JEE Main On 16 April 2018 Question 11
Question: The sum of the intercepts on the coordinate axes of the plane passing through the points $ (-2,-2,2) $ and containing the line joining the points and is? The differential equation representing the family of ellipse having foci either on the x-axis or on the y-axis, centre at the origin and passing through the points $ (1,-1,2) $ and $ (1,1,1) $ is? [JEE Main 16-4-2018]
Options:
A) 12
B) -8
C) -4
D) 4
Show Answer
Answer:
Correct Answer: C
Solution:
- $ \begin{vmatrix} x-x _1 & y-y _1 & z-z _1 \\ z _2-x _1 & y _2-y _1 & z _2-z _1 \\ x _3 & y _3-y _1 & z _3-z _1 \\ \end{vmatrix} =0 $ $ \begin{vmatrix} x+2 & y+2 & z-2 \\ 1+2 & -1+2 & 2-2 \\ 1+2 & 1+2 & 1-2 \\ \end{vmatrix} =0 $ $ \begin{vmatrix} x+2 & y+2 & z+2 \\ 3 & 1 & 0 \\ 3 & 3 & -1 \\ \end{vmatrix} =0 $
$ \Rightarrow $ $ (z-2)(9-3)-1(x+2-3y-6)=0 $
$ \Rightarrow $ $ 6z-12-x-2+3y+6=0 $
$ \Rightarrow $ $ -x+3y+6z-8=0 $
$ \Rightarrow $ $ \frac{x}{8}-\frac{3y}{8}-\frac{6z}{8}+\frac{8}{8}=0 $
$ \Rightarrow $ $ \frac{x}{8}-\frac{y}{\frac{8}{3}}-\frac{z}{\frac{8}{6}}=1 $
$ \Rightarrow $ $ \frac{x}{-8}+\frac{9}{\frac{8}{3}}+\frac{z}{\frac{8}{6}}=1 $ Sum of intercepts $ =-8+\frac{8}{3}+\frac{8}{6} $ $ =-8+\frac{16+8}{6} $ $ =-8+\frac{24}{6} $ $ =-8+4 $ $ =-4 $
