Question

Find the vector and parametric equations for the line through the point P(-2, 3, 1) and orthogonal to the plane 4x−5y+5z=−5.

In vector form and parametric form (parameter t, and passing through P when t = 0):

Answer 1

Please note that the vector v=(4,-5,5) is orthogonal to your plane

Answer 2

Why is that vector orthogonal to the plane?

Answer 3

so your line has to be parallel to the vector v=(4,-5,5)

Answer 4

I don't understand why <4,-5,5> is orthogonal to the plane.

Answer 5

please note that the point (0,0,-1) belongs to your plane.

Answer 6

I also don't understand why (0,0,-1) belongs to the plane.

Answer 7

please substitute those coordinates into the equation of your plane

Answer 8

4(0) - 5(0) +5(-1) = -5, 0 - 0 - 5 = -5, -5 = -5, true statement

Answer 9

Now the generic equation of a plane passing at point P=(x,y,z) and orthogonal to vector N=(a,b,c), is:
\[(X-P) \cdot N=0\]
where X=x,y,z) is the generic point which belongs to the plane

Answer 10

In our case, we have N=(4,-5,5), and P=(0,0,-1), so substituting into my formula, we have:
\[[(x,y,z)-(0,0,-1)] \cdot (4,-5,5)=0\]
from which:
\[(x,y,z+1) \cdot(4,-5,5)=0\]
then:
\[4x-5y+5z+5=0\]
which is the equation of your plane

Answer 11

So I heve showed that your plane is orthogonal to the vector (4,-5,5)

Answer 12

oops...So I have...

Answer 13

im very slow so still not getting it, takes me forever to understand anything

Answer 14

ok! Take your time, if you need please send me a message

Answer 15

ok thank you :)

Answer 16

closing for now so I can post another question, but will use info and post up later if I need help, ty @Michele_Laino