The shortest distance between the lines $\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5}$ and…
The shortest distance between the lines $\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5}$ and $\frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}$ is
- $8 \sqrt{3}$
- $4 \sqrt{3}$
- $5 \sqrt{3}$
- $6 \sqrt{3}$
Solution
$\begin{aligned} & \frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5} \& \frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3} \\ & \text { S.D }=\frac{\left|\left(\overline{\mathrm{a}}_2 \cdot \overline{\mathrm{a}}_1\right) \cdot\left(\overline{\mathrm{b}}_1 \cdot \overline{\mathrm{b}}_2\right)\right|}{\left|\overline{\mathrm{b}}_1 \times \overline{\mathrm{b}}_2\right|} \\ & a_1=3,-15,9 \\ & \mathrm{~b}_1=2,-7,5 \\ & \mathrm{a}_2=-1,1,9 \\ & \mathrm{~b}_2=2,1,-3 \\ & a_2-a_1=-4,16,0 \\ & \overline{\mathrm{b}}_1 \times \overline{\mathrm{b}}_2=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & -7 & 5 \\ 2 & 1 & -3\end{array}\right|=\hat{\mathrm{i}}(16)-\hat{\mathrm{j}}(-16)+\hat{\mathrm{k}}(16) \\ & 16(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}) \\ & \left|\overline{\mathrm{b}}_1 \times \overline{\mathrm{b}}_2\right|=16 \sqrt{3} \\ & \therefore\left(\overline{\mathrm{a}}_2-\overline{\mathrm{a}}_1\right) \cdot\left(\overline{\mathrm{b}}_1-\overline{\mathrm{b}}_2\right)=16[-4+16]=(16)(12) \\ & \text { S.D. }=\frac{(16)(12)}{16 \sqrt{3}}=4 \sqrt{3} \\ & \end{aligned}$
Asked in: JEE Main 2024 (06 Apr Shift 1)
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