Get Answers to all your Questions

header-bg qa

The distance of the point (-1, 2, 3) from the plane\mathrm{\vec{r}\left ( \hat{i}-2\hat{j}+3\hat{k} \right )=10}parallel to the line of the shortest distance
between the lines \mathrm{\vec{r}=\left ( \hat{i}-\hat{j} \right )+\lambda \left ( 2\hat{i}+\hat{k} \right )}and \mathrm{\vec{r}=\left (2 \hat{i}-\hat{j} \right )+\mu \left ( \hat{i}-\hat{j}+\hat{k} \right )} is

Option: 1

2\sqrt{5}


Option: 2

3\sqrt{5}


Option: 3

3\sqrt{6}


Option: 4

2\sqrt{6}


Answers (1)

best_answer

Let \mathrm{L}_1: \overrightarrow{\mathrm{r}}=(\mathrm{i}-\mathrm{j})+\lambda(2 \mathrm{i}+\mathrm{k})
\mathrm{L}_2: \overrightarrow{\mathrm{r}}=(2 \hat{\mathrm{i}}-\hat{\mathrm{j}})+\mu(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})
\overrightarrow{\mathrm{n}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 0 & 1 \\ 1 & -1 & 1\end{array}\right|

\overrightarrow{\mathrm{n}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}-2 \hat{\mathrm{k}}


Equation of line along shortest distance of \mathrm{L}_1$ and $\mathrm{L}_2

\begin{aligned} & \frac{x+1}{1}=\frac{y-2}{-1}=\frac{z-3}{-2}=r \\ & \Rightarrow(x, y, z) \equiv(r-1,2-r, 3-2 r) \\ \end{aligned}

\begin{aligned} & \Rightarrow(r-1)-2(2-r)+3(3-2 r)=10 \\ & \Rightarrow r=-2 \\ \end{aligned}

\begin{aligned} & \Rightarrow Q(x, y, z) \equiv(-3,4,7) \\ & \Rightarrow P Q=\sqrt{4+4+16}=2 \sqrt{6} \end{aligned}

 

Posted by

shivangi.shekhar

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE