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Mathematics Question on Three Dimensional Geometry

The equation of a plane passing through the line of intersection of the planes x+2y+3z=2x + 2y + 3z = 2 and x−y+z=3x - y + z = 3 and at a distance 23\frac{2}{\sqrt 3} from the point (3,1,−1)(3,1, -1) is

A

5x−11y+z=175x-11y+z=17

B

2x+y=32−1\sqrt {2x}+y = 3\sqrt 2-1

C

x+y+z=3 x+ y + z = \sqrt 3

D

x−2y=1−2x -\sqrt {2y} = 1-\sqrt 2

Answer

5x−11y+z=175x-11y+z=17

Explanation

Solution

PLAN
(i) Equation of plane through intersection of two planes,
    i.e,(a1x+b1y+c1z+d1)+λ\, \, \, \, i.e, (a_1 x + b_1 y +c_1 z +d_1)+ \lambda
\hspace30mm (a_2 x +b_2 y +c_2 z +d_2) = 0
(ii) Distance of a point (x1,y1,z1)(x_1,y_1,z_1) from
\hspace15mm ax+by+cz+d = 0
\hspace50mm =\frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}
Equation of plane passing through intersection of two
planes x + 2y + 3z = 2 and x - y + z = 3 is
        (x+2y+3z−2)+λ(x−y+z−3)=0\, \, \, \, \, \, \, \, (x + 2y + 3z - 2) + \lambda (x - y + z - 3) = 0
⇒(1+λ)x+(2+λ)y+(3+λ)z−(2+3λ)=0\Rightarrow (1+ \lambda) x + (2+ \lambda) y +(3+ \lambda)z - (2+3\lambda) = 0
whose distance from (3,1, -1) is 23.\frac{2}{\sqrt 3}.
⇒∣3 (1+λ)+1.(2−λ)−1 (3+λ)−(2+3λ)∣(1+λ)2+(2−λ)2+(3+λ)2=23\Rightarrow \frac{|3\, (1+ \lambda) +1. (2 - \lambda) -1\, (3+ \lambda) - (2+3\lambda)| }{\sqrt {(1+ \lambda)^2 + (2 - \lambda)^2 + (3+ \lambda)^2}} = \frac{2}{\sqrt 3}
⇒\Rightarrow \hspace50mm \frac{| -2 \lambda\, | }{\sqrt {3\lambda^2 + 4 \lambda^2 +14}} = \frac{2}{\sqrt 3}
\Rightarrow \hspace10mm 3\lambda^2=3\lambda^2+4\lambda+14
\Rightarrow \hspace15mm \lambda= -\frac{7}{2}
∴(1−72)x+(2−72)y+(3−72)z−(2−212)=0\therefore \bigg( 1 -\frac{7}{2}\bigg)x +\bigg( 2 -\frac{7}{2}\bigg)y+\bigg( 3 -\frac{7}{2}\bigg)z-\bigg( 2 -\frac{21}{2}\bigg) = 0
\Rightarrow \hspace40mm −5x2+112y−12z+172=0-\frac{5x}{2}+\frac{11}{2}y-\frac{1}{2}z+\frac{17}{2}=0
or \hspace50mm {5x} -{11} y +z+ 17 =0