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Mathematics Question on Shortest Distance between Two Lines

The shortest distance between the lines x−51=y−22=z−4−3\frac{x-5}{1}=\frac{y-2}{2}=\frac{z-4}{-3} and x+31=y+54=z−1−5\frac{x+3}{1}=\frac{y+5}{4}=\frac{z-1}{-5} is

A

737 \sqrt{3}

B

636 \sqrt{3}

C

434 \sqrt{3}

D

535 \sqrt{3}

Answer

636 \sqrt{3}

Explanation

Solution

The correct answer is option (B) : 636 \sqrt{3}

Shortest distance between two lines
x−x1a1=y−y1a2=z−z1a3\frac{x-x_1}{a_1} = \frac{y-y_1}{a_2} = \frac{z-z_1}{a_3} and x−x2b1=y−y2b2=z−z2b3\frac{x-x_2}{b_1} = \frac{y-y_2}{b_2} = \frac{z-z_2}{b_3} is given as
=∣x1−x2y1−y2z1−z2 a1a2a3 b1b2b3∣(a1b3−a3b2)2+(a1b3−a3b1)2+(a1b2−a2b1)2= \frac{\begin{vmatrix} x_1-x_2 & y_1-y_2 & z_1-z_2\\\ a_1 & a_2 & a_3\\\ b_1 & b_2 & b_3 \end{vmatrix}}{\sqrt{(a_1b_3-a_3b_2)^2+(a_1b_3-a_3b_1)^2+(a_1b_2-a_2b_1)^2}}
=∣5−(3)2−(−5)4−1 12−3 14−5∣(−10+12)2+(−5+3)2+(4−2)2= \frac{\begin{vmatrix} 5-(3) & 2-(-5) & 4-1\\\ 1 & 2 & -3\\\ 1 & 4 & -5 \end{vmatrix}}{\sqrt{(-10+12)^2+(-5+3)^2+(4-2)^2}}
=∣873 12−3 14−5∣(2)2+(−2)2+(2)2= \frac{\begin{vmatrix} 8 & 7 & 3\\\ 1 & 2 & -3\\\ 1 & 4 & -5 \end{vmatrix}}{\sqrt{(2)^2+(-2)^2+(2)^2}}
=∣8(−10+12)−7(−5+3)+3(4−2)∣4+4+4= \frac{\begin{vmatrix} 8(-10+12)-7(-5+3)+3(4-2)\end{vmatrix}}{\sqrt{4+4+4}}
=∣16+14+6∣12= \frac{\begin{vmatrix} 16+14+6\end{vmatrix}}{\sqrt{12}}
=3612=3623= \frac{36}{\sqrt{12}} = \frac{36}{2\sqrt{3}}
=183=63= \frac{18}{\sqrt{3}} = 6\sqrt{3}