Question

An equation of the plane through the points $(1,0,0),(0,2,0)$and at a distance $6/7$units from the origin is

• A. $6x+3y+z–6=0$
• B. $6x+3y+2z–6=0$
• C. $6x+3y+z+6=0$
• D. $6x+3y+2z+6=0$
• E. $6x+2y+3z+6=0$

Answer 1

The correct option is B

$6x+3y+2z–6=0$

Explanation for the correct option:

Step1. Apply the given condition :

Let the equation of the plane be$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$

The plane$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ passes through the points$(1,0,0),(0,2,0)$.

$$\begin{gathered} \therefore \frac{1}{a}+0+0=1 \\ \Rightarrow a=1 \\ and0+\frac{2}{b}+0=1 \\ \Rightarrow b=2 \end{gathered}$$

And distance $6/7$units from the origin which is given by

$d=\frac{|A{x}_0+B{y}_0+C{z}_0+D|}{\sqrt{(A^2+B^2+C^2)}}$

where $(x_0,y_0,z_0)$is the given point. and $Ax+By+Cz+D=0$ is the equation of the plane.

Here $(x_0,y_0,z_0)=(0,0,0)$and $A=1/a,B=1/b,C=1/candD=-1$ .

$$\begin{gathered} \Rightarrow d=\frac{\left|\right[(1/a)\times 0+(1/b)\times 0+(1/c)\times 0–1]}{\surd (1/a^2)+(1/b^2)+(1/c^2)} \\ \Rightarrow \frac{6}{7}=\frac{|-1|}{\sqrt{1+1/4+1/c^2}} \\ \Rightarrow \frac{36}{49}=\frac{1}{1+1/4+1/c^2} \\ \Rightarrow 1+1/4+1/c^2=49/36 \\ \Rightarrow 1/c^2=1/9 \\ \Rightarrow c=\pm 3 \end{gathered}$$

Step 2. Calculate the plane equation :

Putting the value of $a,bandc$we get the plane equation

$\begin{array}{rcl}\frac{x}{1}+\frac{y}{2}+\frac{z}{3}& =& 1\\ & \Rightarrow & \mathbf{6}\mathit{x}\mathbf{+}\mathbf{3}\mathit{y}\mathbf{+}\mathbf{2}\mathit{z}\mathbf{-}\mathbf{6}\mathbf{}\mathbf{=}\mathbf{0}\end{array}$

Hence the correct option is B.