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. $x+2y-z=0$
• D. $2x-y+z=5$

Answer 1

The correct option is A

$x–y+z=1$

Explanation for correct option:

Step1. Expressing the given data according to the question.

Given,

Equation of line, $\frac{x-3}{1}=\frac{y-7}{5}=\frac{z-4}{4}$

A plane passing through the point $(3,2,0)$ & line.

Step2. Writing general formulae of a plane

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

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

Step3. Finding equation plane.

The line has direction ratio $(1,5,4)$

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

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

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

$\Rightarrow$ $\mathrm{a}+\mathrm{b}=0...\left(\mathrm{ii}\right)$

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

On solving equation $\left(\mathrm{ii}\right)\&\left(\mathrm{iii}\right)$ we get,

$$\begin{gathered} a=c \\ b=-c \end{gathered}$$

Therefore equation of plane is

$c(x-3)-c(y-2)+cz=0$ From $\left(\mathrm{i}\right)$

$x-y+z=1$

Hence correct option is $\mathbf{\left(}\mathit{A}\mathbf{\right)}\mathbf{.}$