Let the values of p , for which the shortest distance between the lines (x+1)/(3)=(y)/(4)=…

Question

Let the values of p , for which the shortest distance between the lines $\frac{x+1}{3}=\frac{y}{4}=\frac{z}{5}$ and $\overrightarrow{\mathrm{r}}=(\mathrm{p} \hat{i}+2 \hat{j}+\hat{k})+\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})$

Accepted Answer

$\begin{gathered} \frac{x+1}{3}=\frac{y}{4}=\frac{z}{5};(p\hat{i}+2\hat{j}+\hat{k})+\lambda (2\hat{i}+3\hat{j}+4\hat{k}) \ d=|\frac{(\vec{a}-\vec{b})\cdot ({\vec{p}}_{1}	imes {\vec{p}}_{2})}{|{\vec{p}}_{1}	imes {\vec{p}}_{2}|}|=\frac{1}{\sqrt{6}} \ \vec{a}-\vec{b}=(p+1)\hat{i}+2\hat{j}+\hat{k} \ {\vec{p}}_{1}	imes {\vec{p}}_{2}=|\begin{array}{ccc}\hat{i} \quad  \hat{j} \quad  \hat{k} \ 3 \quad  4 \quad  5 \ 2 \quad  3 \quad  4\end{array}|=\hat{i}-2\hat{j}+\hat{k} \ \frac{1}{\sqrt{6}}=|\frac{(p+1)-4+1}{\sqrt{6}}| \ =|p-2|=1\Rightarrow p=3,1 \ a=1,b=3 \ \frac{{x}^{2}}{1}+\frac{{y}^{2}}{9}=1 \end{gathered}$

Length of $LR=\frac{2{a}^{2}}{b}=\frac{2}{3}$

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