Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

For the two lines to be coplanar, the shortest distance between them must be zero.

The first line passes through $A(1,-1,0)$ with direction ratios ${\vec{d}}_1=(2,\lambda ,2)$.

The second line passes through $B(-1,-1,0)$ with direction ratios ${\vec{d}}_2=(5,2,\lambda )$.

The vector joining the points is $\overrightarrow{AB}=(-2,0,0)$.

The condition for coplanarity is $[\overrightarrow{AB},{\vec{d}}_1,{\vec{d}}_2]=0$:

$\left|\begin{array}{ccc}-2& 0& 0\\ 2& \lambda & 2\\ 5& 2& \lambda \end{array}\right|=0$

$-2({\lambda }^2-4)=0\Rightarrow \lambda =\pm 2$

Case 1: $\lambda =2$

${\vec{d}}_1=(2,2,2)$ and ${\vec{d}}_2=(5,2,2)$.

The normal vector to the plane is $\vec{n}={\vec{d}}_1\times {\vec{d}}_2=(0,6,-6)$, which is parallel to $(0,1,-1)$.

The equation of the plane passing through $A(1,-1,0)$ is:

$0(x-1)+1(y+1)-1(z-0)=0\Rightarrow y-z+1=0$

Case 2: $\lambda =-2$

${\vec{d}}_1=(2,-2,2)$ and ${\vec{d}}_2=(5,2,-2)$.

The normal vector to the plane is $\vec{n}={\vec{d}}_1\times {\vec{d}}_2=(0,14,14)$, which is parallel to $(0,1,1)$.

The equation of the plane passing through $A(1,-1,0)$ is:

$0(x-1)+1(y+1)+1(z-0)=0\Rightarrow y+z+1=0$

Combining both cases, the equation of the plane is $y\pm z+1=0$.

Answer: $y\pm z+1=0$