Question

The equation of the plane passing through the origin and containing the line $\frac{x-1}{5}=\frac{y-2}{4}=\frac{z-3}{5}$ is

• A. $x+5y-3z=0$
• B. $x-5y+3z=0$
• C. $3x–10y+5z=0$
• D. $x+5y+3z=0$

Answer 1

The correct option is B

$x-5y+3z=0$

Find the equation of the line based on given information

The equation of the plane containing the given line is given as

$A\left(x-1\right)+B\left(y-2\right)+C\left(z-3\right)=0$ $...\left(i\right)$

Since the line lies on the plane the direction ratios of the plane perpendicular to the line are given as

$5A+4B+5C=0$ $...\left(ii\right)$

Since the origin passes through the plane its co-ordinates satisfy the equation $\left(i\right)$

$\Rightarrow$ $A\left(-1\right)+B\left(-2\right)+C\left(-3\right)=0$

$\Rightarrow$ $A+2B+3C=0$ $...\left(iii\right)$

Solving equations $\left(ii\right),\left(iii\right)$ we get

$\frac{A}{2}=\frac{B}{-10}=\frac{C}{6}$

Hence, the equation of the plane is written as

$2\left(x-1\right)-10\left(y-2\right)+6\left(z-3\right)=0$

$\Rightarrow$ $2x-10y+6z-2+20-18=0$

$\Rightarrow$ $2x-10y+6z=0$

$\Rightarrow$ $x-5y+3z=0$

Hence, option $\left(B\right)$, $x-5y+3z=0$ is the correct answer.