Question

the euation of plane passing through thepoints (-1,1,1),(2,-1,3) and (2,1,0,) is​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

Plane passes through the points (-1,1,1),(2,-1,3) and (2,1,0).

Let assume that the direction ratios of the plane be ( a, b, c ).

Let the equation of plane which passes through the point ( - 1, 1, 1 ) having direction ratios ( a, b, c ) be

$\red{\rm :\longmapsto\:a(x + 1) + b(y - 1) + c(z - 1) = 0 - - (1)}$

Since, plane (1) passes through ( 2, - 1, 3 )

$\rm :\longmapsto\:a(2 + 1) + b( - 1 - 1) + c(3 - 1) = 0$

$\rm :\longmapsto\:3a - 2b + 2c = 0 - - - (2)$

Also, plane (1) passes through ( 2, 1, 0 ).

$\rm :\longmapsto\:a(2 + 1) + b(1 - 1) + c(0 - 1) = 0$

$\rm :\longmapsto\:3a - c = 0$

$\bf\implies \:c = 3a - - - (3)$

Substituting the value of c in equation (2), we get

$\rm :\longmapsto\:3a - 2b + 2(3a) = 0$

$\rm :\longmapsto\:3a - 2b + 6a= 0$

$\rm :\longmapsto\:9a - 2b= 0$

$\bf :\longmapsto\:9a= 2b - - (4)$

From equation (3), on multiply by 3, we get

$\bf\implies \:3c = 9a - - - (5)$

Now, Equating equation (4) and equation (5), we get

$\rm :\longmapsto\:9a = 2b = 3c$

Let assume that,

$\rm :\longmapsto\:9a = 2b = 3c = k$

$\rm :\longmapsto\:a = \dfrac{k}{9}$

$\rm :\longmapsto\:b = \dfrac{k}{2}$

$\rm :\longmapsto\:c = \dfrac{k}{3}$

On substituting the values of a, b, c in equation (1), we get

$\red{\rm :\longmapsto\:\dfrac{k}{9} (x + 1) + \dfrac{k}{2}(y - 1) + \dfrac{k}{3}(z - 1) = 0 - - (1)}$

can be rewritten as

$\red{\rm :\longmapsto\:\dfrac{1}{9} (x + 1) + \dfrac{1}{2}(y - 1) + \dfrac{1}{3}(z - 1) = 0}$

$\red{\rm :\longmapsto\:2(x + 1) + 9(y - 1) + 6(z - 1) = 0}$

$\red{\rm :\longmapsto\:2x +2 + 9y - 9 + 6z - 6 = 0}$

$\red{\rm :\longmapsto\:2x + 9y + 6z - 13 = 0}$