Question

A plane which passes through the point $(3,2,0)$ and the line $\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}$ is:

• A. $x-y+z=1$
• B. $x+y+z=5$
• C. $x+2y-z=1$
• D. $2x-y+z=5$

Answer 1

Answer: (A)

If a plane contains a line, then the normal to the plane is perpendicular to the line.
Any plane passing through $(3,2,0)$ is
$a(x-3)+b(y-2)+c(z-0)=0\dots$ (i)
It passes through $(3,6,4)$
$\therefore 0.a+4b+4c=0\dots$ (ii)
Normal to plane (i) is perpendicular to given line
$\therefore a+5b+4c=0\dots$ (ii)
On solving Eqs. (i) and (ii), we get
$\frac{a}{1}=\frac{b}{-1}=\frac{c}{1}=k$
So, $a=k,b=-k,c=k$
Putting the value of $a,b,c$ in Eq. (i),
we get $x-y+z=1$
In any line of a plane it has only one direction cosines and direction ratios may be more than one.