Question

A plane which passes through the point $(3,2,0)$ and the line $\frac{x-3}{1}=\frac{y-7}{5}=\frac{z-4}{4}$ is?

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

Answer 1

The correct option is A $x-y+z=1$

Let the equation of plane passing through $(3,2,0)$ is

$\Rightarrow$ $a(x-3)+b(y-2)+c(z-0)=0$ ---- ( 1 )

$\Rightarrow$ $\frac{x-3}{1}=\frac{y-7}{5}=\frac{z-4}{4}$ pass through $(3,2,0)$ and has direction ratios as $1,5,4$

If the equation of plane contains this, it should pass from $(3,7,4)$ and parallel to $(1,5,4)$

$\therefore$ $a(3-1)+b(7-5)+c(4-4)=0$

$\Rightarrow$ $2a+2b=0$

$\Rightarrow$ $a+b=0$ ---- ( 2 )

And

$\Rightarrow$ $a+5b+4c=0$ ----- ( 3 )

On solving equation ( 2 ) and ( 3 ) we get,

$\Rightarrow$ $a=c$ and $4b=-c$

Hence, equation of plane will be,

$\Rightarrow$ $c(x-3)-c(y-2)+cz=0$ [ From ( 1 ) ]

$\Rightarrow$ $x-3-y+2+z=0$

$\Rightarrow$ $x-y+z-1=0$

$\Rightarrow$ $x-y+z=1$