MCV 4U3 – CALCULUS AND VECTORS
Unit 6 – Applications of Vectors
Lesson 33 – Scalar And Vector Projections
Mathematically, a projection is formed by dropping a perpendicular from each of the
points of an object onto a line or plane. The shadow of an object, in certain
circumstances, is a physical example of a projection.
The projection of one vector onto another can be pictured as follows. In the diagram,
where u = OA and v = OB, the projection of u onto v is the vector ON. There are two
possibilities for ON depending on whether the angle between the vectors, θ, is less than
90° or greater than 90°.
Case 1: 0° < θ < 90°
A
Case 2: 90° < θ < 180°
A
u
u
θ
O
N
B
N
θ
O
B
As you can see from the given diagrams, the direction of ON is the same as the
direction of v when θ is acute; and opposite to v when θ is obtuse.
The scalar projection of u onto v is derived from the definition of the dot product. The
scalar projection is 𝑢 cos 𝜃, where θ is the angle between u and v.
𝑢∙𝑣 = 𝑢
𝑣 cos 𝜃
If we solve for 𝑢 cos 𝜃 we get:
𝑢∙𝑣
𝑢 cos 𝜃 =
𝑣
Therefore, the the scalar projection of u onto v is:
𝑢∙𝑣
𝑢 cos 𝜃 =
𝑣
The vector projection of u onto v is a vector in the direction of v. We multiply the
above scalar projection by a unit vector in the direction of vector v ie
!
!
.
Therefore, the vector projection of u onto v is:
𝑣
𝑢∙𝑣
𝑣
𝑢 cos 𝜃
=
𝑣
𝑣
𝑣
𝑢 cos 𝜃 𝑣
𝑢∙𝑣 𝑣
=
𝑣 !
𝑣
There is no special symbol for a vector projection.
Some textbooks uses the notation ON = Proj(u onto v).
The vector projection, ON, which is Proj(u onto v) =
! !"# ! !
!
The scalar projection of (u onto v) = 𝑢 cos 𝜃 =
=
!∙! !
! !
!∙!
!
The magnitude of ON, the vector projection of u onto v, is given by
𝑢 cos 𝜃 =
!∙!
!
Note: The scalar projection of u onto v does not equal the scalar projection of v onto u
(unless 𝑢 = 𝑣 ).
Example 1: Find the vector projection of u = (5, 6, -3) onto v = (1, 4, 5).
Solution:
We need to find u • v and ||v||.
u • v = (5)(1) + (6)(4) + (-3)(5)
= 14
||v|| =
=
(1)2 + (4)2 + (5)2
42
Proj(u onto v) =
=
=
14(1,4,5)
42
1
(1, 4, 5)
3
⎛ 1 4 5 ⎞
⎜ , , ⎟
⎝ 3 3 3 ⎠
Direction Angles And Direction Cosines:
In lesson 27 we referred to the angle a vector makes with the coordinate axes. In R3,
there are three such angles – one with each of the x, y, and z axes.
The direction angles of a vector (x, y, z) are the angles α (alpha), β (beta),
and γ (gamma) that the vector makes with the positive x-, y-, and z-axes,
respectively, where 0° ≤ α, β, γ ≤ 180°.
z
P(x, y, z)
γ
α β
y
x
In this context, the components x, y, and z of the vector v are referred to as direction
numbers. In the diagram above, we can see the right triangle that relates v, the
direction number x, and the direction angle α. In fact
x
cos α =
v
The other direction angles are related in the same way.
The direction cosines of a vector are the cosines of the direction angles α,
β, and γ, where
x
y
z
cos α = , cos β = , and cos γ =
v
v
v
Note that if you divide a vector (x, y, z) by its magnitude |v|, you create a unit vector with
⎛ x y z ⎞
components ⎜ , , ⎟ , which is exactly (cosα, cosβ, cosγ). Thus the direction cosines
⎜ v v v ⎟
⎝
⎠
are the components of a unit vector. Consequently,
cos2α + cos2β + cos2γ = 1
It follows from this that the direction cosines, and hence the direction angles, are not all
independent. From any two of them you can find the third.
Homework: Pages 398 – 400
# 1, 6, 11, 13, 15b