Question

An equation of the plane through the points (1, 0, 0) and (0, 2, 0) and at a distance $\frac{6}{7}$ 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

Answer: (B)

The equation of plane passing through (1, 0, 0) is $a(x-1)+b(y-0)+c(z-0)=0$ ...(i) Since, plane also passing through (0, 2, 0).
$\Rightarrow$ $a(0-1)+b(2-0)+c(0-0)=0$
$\Rightarrow$ $-a+2b=0$
$\Rightarrow$ $a=2b$ ...(ii) Given, distance from origin to plane (i) =6/7 $\frac{|a(0-1)+b(0-0)+c(0-0)|}{\sqrt{a^2+b^2+c^2}}=\frac{6}{7}$
$\Rightarrow$ $\left|\frac{-a+0+0}{\sqrt{a^2+b^2+c^2}}\right|=\frac{6}{7}$
$\Rightarrow$ $\frac{a}{\sqrt{a^2+b^2+c^2}}=\frac{6}{7}$
$\Rightarrow$ $\frac{2b}{\sqrt{4b^2+b^2+c^2}}=\frac{6}{7}$ [from Eq.(ii)]
$\Rightarrow$ $14b=6\sqrt{5b^2+c^2}$ Squaring on both sides;
$\Rightarrow$ $196b^2=36(5b^2+c^2)$
$\Rightarrow$ $196b^2=180b^2+36c^2$
$\Rightarrow$ $16b^2=36c^2$
$\Rightarrow$ $4b=6c$ ...(iii) From Eq. (ii) and Eq. (iii),
So, required equation of plane is,
or