Question

The 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

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

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

$\Rightarrow a(x-3)+b(y-2)+c(z-0)=\mathrm{0..............}\left(1\right)$

and

$\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}$

has direction ratios (1,5,4) and passes through the point (3,6,4)

If the equation of the plane contains this ,it should pass from (3,6,4) and parallel to (1,5,4)

$\therefore a(3-3)+b(6-2)+c(4-0)=0$

$\Rightarrow 4b+4c=0$

$\Rightarrow b+c=\mathrm{0............}\left(2\right)$

and

$a_1{a}_2+b_1{b}_2+c_1{c}_2=0$

$\Rightarrow a\cdot 1+b\cdot 5+c\cdot 4=0$

$\Rightarrow a+5b+4c=\mathrm{0.....}\left(3\right)$

Solving eq(2) and eq(3), we get

$\Rightarrow b=-aandc=a$

Hence, the equation of the plane will be,

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

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

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

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

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

$\Rightarrow x-y+z=1$