Three Dimensional Geometry Question 337
Question: The lines $ \frac{x-a+d}{\alpha -\delta }=\frac{y-a}{\alpha }=\frac{z-a-d}{\alpha +\delta } $ and $ \frac{x-b+c}{\beta -\gamma }=\frac{y-b}{\beta }=\frac{z-b-c}{\beta +\gamma } $ are coplanar and then equation to the plane in which they lie, is
Options:
A) $ x+y+z=0 $
B) $ x-y+z=0 $
C) $ x-2y+z=0 $
D) $ x+y-2z=0 $
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Answer:
Correct Answer: C
Solution:
The lines will be coplanar $ | ,\begin{matrix} a-d-b+c & a-b & a+d-b-c \\ \alpha -\delta & \alpha & \alpha +\delta \\ \beta -\gamma & \beta & \beta +\gamma \\ \end{matrix}, |=0 $
Add 3rd column to first and it becomes twice the second and hence the determinant is zero as the two columns are identical. Again the equation of the plane in which they lie is $ | ,\begin{matrix} x-a+d & y-a & z-a-d \\ \alpha -\delta & \alpha & \alpha +\delta \\ \beta -\gamma & \beta & \beta +\gamma \\ \end{matrix}, |=0 $
Adding 1st and 3rd columns and subtracting twice the 2nd, we get $ | ,\begin{matrix} x+z-2y & y-a & z-a-d \\ 0 & \alpha & \alpha +\delta \\ 0 & \beta & \beta +\gamma \\ \end{matrix}, |=0 $
$ \Rightarrow {\alpha ,(\beta +\gamma )-\beta ,(\alpha +\delta )}(x+z-2y)=0 $
$ \Rightarrow x+z-2y=0 $ .
BETA
