Equation of a plane that passes thorugh the line of intersection of [answer check]

[lamerali]

Find the equation of the plane that passes through the line of intersection of the planes 4x - 3y - z - 1 = 0 and 2x + 4y + z - 5 = 0 and passes through A(1, -3, 2).

I have answered the question but I am not sure it is correct; if someone can review it their comments would be appreciated!
Thank you!


The plane that passes through the line of intersection of the two planes has the equation

4x – 3y – z – 1 + k(2x +4y + z – 5) = 0

Substitute the point A(1, -3, 2) into the equation to find k

4(1) – 3(-3) – 2 – 1 + k (2(1) + 4(-3) + 2 – 5) = 0
10 + k(-13) = 0
k = 10 / 13

substitute the value of k into the equation of the plane

4x – 3y – z – 1 + (10/13)(2x +4y + z – 5) = 0
4x – 3y – z – 1 + (20/13)x + (40/13)y + (10/13)z – (50/13) = 0

Therefore the equation of the plane that passes through the line of intersection of the two planes is

(72/13)x + (1/13)y – (3/13)z – (63/13) = 0

 

[Defennder]

Yup it looks ok.

 

[lamerali]

Great! :D Thank you!