JEE Main & Advanced
Mathematics
Three Dimensional Geometry
Question Bank
Critical Thinking
question_answer
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
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 \[\left| \,\begin{matrix} x-a+d & y-a & z-a-d \\ \alpha -\delta & \alpha & \alpha +\delta \\ \beta -\gamma & \beta & \beta +\gamma \\ \end{matrix}\, \right|=0\]
Adding 1st and 3rd columns and subtracting twice the 2nd, we get \[\left| \,\begin{matrix} x+z-2y & y-a & z-a-d \\ 0 & \alpha & \alpha +\delta \\ 0 & \beta & \beta +\gamma \\ \end{matrix}\, \right|=0\]