Question

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

• A. $\sqrt{30}$
• B. $2 \sqrt{30}$
• C. $5 \sqrt{30}$
• D. $3 \sqrt{30}$

Answer 1

The Correct Option is D

$l_1 : \frac{ x - 3}{3} = \frac{y-8}{-1}= \frac{z - 3}{1} $ $l_2 : \frac{ x + 3}{-3} = \frac{y +7}{2}= \frac{z - 6}{4} $ Shortest distance between two lines is, $=\left|\frac{\begin{vmatrix}x_{2}-x_{1}&y_{2}-y_{1}&z_{2}-z_{1}\\ a_{1}&b_{1}&c_{1}\\ a_{2}&b_{2}&c_{2}\end{vmatrix}}{\sqrt{\left(b_{1} c_{2} -b_{2} c_{1}\right)^{2}+\left(c_{1} a_{2} -c_{2}a_{1}\right)^{2}+\left(a_{1}b_{2}-a_{2}b_{1}\right)^{2}}}\right|$ $=\left|\frac{\begin{vmatrix} -3 -3 & -7 - 8 & 6 - 3\\ 3 & -1 & 1 \\ -3 & 2 & 4 \end{vmatrix}}{\sqrt{\left(- 4 - 2 \right)^{2}+\left(-3 -12\right)^{2}+\left(6 - 3\right)^{2}}}\right|$ $=\left|\frac{\begin{vmatrix} -6 & - 15 & 3\\ 3 & -1 & 1 \\ -3 & 2 & 4 \end{vmatrix}}{\sqrt{\left(-6 \right)^{2}+\left(-15\right)^{2}+\left(3\right)^{2}}}\right|$ $= \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}} $ $=\frac{270}{\sqrt{270}}=\sqrt{270}=3\sqrt{30}$