Three Dimensional Geometry Question 31
Question 31 - 01 February - Shift 1
The shortest distance between the lines $\frac{x-5}{1}=\frac{y-2}{2}=\frac{z-4}{-3}$ and $\frac{x+3}{1}=\frac{y+5}{4}=\frac{z-1}{-5}$ is
(1) $7 \sqrt{3}$
(2) $5 \sqrt{3}$
(3) $6 \sqrt{3}$
(4) $4 \sqrt{3}$
Show Answer
Answer: (3)
Solution:
Formula: Skew Lines
Shortest distance between two lines
$ \begin{aligned} & \frac{x-x_1}{a_1}=\frac{y-y_1}{a_2}=\frac{z-z_1}{a_3} \& \\ & \frac{x-x_2}{b_1}=\frac{y-y_2}{b_2}=\frac{z-z_2}{b_3} \text{ is given as } \\ & \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_1 b_3-a_3 b_2)^{2}+(a_1 b_3-a_3 b_1)^{2}+(a_1 b_2-a_2 b_1)^{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}}} \\ & \frac{\begin{vmatrix} 8 & 7 & 3 \\ 1 & 2 & -3 \\ 1 & 4 & -5 \end{vmatrix} }{\sqrt{(2)^{2}+(2)^{2}+(2)^{2}}} \\ & =\frac{|8(-10+12)-7(-5+3)+3(4-2)|}{\sqrt{4+4+4}} \\ & =\frac{16+14+6}{\sqrt{12}}=\frac{36}{\sqrt{12}}=\frac{36}{2 \sqrt{3}} \\ & =\frac{18}{\sqrt{3}}=6 \sqrt{3} \end{aligned} $
