# Thread: 3d vector perpindicular question

1. ## 3d vector perpindicular question

Explain why the line (x,y,z) = (4,2,1) + t(2,3,4) is perpendicular to 3x+4y+5z = 10.
I know that this is vector form of an equation, well actually I have no clue how to find perpendicular vectors. Something to do with dot product equaling to 0. help?

2. Originally Posted by hellojellojw
Explain why the line (x,y,z) = (4,2,1) + t(2,3,4) is perpendicular to 3x+4y+5z = 10.
I know that this is vector form of an equation, well actually I have no clue how to find perpendicular vectors. Something to do with dot product equaling to 0. help?
A vector in the direction of the given line is <2, 3, 4>. A vector perpendicular to the given plane is <3, 4, 5>. The given line is NOT perpendicular to the given plane.

3. Originally Posted by mr fantastic
A vector in the direction of the given line is <2, 3, 4>. A vector perpendicular to the given plane is <3, 4, 5>. The given line is NOT perpendicular to the given plane.
haha right, the line was actually (8,9,10) + t(3,4,5) is perpendicular to the blah blah.....

So yeah, my bad, but I think it has something to do with proportions or something, 8-3, 9-4, 10-5
yeah can someone explain?

4. First, the problem does NOT ask you to find a perpendicular. It only asks you to verify that the given line is perpendicular to the given plane. You should know two things before you attempt such a problem:
1) If a plane is given in the form Ax+ By+ Cz= D, then the vector <A, B, C> is perpendicular to the plane.

2) If a line is given in the form (a, b, c)+ t(X, y, Z) then the vector <X, Y, Z> is in the same direction as the line.

Those two together say that if line (a, b, c)+ t(X, Y, Z) is perpendicular to the plane Ax+ By+ Cz= D, then the vectors <A, B, C> and <X, Y, Z> are parallel and so one is a multiple of the other. In your case, <A, B, C> and <X, Y, Z> are the same. The "multiple" is 1.