UPSC Maths 2025 Paper 1 Q2c-ii — Solution

10 marks · Section A

Question

Find the shortest distance between the straight lines $\dfrac{x-3}{3} = \dfrac{y-8}{-1} = \dfrac{z-3}{1}$ and $\dfrac{x+3}{-3} = \dfrac{y+7}{2} = \dfrac{z-6}{4}$ .

Technique

Use the skew-line formula $d=\dfrac{|(\vec{P_2}-\vec{P_1})\cdot(\vec d_1\times\vec d_2)|}{|\vec d_1\times\vec d_2|}$ .

Solution

Step 1 — Identify points and directions.

Line 1: point $P_1=(3,8,3)$ , direction $\vec d_1=(3,-1,1)$ . Line 2: point $P_2=(-3,-7,6)$ , direction $\vec d_2=(-3,2,4)$ .

Step 2 — Cross product $\vec d_1\times\vec d_2$ .

=\hat i\big((-1)(4)-(1)(2)\big)-\hat j\big((3)(4)-(1)(-3)\big)+\hat k\big((3)(2)-(-1)(-3)\big).$$ $$=\hat i(-4-2)-\hat j(12+3)+\hat k(6-3)=(-6,\,-15,\,3).$$ Magnitude: $$|\vec d_1\times\vec d_2|=\sqrt{(-6)^2+(-15)^2+3^2}=\sqrt{36+225+9}=\sqrt{270}=3\sqrt{30}.$$ **Step 3 — Vector between the points.** $$\vec{P_2}-\vec{P_1}=(-3-3,\,-7-8,\,6-3)=(-6,\,-15,\,3).$$ **Step 4 — Scalar triple product and distance.** $$(\vec{P_2}-\vec{P_1})\cdot(\vec d_1\times\vec d_2)=(-6)(-6)+(-15)(-15)+(3)(3)=36+225+9=270.$$ $$d=\frac{|270|}{3\sqrt{30}}=\frac{270}{3\sqrt{30}}=\frac{90}{\sqrt{30}}=\frac{90\sqrt{30}}{30}=3\sqrt{30}.$$ (The triple product $\neq0$ confirms the lines are skew.) ## Answer $$\boxed{\;d=3\sqrt{30}\ \approx\ 16.43\ \text{units}.\;}$$