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UPSC 2017 Maths Optional Paper 1 Q1e — Step-by-Step Solution

10 marks · Section A

Shortest distance between two skew lines · Analytic Geometry · asked 4× in 13 yrs · Read the full method →

Question

Find the shortest distance between the skew lines

x−33=8−y1=z−31andx+3−3=y+72=z−64.\frac{x-3}{3}=\frac{8-y}{1}=\frac{z-3}{1}\quad\text{and}\quad \frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}.

Technique

Skew-line distance =∣(P2−P1)⋅(d1×d2)∣∣d1×d2∣=\dfrac{|(\mathbf P_2-\mathbf P_1)\cdot(\mathbf d_1\times\mathbf d_2)|}{|\mathbf d_1\times\mathbf d_2|} (scalar triple product over cross-product norm).

Solution

Step 1 — Put both lines in standard symmetric form

In the first line the middle ratio is 8−y1=y−8−1\dfrac{8-y}{1}=\dfrac{y-8}{-1}, so the direction ratio for yy is −1-1 (not +1+1). Thus

L1: x−33=y−8−1=z−31,point P1=(3,8,3), d1=(3,−1,1).L_1:\ \frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1},\qquad \text{point }P_1=(3,8,3),\ \mathbf d_1=(3,-1,1). L2: x+3−3=y+72=z−64,point P2=(−3,−7,6), d2=(−3,2,4).L_2:\ \frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4},\qquad \text{point }P_2=(-3,-7,6),\ \mathbf d_2=(-3,2,4).

Step 2 — Common perpendicular direction d1×d2\mathbf d_1\times\mathbf d_2

d1×d2=∣ijk3−11−324∣=((−1)(4)−(1)(2), (1)(−3)−(3)(4), (3)(2)−(−1)(−3))=(−6,−15,3).\mathbf d_1\times\mathbf d_2=\begin{vmatrix}\mathbf i & \mathbf j & \mathbf k\\ 3 & -1 & 1\\ -3 & 2 & 4\end{vmatrix}=\big((-1)(4)-(1)(2),\ (1)(-3)-(3)(4),\ (3)(2)-(-1)(-3)\big)=(-6,-15,3).

Its magnitude:

∣d1×d2∣=(−6)2+(−15)2+32=36+225+9=270=330.|\mathbf d_1\times\mathbf d_2|=\sqrt{(-6)^2+(-15)^2+3^2}=\sqrt{36+225+9}=\sqrt{270}=3\sqrt{30}.

Step 3 — Project the joining vector onto the common perpendicular

P2−P1=(−3−3, −7−8, 6−3)=(−6,−15,3).\mathbf P_2-\mathbf P_1=(-3-3,\,-7-8,\,6-3)=(-6,-15,3).

The shortest distance is

d=∣(P2−P1)⋅(d1×d2)∣∣d1×d2∣.d=\frac{\big|(\mathbf P_2-\mathbf P_1)\cdot(\mathbf d_1\times\mathbf d_2)\big|}{|\mathbf d_1\times\mathbf d_2|}.

Numerator: (−6)(−6)+(−15)(−15)+(3)(3)=36+225+9=270(-6)(-6)+(-15)(-15)+(3)(3)=36+225+9=270. Therefore

d=270330=9030=903030=330.d=\frac{270}{3\sqrt{30}}=\frac{90}{\sqrt{30}}=\frac{90\sqrt{30}}{30}=3\sqrt{30}.

Answer

  d=330≈16.43.  \boxed{\;d=3\sqrt{30}\approx 16.43.\;}
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