A) \[x+y\,\text{tan}\,\theta =0\]
B) \[y+x\,\text{tan}\,\theta =0\]
C) \[x\,\cos \theta -y\,\sin \theta =0\]
D) \[x\,\sin \theta -y\,\cos \theta =0\]
Correct Answer: A
Solution :
[a] The plane is perpendicular to the line \[\frac{x-a}{\cos \theta }=\frac{y+2}{\sin \theta }=\frac{z-3}{0}\] Hence, the direction ratios of the normal of the plane are\[\cos \theta \], \[\sin \theta \] and 0. ...(i) Now, the required plane passes through the z-axis. Hence, the point (0, 0, 0) lies on the plane. From Eqs. (i) and (ii), we get equation of the plane as \[\cos \theta (x-0)+\sin \theta (y-0)+0(z-0)=0\] \[\cos \theta x+\sin \theta y=0\] \[x+y\tan \theta =0\]
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