Question

If two lines
$\frac{x-\alpha }{l}=\frac{y-\beta }{m}=\frac{z-\gamma }{n}$ and $\frac{x-{\alpha }^{\prime }}{l^{\prime }}=\frac{y-{\beta }^{\prime }}{m^{\prime }}=\frac{z-{\gamma }^{\prime }}{n^{\prime }}$
lie in the same plane then the equation of the plane is -

• A. $\left|\begin{array}{ccc}x& y& z\\ l& m& n\\ l^{\prime }& m^{\prime }& n^{\prime }\end{array}\right|=0$
• B. $\left|\begin{array}{ccc}x-\alpha & y-\beta & z-\gamma \\ l& m& n\\ l^{\prime }& m^{\prime }& n^{\prime }\end{array}\right|=0$
• C. $\left|\begin{array}{ccc}l-l^{\prime }& m-m^{\prime }& n-n^{\prime }\\ l& m& n\\ l^{\prime }& m^{\prime }& n^{\prime }\end{array}\right|=0$
• D. None of these

Answer 1

The correct option is B $\left|\begin{array}{ccc}x-\alpha & y-\beta & z-\gamma \\ l& m& n\\ l^{\prime }& m^{\prime }& n^{\prime }\end{array}\right|=0$

The vector along the first line can be written as $l\hat{i}+m\hat{j}+n\hat{k}$ and for the second line $l\hat{i}+m^{\prime }\hat{j}+n^{\prime }\hat{k}$. The vector joining points $(\alpha ,\beta ,\gamma )$ and $(x,y,x)$ will be $(x-\alpha ,y-\beta ,z-\gamma ),$ where $(x,y,z)$ is any general point on the plane.
Now, if these two lines lie in the same plane then the volume of parallelepiped formed by these three vectors should be Zero.
We know from the vectors, the volume of parallelepiped joining these three vectors can be written in the determinant form in this way.
$\left|\begin{array}{ccc}x-\alpha & y-\beta & z-\gamma \\ l& m& n\\ l^{\prime }& m^{\prime }& n^{\prime }\end{array}\right|=0$