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:

• A. Coplanar
• B. Parallel
• C. Skew
• D. None of these

Answer 1

The correct option is A Coplanar

Any points on the two given lines $\frac{x-a+d}{\alpha -\delta }=\frac{y-a}{\alpha }=\frac{z-a-d}{\alpha +\delta }$ and $\frac{x-b+c}{\beta -\gamma }=\frac{y-b}{\beta }=\frac{z-b-c}{\beta +\gamma }$ can be taken as $(x_1,y_1,z_1)=(a-d,a,a+d)$ and $(x_2,y_2,z_2)=(b-c,b,b+c)$ respectively.
The corresponding $D.R^{\prime }s$ are $(a_1,b_1,c_1)=(\alpha -\delta ,\alpha ,\alpha +\delta )$ and $(a_2,b_2,c_2)=(\beta -\gamma ,\beta ,\beta +\gamma )$
Now, let us verify the determinant, $D=\left|\begin{array}{ccc}{x}_1-x_2& y_1-y_2& z_1-z_2\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|$
$\Rightarrow D=\left|\begin{array}{ccc}a-d-b+c& a-b& a+d-b-c\\ \alpha -\delta & \alpha & \alpha +\delta \\ \beta -\gamma & \beta & \beta +\gamma \end{array}\right|$
Using $C_1\to C_1+C_2+C_3$
$\Rightarrow D=\left|\begin{array}{ccc}3(a-b)& a-b& a+d-b-c\\ 3\alpha & \alpha & \alpha +\delta \\ 3\beta & \beta & \beta +\gamma \end{array}\right|$
Clearly, two columns are proportional. Hence $D=0$
$\therefore$ Given lines are coplanar.