The shortest distance between the lines (x+2)/(1)=(y)/(-2)=(z-5)/(2) and (x-4)/(1)=(y-1)/(…

Question

The shortest distance between the lines $\frac{x+2}{1}=\frac{y}{-2}=\frac{z-5}{2}$ and $\frac{x-4}{1}=\frac{y-1}{2}=\frac{z+3}{0}$ is :

Accepted Answer

Given, the lines are

$$
\begin{aligned}
\frac{x+2}{1} & =\frac{y}{-2}=\frac{z-5}{2} ~~~~..........(i)\\
\text { and } \frac{x-4}{1} & =\frac{y-1}{2}=\frac{z+3}{0} ~~~~..........(ii)
\end{aligned}
$$

Formula for shortest distance between two skew-lines,

$$
\begin{aligned}
S D & =\left|\frac{\left|\[\begin{array}{ccc} 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{array}\end{aligned}$$\right|}{\left|$$\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \ a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \end{array}$$\right|}\right|=\left|\frac{\left|$$\begin{array}{ccc} 6 & 1 & -8 \ 1 & -2 & 2 \ 1 & 2 & 0 \end{array}$$\right|}{\left|$$\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \ 1 & -2 & 2 \ 1 & 2 & 0 \end{array}$$\right|}\right| \\
\& =\left|\frac{6(-4)-1(-2)-8(4)}{|-4 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}|}\right| \\
\& =\left|\frac{-24+2-32}{\sqrt{36}}\right| \\
\& =\left|\frac{-54}{6}\right|=|-9|=9

\]

Solutions · 1