A) \[x-y+z=1\]
B) \[x+y+z=5\]
C) \[x+2y-z=0\]
D) \[2x-y+z=5\]
Correct Answer: A
Solution :
Plane passing through (3, 2, 0) is \[A(x-3)+B(y-2)+c(z-0)=0\] ?..(i) Plane (i) is passing through the line, \[\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}\] \[\therefore \] \[A(3-3)+B(6-2)+C(4-0)=0\] \[0.A+4B+4C=0\] ?..(ii) and also 1.A + 5B + 4C = 0 ?..(iii) Solving (ii) and (iii), we get \[x-y+z=1\]. Trick: Required plane is \[\left| \,\begin{matrix} x-3 & y-6 & z-4 \\ 3-3 & 2-6 & 0-4 \\ 1 & 5 & 4 \\ \end{matrix}\, \right|\,=0\] Solving, we get \[x-y+z=1.\]You need to login to perform this action.
You will be redirected in
3 sec