Answer to Question #91549 in Analytic Geometry for Ra

Question

Find the equation of the plane which passes through the line of intersection of the planes x+y-2z=1 and 2x+y-4z=3 and which is perpendicular to the plane x+y+z=1.

Answer 1

Solution:

Equation of the plane passing through the line of intersection of the plane

A₁x + B₁y + C₁z = d₁ and -------------------- (1)

A₂x + B₂y + C₂z = d2 -------------------- (2)

(A₁x + B₁y + C₁z - d₁ ) + $\lambda$ ( (A₂x + B₂y + C₂z - d₂ ) ------------------ (G)

Given equation of the planes

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

and 2x+y-4z=3 ----------------------- (4)

Comparing equation (1) and (3)

A₁ = 1 , B₁ = 1 , C₁ = -2 , d₁ = 1

Comparing equation (2) and (4)

A₂ = 2, B₂= 1 , C₂ = -4 , d₂ = 3

Putting the values in equation (G)

( x + y -2z -1 ) + λ(2x + y -4z -3 )

x + y -2z -1 + 2λx + λy -4λz -3λ = 0

(1+2λ)x + (1+λ)y + (-2 -4λ) z + (-1-3λ) = 0 ---------------- (5)

Also, the plane is perpendicular to the plane x + y + z = 1 ------------- (6)

So the normal vector $\overrightarrow{N}$ to be the plane is perpendicular to the normal vector of x + y + z =1

$\overrightarrow{N}$ = (1+2λ)$\widehat{i}$ + (1+λ)$\widehat{j}$ + (-2 -4λ) $\widehat{j}$ --------------------- (7)

and $\overrightarrow{n}$ = 1$\widehat{i}$ + 1$\widehat{j}$ + 1$\widehat{k}$ ----------------------(8)

We know that two lines with direction ratio a₁, b₁,c₁ and a₂,b₂,c₂ are perpendicular if

a₁a₂ + b₁b₂ + c₁c₂ = 0 ----------------------(9)

From equation (7) and (8) we have

a₁= (1+2λ)

b₁ = (1+λ)

c₁ = (-2 -4λ)

a₂ = 1

b₂ = 1

c₂ = 1

Putting the values in equation (9)

(1+2λ) (1) + (1+λ) (1) + (-2 -4λ)(1) = 0

1+2λ + 1+λ -2 -4λ = 0

3λ - 4λ = 0

So we get λ = 0

Put the value of λ in equation (5)

(1+2λ)x + (1+λ)y + (-2 -4λ) z + (-1-3λ) = 0

(1+2*0)x + (1+0)y + (-2 -4*0) z + (-1-3*0) = 0

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

x + y-2z -1 =0 (Answer)