3. (30 pts) Given the triangular plane shown below,
a) Write each side of the plane as a vector.
I’ve labeled and drawn vectors on each side of the plane in the figure below. You might have chosen
opposite directions, just multiply your vector components by −1to swap the direction if you want your
answers to exactly match mine. In any case, the way I have drawn the vectors gives
~
A=−ˆ+ 1.5ˆ
k
~
B= 2ˆı−ˆ
~
C= 2ˆı−1.5ˆ
k
b) Work out a vector ~n that is perpendicular to the plane in the direction shown. (Hint: A vector that
is perpendicular to the plane is also perpendicular to any pair of edges of the plane.)
If we want a vector perpendicular to the plane, it will also be perpendicular to all three of the sides.
The cross product of two vectors gives a vector that is perpendicular to both, so maybe we can use that.
But make sure we take the product in an order that will give a result in the direction we want. In my
example below, the cross product of ~
Awith ~
Bshould give what we want.
~
A×~
B=
ˆıˆˆ
k
0−1 1.5
2−1 0
= ˆı(1.5) −ˆ(−3) + ˆ
k(2)
~n = 1.5ˆı+ 3ˆ+ 2ˆ
k
c) Come up with the unit vector ˆnthat is perpendicular to the plane.
We discussed how to find a unit vector in any direction. Just divide the vector by its magnitude,
ˆn=~n
|~n|
ˆn=1.5ˆı+ 3ˆ+ 2ˆ
k
√1.52+ 32+ 22= 0.384ˆı+ 0.768ˆ+ 0.512ˆ
k
x
y
z
z = 1.5
x = 2
y = 1
n
CA
B
