Let L be the line of intersection between the planes x + 3y = 7 and 2y + z = 4, and let P be the point (3, 1, 1). Determine the equation (in normal form) of the plane containing L and P. I struggle with getting vectors from the two planes to multiply, can anyone show me how?

dannyboi2006tk

dannyboi2006tk

Answered question

2022-10-08

Let L be the line of intersection between the planes x + 3y = 7 and 2y + z = 4, and let P be the point (3, 1, 1).
Determine the equation (in normal form) of the plane containing L and P.
I struggle with getting vectors from the two planes to multiply, can anyone show me how?
L = ( 1 , 2 , 0 ) + t ( 3 , 1 , 2 ). But i somehow fail to see how my book got that.

Answer & Explanation

Emmanuel Russo

Emmanuel Russo

Beginner2022-10-09Added 9 answers

To obtain L, parameterize it with y = 2 + t and, from the two given planes, we have x = 1 3 t and z = 2 t, i.e.
L = ( 1 , 2 , 0 ) + t ( 3 , 1 , 2 )
with Q ( 1 , 2 , 0 ) and d = ( 3 , 1 , 2 ). Let A ( x , y , z ) be any point on the plane and P ( 3 , 1 , 1 ). Then, the normal vector of plane is
n = Q A × Q P = ( x 1 , y 2 , z ) × ( 2 , 1 , 1 ) = ( 2 + y + z , 1 x + 2 z , 5 x 2 y )
which satisfies n d = 0, i.e.
( 2 + y + z , 1 x + 2 z , 5 x 2 y ) ( 3 , 1 , 2 ) = 0
which yields the equation of the plane
x + y z = 3

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?