The shortest distance between lines L_1 and L_2, where L_1: x-12=y+1-3=z+42 and L_2 is the…

Question

The shortest distance between lines $L_1$ and $L_2$, where $L_1:\ \dfrac{x-1}{2}=\dfrac{y+1}{-3}=\dfrac{z+4}{2}$ and $L_2$ is the line passing through the points $A(-4,4,3)$, $B(-1,6,3)$ (and perpendicular to the line $\dfrac{x-3}{-2}=\dfrac{y}{3}=\dfrac{z-1}{1}$), is:

Accepted Answer

$$\begin{aligned}
& \mathrm{L}_2=\frac{\mathrm{x}+4}{3}=\frac{\mathrm{y}-4}{2}=\frac{\mathrm{z}-3}{0} \
& \therefore \mathrm{S} . \mathrm{D}=\frac{\left|\[\begin{array}{ccc} \mathrm{x}_2-\mathrm{x}_1 & \mathrm{y}_2-\mathrm{y}_1 & \mathrm{z}_2-\mathrm{z}_1 \ 2 & -3 & 2 \ 3 & 2 & 0 \end{array}\end{aligned}$$\right|}{\left|\overrightarrow{\mathrm{n}\_1} \times \overrightarrow{\mathrm{n}\_2}\right|} \
\& =\frac{\left|$$\begin{array}{ccc} 5 & -5 & -7 \ 2 & -3 & 2 \ 3 & 2 & 0 \end{array}$$\right|}{\left|\overrightarrow{\mathrm{n}\_1} \times \overrightarrow{\mathrm{n}\_2}\right|} \
\& =\frac{141}{|-4 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+13 \hat{\mathrm{k}}|} \
\& =\frac{141}{\sqrt{16+36+169}} \
\& =\frac{141}{\sqrt{221}}
\]

Solutions · 1