A plane has parametric equations
x = 3 + 4t - s
y = 1 + 2t
z = 1 - 5t + 4s
So we have to find the direction vector for the plane and I'm thinking it's (3,1,1). Can somebody please double check?
2 answers
nope. (3,1,1) is just a point in the plane, right?
Do you mean the normal vector to the plane?
If we know 3 points in the plane we can draw two vectors and find their vector product
we have one point (3,1,1)
now pick a t and s
for example t = 1 and s =0
that gives x = 7, y = 3 and z = -4 so (7,3,-4) is in the plane
now maybe t = 0 and s = 1
then x = 2 , y = 1 , z = 5 so (2,1,5)
now get vector from (3,1,1) to (7, 3, -4)
dx = 4, dy = 2 , dz = -5
so one vector in the plane is V1 = 4 i + 2 j - 5 k
a second is from (3,1,1) to (2,1,5)
dx = -1 , dy = 0 , dz = 4 so V2 = -1 i + 0 j + 4 k
Now we want a vector perpendicular to those two
V = V1 cross V2 = x i + y j + z k
i j k
4 2 -5
-1 0 4
= (8-0) i + (5-16) j + (0 + 2) k
= 8 i -11 j + 2 k
If we know 3 points in the plane we can draw two vectors and find their vector product
we have one point (3,1,1)
now pick a t and s
for example t = 1 and s =0
that gives x = 7, y = 3 and z = -4 so (7,3,-4) is in the plane
now maybe t = 0 and s = 1
then x = 2 , y = 1 , z = 5 so (2,1,5)
now get vector from (3,1,1) to (7, 3, -4)
dx = 4, dy = 2 , dz = -5
so one vector in the plane is V1 = 4 i + 2 j - 5 k
a second is from (3,1,1) to (2,1,5)
dx = -1 , dy = 0 , dz = 4 so V2 = -1 i + 0 j + 4 k
Now we want a vector perpendicular to those two
V = V1 cross V2 = x i + y j + z k
i j k
4 2 -5
-1 0 4
= (8-0) i + (5-16) j + (0 + 2) k
= 8 i -11 j + 2 k
1 answer