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 :

• A. \(\frac{1}{19}\) units
• B. \(\sqrt{19}\) units
• C. \(\frac{3}{\sqrt{19}}\) units
• D. \(\frac{1}{\sqrt{19}}\) units

Hint:

Use \(d=\frac{|(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2})|}{|\vec{b_1}\times\vec{b_2}|}\).

Answer 1

The Correct Option is C

The distance is the length of the projection of the join of the two points on the common perpendicular direction \(\vec n=\vec{b_1}\times\vec{b_2}\).
Cross product components: \(\hat i:(-3)(1)-(2)(3)=-9\); \(\hat j:-[(1)(1)-(2)(2)]=3\); \(\hat k:(1)(3)-(-3)(2)=9\). So \(\vec n=(-9,3,9)\), \(|\vec n|=3\sqrt{19}\).
Join vector \((3,3,3)\): \(\vec n\cdot(3,3,3)=9\).
\(d=\frac{9}{3\sqrt{19}}=\frac{3}{\sqrt{19}}\) units.
Answer: (3) \(\frac{3}{\sqrt{19}}\) units