How do you find the equation of the plane in

licencegpopc

licencegpopc

Answered question

2022-01-22

How do you find the equation of the plane in xyz-space through the p=(4,5,4) and perpendicular to the n=(−5,−3,−4)?

Answer & Explanation

Sean Becker

Sean Becker

Beginner2022-01-23Added 16 answers

If the repair is orthonormal, then the plane has 5x3y4z=d where d is a real number
Because the plane contains p=(x=4,y=5,z=4), you can write
5×43×54×4=d
So you find d=51 Conclusion an (not "the" !!) equation of your plane is 5x3y4z=51
Another one, easier, is 5x+3y+4z=51.
Tapanuiwp

Tapanuiwp

Beginner2022-01-24Added 13 answers

The answer is: 5x+3y+4z51=0
Given a point P(xp,yp,zp)andrv(a,b,c) perpendicular to the plane, the equation is:
a(xxp)+b(yyp)+c(zzp)=0
So:
5(x4)3(y5)4(z4)=0
5x+203y+154z+16=0
5x+3y+4z51=0

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Linear algebra

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?