If the shortest distance between the lines [[arrayll L_1: r=(2+) i+(1-3 ) j+(3+4 ) k, & R …

Question

If the shortest distance between the lines

$$\[\begin{array}{ll} L_1: \vec{r}=(2+\lambda) \hat{i}+(1-3 \lambda) \hat{j}+(3+4 \lambda) \hat{k}, & \lambda \in \mathbb{R} \ L_2: \vec{r}=2(1+\mu) \hat{i}+3(1+\mu) \hat{j}+(5+\mu) \hat{k}, & \mu \in \mathbb{R} \end{array}$$\]

is $$\frac{m}{\sqrt{n}}$$, where $$\operatorname{gcd}(m, n)=1$$, then the value of $m+n$ equals

Accepted Answer

$$\begin{aligned}
& L_1: \vec{r}=(2+\lambda) \hat{i}+(1-3 \lambda) \hat{j}+(3+4 \lambda) \hat{k} \
& L_1=2 \hat{i}+\hat{j}+3 \hat{k}+\lambda(\hat{i}-3 \hat{j}+4 \hat{k}) \
& L_2: \vec{r}=2 \hat{i}+3 \hat{j}+5 \hat{k}+\mu(2 \hat{i}+3 \hat{j}+\hat{k}) \
& \vec{a}_1=2 \hat{i}+\hat{j}+3 \hat{k} \
& \vec{a}_2=2 \hat{i}+3 \hat{j}+5 \hat{k} \
& \vec{a}_2-\vec{a}_1=2 \hat{j}+2 \hat{k} \
& \vec{b}_1=\hat{i}-3 \hat{j}+4 \hat{k}, \vec{b}_2=2 \hat{i}+3 \hat{j}+\hat{k} \
& \vec{b}_1 \times \vec{b}_2=\left|\[\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \ 1 & -3 & 4 \ 2 & 3 & 1 \end{array}\end{aligned}$$\right|
\]

$$\begin{aligned}
& \hat{i}(-3-12)-\hat{j}(1-8)+\hat{k}(3+6) \
& =-15 \hat{i}+7 \hat{j}+9 \hat{k} \
& \left|\vec{b}_1 \times \vec{b}_2\right|=\sqrt{225+49+81} \
& \left|\frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right|=\frac{14+18}{\sqrt{355}}=\frac{32}{\sqrt{355}} \
& m+n=387
\end{aligned}$$

Solutions · 1