Question
What are the direction ratios of the line determined by the planes $$x - y + 2z = 1$$ and $$x + y - z = 3\,?$$
A.
$$\left( { - 1,\,3,\,2} \right)$$
B.
$$\left( { - 1,\, - 3,\,2} \right)$$
C.
$$\left( {2,\,1,\,3} \right)$$
D.
$$\left( {2,\,3,\,2} \right)$$
Answer :
$$\left( { - 1,\,3,\,2} \right)$$
Solution :
The intersection of given plane is
$$\eqalign{
& x - y + 2z - 1 + \lambda \left( {x + y - z - 3} \right) = 3 \cr
& \Rightarrow x\left( {1 + \lambda } \right) + y\left( {\lambda - 1} \right) + z\left( {2 - \lambda } \right) - 3\lambda - 1 = 0 \cr} $$
DR’s of normal to the above plane is $$\left( {1 + \lambda ,\,\lambda - 1,\,2 - \lambda } \right)$$
By taking option (A)
$$\eqalign{
& - 1\left( {1 + \lambda } \right) + 3\left( {\lambda - 1} \right) + 2\left( {2 - \lambda } \right) = 0 \cr
& \Rightarrow - 1 - \lambda + 3\lambda - 3 + 4 - 2\lambda = 0 \cr
& \Rightarrow 0 = 0{\text{ which is true}}{\text{.}} \cr} $$