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Mathematics Question on Shortest Distance between Two Lines

The shortest distance between the line:
x−34=y+7−11=z−15\frac{x-3}{4} = \frac{y+7}{-11} = \frac{z-1}{5} and x−53=y−9−6=z+21\frac{x-5}{3} = \frac{y-9}{-6} = \frac{z+2}{1} is:

A

187563\frac{187}{\sqrt{563}}

B

178563\frac{178}{\sqrt{563}}

C

185563\frac{185}{\sqrt{563}}

D

179563\frac{179}{\sqrt{563}}

Answer

187563\frac{187}{\sqrt{563}}

Explanation

Solution

Step 1: Represent the lines in vector form The first line can be written as:

r1⃗=a1⃗+λp⃗,where a1⃗=3i^−7j^+k^,p⃗=4i^−11j^+5k^.\vec{r_1} = \vec{a_1} + \lambda \vec{p}, \quad \text{where } \vec{a_1} = 3\hat{i} - 7\hat{j} + \hat{k}, \quad \vec{p} = 4\hat{i} - 11\hat{j} + 5\hat{k}.

The second line can be written as:

r2⃗=a2⃗+μq⃗,where a2⃗=5i^+9j^−2k^,q⃗=3i^−6j^+k^.\vec{r_2} = \vec{a_2} + \mu \vec{q}, \quad \text{where } \vec{a_2} = 5\hat{i} + 9\hat{j} - 2\hat{k}, \quad \vec{q} = 3\hat{i} - 6\hat{j} + \hat{k}.

Step 2: Find the direction vector perpendicular to both lines The direction vector perpendicular to both lines is:

n⃗=p⃗×q⃗.\vec{n} = \vec{p} \times \vec{q}.

Using the determinant method for the cross product:

n⃗=∣i^j^k^ 4−115 3−61∣.\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\\ 4 & -11 & 5 \\\ 3 & -6 & 1 \end{vmatrix}.

Expanding the determinant:

n⃗=i^((−11)(1)−(−6)(5))−j^((4)(1)−(3)(5))+k^((4)(−6)−(3)(−11)).\vec{n} = \hat{i}((-11)(1) - (-6)(5)) - \hat{j}((4)(1) - (3)(5)) + \hat{k}((4)(-6) - (3)(-11)).

n⃗=i^(−11+30)−j^(4−15)+k^(−24+33).\vec{n} = \hat{i}(-11 + 30) - \hat{j}(4 - 15) + \hat{k}(-24 + 33).

n⃗=19i^+11j^+9k^.\vec{n} = 19\hat{i} + 11\hat{j} + 9\hat{k}.

Step 3: Find AB⃗\vec{AB} The vector AB⃗\vec{AB} is:

AB⃗=a2⃗−a1⃗=(5−3)i^+(9−(−7))j^+(−2−1)k^.\vec{AB} = \vec{a_2} - \vec{a_1} = (5 - 3)\hat{i} + (9 - (-7))\hat{j} + (-2 - 1)\hat{k}.

AB⃗=2i^+16j^−3k^.\vec{AB} = 2\hat{i} + 16\hat{j} - 3\hat{k}.

Step 4: Shortest distance formula The shortest distance between two skew lines is:

S.D.=∣AB⃗⋅n⃗∣∣n⃗∣.\text{S.D.} = \frac{|\vec{AB} \cdot \vec{n}|}{|\vec{n}|}.

Dot product AB⃗⋅n⃗\vec{AB} \cdot \vec{n}:

AB⃗⋅n⃗=(2)(19)+(16)(11)+(−3)(9).\vec{AB} \cdot \vec{n} = (2)(19) + (16)(11) + (-3)(9).

AB⃗⋅n⃗=38+176−27=187.\vec{AB} \cdot \vec{n} = 38 + 176 - 27 = 187.

Magnitude of n⃗\vec{n}:

∣n⃗∣=192+112+92.|\vec{n}| = \sqrt{19^2 + 11^2 + 9^2}.

∣n⃗∣=361+121+81=563.|\vec{n}| = \sqrt{361 + 121 + 81} = \sqrt{563}.

Shortest distance:

S.D.=∣AB⃗⋅n⃗∣∣n⃗∣=187563.\text{S.D.} = \frac{|\vec{AB} \cdot \vec{n}|}{|\vec{n}|} = \frac{187}{\sqrt{563}}.

Final Answer: Option (1).