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. $x+ytan\theta =0$
• B. $y+xtan\theta =0$
• C. $xcos\theta -ysin\theta =0$
• D. $xsin\theta -ycos\theta =0$

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
$cos\theta (x-0)+sin\theta (y-0)+0(z-0)=0$
$cos\theta x+sin\theta y=0$
$x+ytan\theta =0$