Question:
The shortest distance between the lines \(\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-3 \hat{j}+2 \hat{k})\) and \(\vec{r}=(4 \hat{i}+5 \hat{j}+6 \hat{k})+\mu(2 \hat{i}+3 \hat{j}+\hat{k})\) is :
The shortest distance between the lines \(\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-3 \hat{j}+2 \hat{k})\) and \(\vec{r}=(4 \hat{i}+5 \hat{j}+6 \hat{k})+\mu(2 \hat{i}+3 \hat{j}+\hat{k})\) is :
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Use \(d=\frac{|(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2})|}{|\vec{b_1}\times\vec{b_2}|}\).
Updated On: Nov 12, 2026
- \(\frac{1}{19}\) units
- \(\sqrt{19}\) units
- \(\frac{3}{\sqrt{19}}\) units
- \(\frac{1}{\sqrt{19}}\) units
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The Correct Option is C
Solution and Explanation
Here \(\vec{a_2}-\vec{a_1}=3\hat{i}+3\hat{j}+3\hat{k}\), \(\vec{b_1}=\hat{i}-3\hat{j}+2\hat{k}\), \(\vec{b_2}=2\hat{i}+3\hat{j}+\hat{k}\).
\(\vec{b_1}\times\vec{b_2}=\begin{vmatrix}\hat i&\hat j&\hat k\\1&-3&2\\2&3&1\end{vmatrix}=-9\hat{i}+3\hat{j}+9\hat{k}\), with magnitude \(\sqrt{81+9+81}=\sqrt{171}=3\sqrt{19}\).
Numerator: \(|3(-9)+3(3)+3(9)|=|-27+9+27|=9\).
\(d=\frac{9}{3\sqrt{19}}=\frac{3}{\sqrt{19}}\) units.
Answer: (3) \(\frac{3}{\sqrt{19}}\) units
\(\vec{b_1}\times\vec{b_2}=\begin{vmatrix}\hat i&\hat j&\hat k\\1&-3&2\\2&3&1\end{vmatrix}=-9\hat{i}+3\hat{j}+9\hat{k}\), with magnitude \(\sqrt{81+9+81}=\sqrt{171}=3\sqrt{19}\).
Numerator: \(|3(-9)+3(3)+3(9)|=|-27+9+27|=9\).
\(d=\frac{9}{3\sqrt{19}}=\frac{3}{\sqrt{19}}\) units.
Answer: (3) \(\frac{3}{\sqrt{19}}\) units
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