Question
What is the equation of the plane through $$z$$-axis and parallel to the line $$\frac{{x - 1}}{{\cos \,\theta }} = \frac{{y + 2}}{{\sin \,\theta }} = \frac{{z - 3}}{0}\,?$$
A.
$$x\,\cot \,\theta + y = 0$$
B.
$$x\,\tan \,\theta - y = 0$$
C.
$$x + y\,\cot \,\theta = 0$$
D.
$$x - y\,\tan \,\theta = 0$$
Answer :
$$x\,\tan \,\theta - y = 0$$
Solution :
Let equation of plane through $$z$$-axis is $$ax + by = 0$$
It is given that this plane is parallel to the line $$\frac{{x - 1}}{{\cos \,\theta }} = \frac{{y + 2}}{{\sin \,\theta }} = \frac{{z - 3}}{0}$$
Since the plane parallel to the line
$$\eqalign{
& \therefore \,a\,\cos \,\theta + b\,\sin \,\theta = 0 \cr
& \Rightarrow a\,\cos \,\theta = - b\,\sin \,\theta \cr
& \Rightarrow a = - b\,\tan \,\theta \cr
& \therefore \, - b\,\tan \,\theta \,x + by = 0 \cr
& \Rightarrow x\,\tan \,\theta - y = 0\,\,\,\,\,\,\,\left( {\because \,b \ne 0} \right) \cr} $$
which is required equation of plane.