The shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and…
The shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ is
- $\frac{1}{\sqrt{14}}$ units.
- $\frac{1}{\sqrt{5}}$ units.
- $\frac{1}{\sqrt{11}}$ units.
- $\frac{1}{\sqrt{6}}$ units.
Solution
The lines are
$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} \text { and } \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$
Comparing equations with
$\frac{x-x_1}{\mathrm{a}_1}=\frac{y-y_1}{\mathrm{~b}_1}=\frac{\mathrm{z}-\mathrm{z}_1}{\mathrm{c}_1} \text { and } \frac{x-x_2}{\mathrm{a}_2}=\frac{y-y_2}{\mathrm{~b}_2}=\frac{\mathrm{z}-\mathrm{z}_2}{\mathrm{c}_2} \text {, }$
we get
$\begin{aligned}
& x_1=1, y_1=2, \mathrm{z}_1=3 \quad x_2=2, y_2=4, \mathrm{z}_2=5 \\
& \mathrm{a}_1=2, \mathrm{~b}_1=3, \mathrm{c}_1=4 \quad \mathrm{a}_2=3, \mathrm{~b}_2=4, \mathrm{c}_2=5 \\
& \left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & z_2-z_1 \\
\mathrm{a}_1 & \mathrm{~b}_1 & \mathrm{c}_1 \\
\mathrm{a}_2 & \mathrm{~b}_2 & \mathrm{c}_2
\end{array}\right|=\left|\begin{array}{ccc}
1 & 2 & 2 \\
2 & 3 & 4 \\
3 & 4 & 5
\end{array}\right| \\
& =1(15-16)-2(10-12)+2(8-9) \\
& =1 \\
& \therefore \quad \sqrt{\left(a_1 b_2-a_2 b_1\right)^2+\left(b_1 c_2-b_2 c_1\right)^2+\left(c_1 a_2-c_2 a_1\right)^2} \\
& =\sqrt{(2 \times 4-3 \times 3)^2+(3 \times 5-4 \times 4)^2+(4 \times 3-5 \times 2)^2} \\
& =\sqrt{1+1+4} \\
& =\sqrt{6} \\
&
\end{aligned}$
Shortest distance between line is $\mathrm{d}$
$\begin{aligned}
\mathrm{d} & =\frac{\left|\begin{array}{ccc}
x_2-x_1 & y_2-y_1 & \mathrm{z}_2-\mathrm{z}_1 \\
\mathrm{a}_1 & \mathrm{~b}_1 & \mathrm{c}_1 \\
\mathrm{a}_2 & \mathrm{~b}_2 & \mathrm{c}_2
\end{array}\right|}{\sqrt{\left(\mathrm{a}_1 \mathrm{~b}_2-\mathrm{a}_2 \mathrm{~b}_1\right)^2+\left(\mathrm{b}_1 \mathrm{c}_2-\mathrm{b}_2 \mathrm{c}_1\right)^2+\sqrt{\left(\mathrm{c}_1 \mathrm{a}_2-\mathrm{c}_2 \mathrm{a}_1\right)^2}}} \\
& =\frac{1}{\sqrt{6}}
\end{aligned}$
Asked in: MHT CET 2023 (09 May Shift 2)
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