Question

Condition for two lines $\frac{x-\alpha }{l}=\frac{y-\beta }{m}=\frac{z-\gamma }{n}$ and $\frac{x-{\alpha }^{\prime }}{l^{\prime }}=\frac{y-{\beta }^{\prime }}{m^{\prime }}=\frac{z-{\gamma }^{\prime }}{n^{\prime }}$ to be co - planar is -

• A.
• B.
• C.
• D. None of these

Answer 1

The correct option is A

The vector along the first line can be written as li + mj + nk and for the second line $l^{\prime }l+m^{\prime }j+n^{\prime }k$. The vector joining points $(\alpha ,\beta ,\gamma )$ and $({\alpha }^{\prime },{\beta }^{\prime },\gamma )$ will be $(\alpha -{\alpha }^{\prime },\beta -{\beta }^{\prime },\gamma -{\gamma }^{\prime }).$
We know from the vectors, the volume of parallelepiped joining these three vectors can be written in the determinant form in this way.
I.e, Volume of the parallelepiped = $\left|\begin{array}{ccc}\alpha -{\alpha }^{\prime }& \beta -{\beta }^{\prime }& \gamma -{\gamma }^{\prime }\\ l& m& n\\ l^{\prime }& m^{\prime }& n^{\prime }\end{array}\right|=0$
Now if these two lines lie in the same plane.Then the area of parallelepiped formed by these three vectors should be Zero.
So, we get
$\left|\begin{array}{ccc}\alpha -{\alpha }^{\prime }& \beta -{\beta }^{\prime }& \gamma -{\gamma }^{\prime }\\ l& m& n\\ l^{\prime }& m^{\prime }& n^{\prime }\end{array}\right|=0$