What are the parametric equations for the intersection of the planes x-y-z=1 and 2x + 3y + z = 2 ?

Carol Gates

Carol Gates

Answered question

2021-03-29

What are the parametric equations for the intersection of the planes x-y-z=1 and 2x + 3y + z = 2 ?

Answer & Explanation

Malena

Malena

Skilled2021-03-31Added 83 answers

To find a point on this line we can for instance set z = 0 and then use the given equations to solve for x and y.
x-y=1
2x+3y=2
After we have x = 1, y = 0. So the point on the line is (1, 0, 0).
The direction of the line lives in both the planes and so, in particular, is perpendicular to both normal vectors, therefore a vector which is parallel to the line is given by
(1,1,1)×(2,3,1)=|ijk111231|=i(1+3)j(1+2)+k(3+2)
=2i-3j+5k=(2,-3,5)
The equation of the line is given by the vector equation
(x,y,z)=(1,00)+t(2,-3,5)=(1+2t,-3t, 5t)
Hence the parametric equations are x = 1 + 2t, y = -3t, and z = 5t.

Jeffrey Jordon

Jeffrey Jordon

Expert2021-11-20Added 2605 answers

Answer is given below (on video)

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