Question

What is the equation of the plane which passes through the $z$-axis and is perpendicular to the line
$\frac{x-a}{cos\, \theta}= \frac{y+2}{sin \,\theta} =\frac{z-3}{0}?$

• A. $x + y\, tan \,\theta=0$
• B. $y + x\, tan\, \theta=0$
• C. $x \,cos \,\theta -y\, sin\, \theta=0$
• D. $x \,sin\, \theta -y \,cos \,\theta=0$

Answer 1

Answer: (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
$cos \,\,\theta (x-0) + sin \,\,\theta (y - 0) + 0 (z - 0) = 0$
$cos \,\,\theta \,x + sin \,\,\theta\,y = 0$
$x+ y\, tan\,\,\theta=0$