# Thread: Vector Algebra (line perpendicular to plane)

1. ## Vector Algebra (line perpendicular to plane)

1. The problem statement, all variables and given/known data

Find the parametric equations for the line which passes through the point (3,5,7)
and parallel to the vector <2,6,8>. Is the line perpendicular to the plane 5x + 6y +7z = 10?

2. Relevant equations

line equation, r = Ro + tV

3. The attempt at a solution

line equation, r = Ro + tV
-------------> = <3,5,7> + t<2,6,8>

hence, x = 3 + 2t ; y = 5 + 6t ; z = 7 + 8t

the problem is.. how am i going to prove that the line perpendicular
to the plane 5x + 6y +7z = 10 or not..?

2. Originally Posted by nameck
1. The problem statement, all variables and given/known data

Find the parametric equations for the line which passes through the point (3,5,7)
and parallel to the vector <2,6,8>. Is the line perpendicular to the plane 5x + 6y +7z = 10?

2. Relevant equations

line equation, r = Ro + tV

3. The attempt at a solution

line equation, r = Ro + tV
-------------> = <3,5,7> + t<2,6,8>

hence, x = 3 + 2t ; y = 5 + 6t ; z = 7 + 8t

the problem is.. how am i going to prove that the line perpendicular
to the plane 5x + 6y +7z = 10 or not..?
You can rewrite the equation of the plane as 5x-10+ 6y+ 7z= 5(x-2)+ 6y+ 7z= 0 so it is obvious that the point (2, 0, 0) is in that plane. Further, if (x,y,z) is any point in that plane, then the vector <x-2, y-0, z- 0>= <x- 2, y, z>, the vector from (2, 0, 0) to (x, y, z) is also in that plane. But $5(x-2)+ 6y+ 7z= <5, 6, 7>\cdot= 0$. That is, the vector <5, 6, 7>, the vector of the coefficients of the plane, is perpendicular to every vector in the plane and so is a normal vector to the plane (I'll bet you already knew that). To determine if the line x= 3+ 2t, y= 5+ 6t, z= 7+ 8t is perpendicular to the plane, you just need to determine if its "direction vector", <2, 6, 8> is parallel to <5, 6, 7> (which should be obvious).