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}}, be a, b,(a

Question

Accepted Answer

Answer: D. \frac{2}{3} \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 	imes \vec{p}_2ight)}{\left|\vec{p}_1 	imes \vec{p}_2ight|}ight|=\frac{1}{\sqrt{6}} \ & \vec{a}-\vec{b}=(p+1) \hat{i}+2 \hat{j}+\hat{k} \ & \vec{p}_1 	imes \vec{p}_2=\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \ 3 & 4 & 5 \ 2 & 3 & 4 \end{array}ight|=\hat{i}-2 \hat{j}+\hat{k} \ & \frac{1}{\sqrt{6}}=\left|\frac{(p+1)-4+1}{\sqrt{6}}ight| \ & =|p-2|=1 \Rightarrow p=3,1 \ & a=1, b=3 \ & \frac{x^2}{1}+\frac{y^2}{9}=1 \end{aligned} Length of L R=\frac{2 a^2}{b}=\frac{2}{3}

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 $\frac{1}{\sqrt{6}}$, be $a, b,(a

Answer

Explanation

More JEE Math questions