Question

The lines $\frac{x-a+d}{\alpha -\delta }=\frac{y-a}{\alpha }=\frac{z-a-d}{\alpha +\delta }$ and $\frac{x-b+c}{\beta -r}=\frac{y-b}{\beta }=\frac{z-b-c}{\beta +r}$ are coplanar and then equation to the plane in which they lie, is?

• A. $x+y+z=0$
• B. $x-y+z=0$
• C. $x-2y+z=0$
• D. $x+y-2z=0$

Answer 1

Answer: (C)

$x-2y+z=0$

The lines will be coplanar, if
$\left|\begin{array}{ccc}a-d-b+c& a-b& a+d-b-c\\ \alpha +\delta & \alpha & \alpha +\delta \\ \beta +r& \beta & \beta +r\end{array}\right|=0$
Add $3$rd 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
$\left|\begin{array}{ccc}x-a+d& y-a& z-a-d\\ \alpha -\delta & \alpha & \alpha +\delta \\ \beta -r& \beta & \beta +r\end{array}\right|=0$
On adding $1$st and $3$rd columns and subtracting twice the $2$nd, we get
$\left|\begin{array}{ccc}x+z-2y& y-a& z-a-d\\ 0& \alpha & \alpha +\delta \\ 0& \beta & \beta +r\end{array}\right|=0$
$\Rightarrow \left[\alpha \right(\beta +r)-\beta (\alpha +\delta \left)\right](x+z-2y)=0$
$\Rightarrow (x+z-2y)=0$.