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 $\overrightarrow{\mathrm{r}}=(\mathrm{p} \hat{i}+2 \hat{j}+\hat{k})+\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})$ is $\frac{1}{\sqrt{6}}$

Accepted Answer

$$\begin{aligned}
& \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=\left|\frac{(\vec{a}-\vec{b}) \cdot\left(\vec{p}_1 \times \vec{p}_2\right)}{\left|\vec{p}_1 \times \vec{p}_2\right|}\right|=\frac{1}{\sqrt{6}} \
& \vec{a}-\vec{b}=(p+1) \hat{i}+2 \hat{j}+\hat{k} \
& \vec{p}_1 \times \vec{p}_2=\left|\[\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \ 3 & 4 & 5 \ 2 & 3 & 4 \end{array}\end{aligned}$$\right|=\hat{i}-2 \hat{j}+\hat{k} \
\& \frac{1}{\sqrt{6}}=\left|\frac{(p+1)-4+1}{\sqrt{6}}\right| \
\& =|p-2|=1 \Rightarrow p=3,1 \
\& a=1, b=3 \
\& \frac{x\^2}{1}+\frac{y\^2}{9}=1
\]

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

Solutions · 1