## The equation of the plane passing through the line of intersection of planes 2x + 3y – 4z = 1 , 3x – y + z + 2 = 0 and the point (0, 1, 1) i

Question

The equation of the plane passing through the line of intersection of planes 2x + 3y – 4z = 1 , 3x – y + z + 2 = 0 and the point (0, 1, 1) is

in progress 0
8 months 2021-10-03T07:44:40+00:00 2 Answers 0 views 0

1. ### Solution:

Equation of required plane is 5x+2y-3z+1=0 in Cartesian form.

### Explanatìon:

There are two methods to solve the question

1. Vector form
2. Cartesian form

I’m solving question, in Cartesian form.We can convert Cartesian form in vector form easily . We have to find ,Equation of plane passing through the intersection of two planes Equation of plane passing through the line of intersection of these planes is given by :

2x+3y -4z-1+λ (3x-y+z+2)=0

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

Since,This plane is passing through point

(0, 1 , 1) also ,then this point satisfy the equation of required plane .

(2+3λ)×0+(3-λ)× 1+(-4+λ)× 1+2λ-1= 0

⇒ 3 -λ -4+λ +2λ-1= 0

⇒-2 + 2λ= 0

⇒ 2λ = 2

⇒ λ = 1

Now , Put λ = 1 in equation (1)

⇒2x+3y-4z-1+( 1)× (3x-y+z+2)=0

⇒2x+3x +3y -y -4z +z -1+ 2= 0

⇒5x +2y -3z +1 = 0

Therefore,The Equation of required plane is 5x +2y -3z +1 = 0

### Note :

In vector form,the equation of required plane is : 2. Solution :

Equation of required plane is 5x+2y-3z+1=0 in Cartesian form.

Explanatìon:

There are two methods to solve the question

Vector form

Cartesian form

I’m solving question, in Cartesian form.We can convert Cartesian form in vector form easily .

$$\rule{200}2$$

We have to find ,Equation of plane passing through the intersection of two planes

$$\sf\:2x+3y-4z=1\:\:and\:\:3x-y+z+2=0$$

Equation of plane passing through the line of intersection of these planes is given by :

2x+3y -4z-1+λ (3x-y+z+2)=0

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

Since,This plane is passing through point

(0, 1 , 1) also ,then this point satisfy the equation of required plane .

(2+3λ)×0+(3-λ)× 1+(-4+λ)× 1+2λ-1= 0

⇒ 3 -λ -4+λ +2λ-1= 0

⇒-2 + 2λ= 0

⇒ 2λ = 2

⇒ λ = 1

Now , Put λ = 1 in equation (1)

⇒2x+3y-4z-1+( 1)× (3x-y+z+2)=0

⇒2x+3x +3y -y -4z +z -1+ 2= 0

⇒5x +2y -3z +1 = 0

Therefore,The Equation of required plane is 5x +2y -3z +1 = 0

Note :

In vector form,the equation of required plane is :

$$\sf\vec{r}\cdot\:(5\hat{i}+2\hat{j}-3\hat{k})+1=0$$

Step-by-step explanation: