Question

Show that the lines $\frac{x-a+d}{\alpha -\delta }=\frac{y-a}{\alpha }=\frac{z-a-d}{\alpha +\delta }$
$\frac{x-b+c}{\beta -\gamma }=\frac{y-b}{\beta }=\frac{z-b-c}{\beta +\gamma }$ are coplanar.

Answer 1

We have

$L_1=\frac{x-a+d}{\alpha -\delta }=\frac{y-a}{\alpha }=\frac{z-a-d}{\alpha +\delta }$

$L_2=\frac{x-b+c}{\beta -\gamma }=\frac{y-b}{\beta }=\frac{z-a-d}{\beta +\gamma }$

Now by using determinants methods

$\left|\begin{array}{ccc}a-d-(b-c)& a-b& \left(a+d\right)+\left(b+c\right)\\ \alpha -\delta & \alpha & \alpha +\delta \\ \beta -\gamma & \beta & \beta +\gamma \end{array}\right|$

$C_1\to C_1-C_2{C}_3\to C_3-C_2$

$=\left|\begin{array}{ccc}-d+c& a-b& -c\\ -\delta & \beta & \delta \\ -\gamma & \beta & \gamma \end{array}\right|$

$C_1\to C_1+C_3$

$=\left|\begin{array}{ccc}0& a-b& d-c\\ 0& \alpha & \delta \\ 0& \beta & \gamma \end{array}\right|=0$

Hence, the following lines are coplaner.