The distance between the lines $\vec{r}=\hat{i}-2 \hat{j}+3 \hat{k}+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})$ and $\vec{r}=3 \hat{i}-2 \hat{j}+1 \hat{k}+\mu(4 \hat{i}+6 \hat{j}+12 \hat{k})$ is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $\frac{\sqrt{328}}{7}$
$\vec{r}=\hat{i}-2 \hat{j}+3 \hat{k}+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})$
$\vec{r}=3 \hat{i}-2 \hat{j}+1 \hat{k}+\mu(4 \hat{i}+6 \hat{j}+12 \hat{k})$
$\vec{a_2}=3 \hat{i}-2 \hat{j}+\hat{k},\vec{a_1}=\hat{i}-2 \hat{j}+3 \hat{k}$
$\vec{b_1}=2 \hat{i}+3 \hat{j}+6 \hat{k},\vec{b_2}=4 \hat{i}+6 \hat{j}+12 \hat{k}$
$\vec{b_1}=\vec{b_2}/2$
so $\vec{b_1}||\vec{b_2}$
$\vec b=\vec{b_1}$
shortest distance = $\frac{|\vec b×(\vec{a_2}-\vec{a_1})|}{|\vec b|}$
$\frac{(2\hat i+3\hat j+6\hat k)×(2\hat i-2\hat k)}{\sqrt{4+9+36}}$
$=\frac{|-6\hat i+16\hat j-6\hat k|}{\sqrt{49}}=\frac{\sqrt{328}}{7}$