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

Correct Answer is : 6x+3y+2z−6=06x + 3y + 2 z - 6 = 06x+3y+2z−6=0

Correct Answer is : 6x+3y+2z−6=06x + 3y + 2 z - 6 = 06x+3y+2z−6=0

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KEAM 2011 Math Paper

Question : 28 of 120

Marks: +1, -0

(1, 0, 0)

(0, 2, 0)

\frac{6}{7}

units from the origin is

Solution:

The equation of plane passing through

(1,0,0)

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+2 b=0

\Rightarrow \;\; a=2 b

....(ii)

Given, distance from origin to plane (i)

= \frac{6}{7}

\; \frac{\left|a(0-1)+b(0-0)+c(0-0)\right|}{\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{2 b}{\sqrt{4 b^{2}+b^{2}+c^{2}}} = \frac{6}{7}

[from Eq.(ii)]

\Rightarrow \;\; 14 b=6 \sqrt{5 b^{2}+c^{2}}

Squaring on both sides;

\Rightarrow \;\; 196 b^{2}=36(5 b^{2}+c^{2})

\Rightarrow \;\; 196 b^{2}=180 b^{2}+36 c^{2}

\Rightarrow \;\; 16 b^{2}=36 c^{2}

\Rightarrow \;\; 4 b=6 c

....(iii)

From Eq. (ii) and Eq. (iii),

2 a=4 b=6 c

\Rightarrow \;\; \frac{a}{6} = \frac{b}{3} = \frac{c}{2}

So, required equation of plane is,

6(x-1)+3(y-0)+2(z-0)=0

\;\; 6 x+3 y+2 z-6=0

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