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

Answers ( )

    0
    2021-10-03T07:46:03+00:00

    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 .

    \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

  1. Emma
    0
    2021-10-03T07:46:17+00:00

    Answer:

    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:

    hope it helped you friend please mark me as brainliest and follow me I need more thanks please

Leave an answer

Browse
Browse

18:9+8+9*3-7:3-1*13 = ? ( )