − 2 + 6 ) − j ^ ( 1 − 7 ) + k ^ ( − 6 + 14 ) = ( 4 , 6 , 8 ) , ∣ d ⃗ 1 × d ⃗ 2 ∣ = 16 + 36 + 64 = 116 = 2 29 . |\vec d_1\times\vec d_2|=\sqrt{16+36+64}=\sqrt{116}=2\sqrt{29}. ∣ d 1 × d 2 ∣ = 16 + 36 + 64 = 116 = 2 29 . a ⃗ 2 − a ⃗ 1 = ( − 4 , − 6 , − 8 ) , ( a ⃗ 2 − a ⃗ 1 ) ⋅ ( d ⃗ 1 × d ⃗ 2 ) = − 16 − 36 − 64 = − 116. \vec a_2-\vec a_1=(-4,-6,-8),\qquad (\vec a_2-\vec a_1)\cdot(\vec d_1\times\vec d_2)=-16-36-64=-116. a 2 − a 1 = ( − 4 , − 6 , − 8 ) , ( a 2 − a 1 ) ⋅ ( d 1 × d 2 ) = − 16 − 36 − 64 = − 116. S D = ∣ − 116 ∣ 2 29 = 116 2 29 = 58 29 = 58 29 29 = 2 29 . SD=\frac{|-116|}{2\sqrt{29}}=\frac{116}{2\sqrt{29}}=\frac{58}{\sqrt{29}}=\frac{58\sqrt{29}}{29}=2\sqrt{29}. S D = 2 29 ∣ − 116∣ = 2 29 116 = 29 58 = 29 58 29 = 2 29 . The two independent routes agree. Note also that P Q ⃗ = ( − 4 , − 6 , − 8 ) = − ( 4 , 6 , 8 ) \vec{PQ}=(-4,-6,-8)=-(4,6,8) P Q = ( − 4 , − 6 , − 8 ) = − ( 4 , 6 , 8 ) d ⃗ 1 × d ⃗ 2 \vec d_1\times\vec d_2 d 1 × d 2
Step 7 — Equation of the line of shortest distance.
It is the line through P ( 3 , 5 , 7 ) P(3,5,7) P ( 3 , 5 , 7 ) Q ( − 1 , − 1 , − 1 ) Q(-1,-1,-1) Q ( − 1 , − 1 , − 1 ) P Q ⃗ ∥ ( 4 , 6 , 8 ) ∥ ( 2 , 3 , 4 ) \vec{PQ}\parallel(4,6,8)\parallel(2,3,4) P Q ∥ ( 4 , 6 , 8 ) ∥ ( 2 , 3 , 4 ) ∥ d ⃗ 1 × d ⃗ 2 \parallel \vec d_1\times\vec d_2 ∥ d 1 × d 2
x − 3 2 = y − 5 3 = z − 7 4 . \frac{x-3}{2}=\frac{y-5}{3}=\frac{z-7}{4}. 2 x − 3 = 3 y − 5 = 4 z − 7 . Check that Q Q Q putting x = − 1 , y = − 1 , z = − 1 x=-1,y=-1,z=-1 x = − 1 , y = − 1 , z = − 1 − 4 2 = − 6 3 = − 8 4 = − 2 \dfrac{-4}{2}=\dfrac{-6}{3}=\dfrac{-8}{4}=-2 2 − 4 = 3 − 6 = 4 − 8 = − 2
(Equivalently, referred to Q Q Q x + 1 2 = y + 1 3 = z + 1 4 \dfrac{x+1}{2}=\dfrac{y+1}{3}=\dfrac{z+1}{4} 2 x + 1 = 3 y + 1 = 4 z + 1
Answer
Nearest points: P ( 3 , 5 , 7 ) on L 1 , Q ( − 1 , − 1 , − 1 ) on L 2 , Shortest distance = 116 = 2 29 ≈ 10.77 , Line of shortest distance: x − 3 2 = y − 5 3 = z − 7 4 . \boxed{\;
\begin{aligned}
&\text{Nearest points: } P(3,5,7)\ \text{on } L_1,\quad Q(-1,-1,-1)\ \text{on } L_2,\\[2pt]
&\text{Shortest distance } = \sqrt{116}=2\sqrt{29}\ \approx 10.77,\\[2pt]
&\text{Line of shortest distance: } \frac{x-3}{2}=\frac{y-5}{3}=\frac{z-7}{4}.
\end{aligned}\;} Nearest points: P ( 3 , 5 , 7 ) on L 1 , Q ( − 1 , − 1 , − 1 ) on L 2 , Shortest distance = 116 = 2 29 ≈ 10.77 , Line of shortest distance: 2 x − 3 = 3 y − 5 = 4 z − 7 .