JEE Main On 8 April 2017 Question 21
Question: The line of intersection of the planes $ \vec{r}.( 3\hat{i}-\hat{j}+\hat{k} )=1 $ and $ \vec{r}.( \hat{i}+4\hat{j}-2\hat{k} )=2, $ is: [JEE Online 08-04-2017]
Options:
A) $ \frac{x-\frac{6}{13}}{2}=\frac{y-\frac{5}{13}}{7}=\frac{z}{-13} $
B) $ \frac{x-\frac{7}{7}}{2}=\frac{y}{-7}=\frac{z+\frac{5}{7}}{13} $
C) $ \frac{x-\frac{6}{13}}{2}=\frac{y-\frac{5}{13}}{-7}=\frac{z}{-13} $
D) $ \frac{x-\frac{4}{7}}{-2}=\frac{y}{7}=\frac{z-\frac{5}{7}}{13} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \vec{n}={{\vec{n}}_1}\times {{\vec{n}}_1} $
$ \begin{vmatrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 3 & -1 & 1 \\ 1 & 4 & -2 \\ \end{vmatrix} $
$ =\hat{i}(-2)-\hat{j}(-7)+\hat{k}(13) $
$ \overline{n}=2\hat{i}+7\hat{j}+13\hat{k} $
Now $ 3x-y+z=1 $
$ x+4y-2z=2 $ But z = 0
$ 3x-y=1\times 4 $
$ x+4y=2 $
$ 13x=6 $ $ x=6/13 $
$ y=5/13 $
