Find a plane with the P(3,-4,5) and contains the line L:

[john458776]

I found that the normal vector is <2,-3,5> plane is 2x-3y-5z-43=0

I'm not sure about my solution or the approach I used can you guide me if I'm wrong or ?

 

[Dr.Peterson]

I found that the normal vector is <2,-3,5> plane is 2x-3y-5z-43=0

I'm not sure about my solution or the approach I used can you guide me if I'm wrong or ?

What you are calling "the normal vector" is the direction vector of the given line. The plane you want contains that line; it is not normal to it!

Find a second point in the desired plane; that will give you a second vector to work with.

 

[john458776]

What you are calling "the normal vector" is the direction vector of the given line. The plane you want contains that line; it is not normal to it!

Find a second point in the desired plane; that will give you a second vector to work with.

Okay I think I get it now.
a(x-3)+b(y+4)+c(z-5)=0
The plane contain the point (2,-1,-3)
Then a(2-3)+b(-1+4)+c(-3-5)=0
This is how I'll approach it let me write it on the paper and post it.

 

[john458776]

What you are calling "the normal vector" is the direction vector of the given line. The plane you want contains that line; it is not normal to it!

Find a second point in the desired plane; that will give you a second vector to work with.

 

[Dr.Peterson]

You didn't check that the two known points lie on this plane, did you?

Your method is different from mine, and I'm not sure exactly where the first line below came from, but the second line contains a simple arithmetic error:

Fix that, and you should get the right answer.

 

[john458776]

You didn't check that the two known points lie on this plane, did you?

Your method is different from mine, and I'm not sure exactly where the first line below came from, but the second line contains a simple arithmetic error:
View attachment 29496
Fix that, and you should get the right answer.

Yeah I didnt check it but I realised that it's simple if I would consider that i tried that approach and i came up with this solution.

Solution

Line has direction vector <2,-3,5>

Line passes through point (2,-1,-3)

Vector from (2,-1,-3) to (3,-4,5) is equal to <1,-3,8>

Normal Vector = <2,-3,5> × <1,-3,8> = <-9,11,-3>

Then the equation of the plane is
-9(x-2)+11(y+1)-3(z+3)=0

I hope I'm on a good lane now.