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Mathematics Question on Straight lines

The shortest distance between the lines x−33=y−8−1=z−31\frac{ x - 3}{3} = \frac{y-8}{-1}= \frac{z - 3}{1} and x+3−3=y+72=z−64\frac{ x + 3}{-3} = \frac{y +7}{2}= \frac{z - 6}{4} is

A

30\sqrt{30}

B

2302 \sqrt{30}

C

5305 \sqrt{30}

D

3303 \sqrt{30}

Answer

3303 \sqrt{30}

Explanation

Solution

l1:x−33=y−8−1=z−31l_1 : \frac{ x - 3}{3} = \frac{y-8}{-1}= \frac{z - 3}{1} l2:x+3−3=y+72=z−64l_2 : \frac{ x + 3}{-3} = \frac{y +7}{2}= \frac{z - 6}{4} Shortest distance between two lines is, =∣∣x2−x1y;2−y1z;2−z1 a1b;1c;1 a2b;2c;2∣(b1c2−b2c1)2+(c1a2−c2a1)2+(a1b2−a2b1)2∣=\left|\frac{\begin{vmatrix}x_{2}-x_{1}&y;_{2}-y_{1}&z;_{2}-z_{1}\\\ a_{1}&b;_{1}&c;_{1}\\\ a_{2}&b;_{2}&c;_{2}\end{vmatrix}}{\sqrt{\left(b_{1} c_{2} -b_{2} c_{1}\right)^{2}+\left(c_{1} a_{2} -c_{2}a_{1}\right)^{2}+\left(a_{1}b_{2}-a_{2}b_{1}\right)^{2}}}\right| =∣∣−3−3−7−86−3 3−11 −324∣(−4−2)2+(−3−12)2+(6−3)2∣=\left|\frac{\begin{vmatrix} -3 -3 & -7 - 8 & 6 - 3\\\ 3 & -1 & 1 \\\ -3 & 2 & 4 \end{vmatrix}}{\sqrt{\left(- 4 - 2 \right)^{2}+\left(-3 -12\right)^{2}+\left(6 - 3\right)^{2}}}\right| =∣∣−6\-153 3−11 −324∣(−6)2+(−15)2+(3)2∣=\left|\frac{\begin{vmatrix} -6 & \- 15 & 3\\\ 3 & -1 & 1 \\\ -3 & 2 & 4 \end{vmatrix}}{\sqrt{\left(-6 \right)^{2}+\left(-15\right)^{2}+\left(3\right)^{2}}}\right| =3(6−3)−1(−12−45)+4(6+45)36+225+9=9+57+204270= \frac{3\left(6-3\right)-1\left(-12-45\right)+4\left(6+45\right)}{\sqrt{36+225+9}} =\frac{9+57+204}{\sqrt{270}} =270270=270=330=\frac{270}{\sqrt{270}}=\sqrt{270}=3\sqrt{30}