1. Angles Between Vectors Orthogonal Vectors
The Dot Product
Angles Between Vectors
Orthogonal Vectors
Section 6.2a
HW: p odd

2. Definition: Dot Product
The dot product or inner product of u = u , u
and v = v , v is 1 2 1 2
u v = u v + u v 1 1 2 2
The sum of two vectors is a…
vector!
The product of a scalar and a vector is a…
vector!
The dot product of two vectors is a…
scalar!

3. Properties of the Dot Product
Let u, v, and w be vectors and let c be a scalar.
1. u v = v u 2
2. u u = |u|
3. 0 u = 0
4. u (v + w) = u v + u w
(u + v) w = u w + v w
5. (cu) v = u (cv) = c(u v)

4. Finding the Angle Between Two Vectors (Law of cosines)
v – u v u

5. Angle Between Two Vectors
Theorem:
Angle Between Two Vectors
v – u
If 0 is the angle between nonzero
vectors u and v, then v u
and

6. Definition: Orthogonal Vectors
The vectors u and v are orthogonal if
and only if u v = 0.
The terms “orthogonal” and “perpendicular”
are nearly synonymous (with the exception
of the zero vector)

7. Guided Practice Find each dot product = 23 = –10 = 11 1. 3, 4 5, 2
1. 3, , 2
= –10
2. 1, –2 –4, 3
3. (2i – j) (3i – 5j)
= 11

8. Guided Practice Use the dot product to find the length of
vector v = 4, –3
(hint: use property 2!!!)
Length = 5

9. Guided Practice Find the angle between vectors u and v
u = 2, 3 , v = –2, 5
0 =

10. Guided Practice
Find the angle between vectors u and v

11. Guided Practice Find the angle between vectors u and v 0 = 90
Is there an easier way to solve this???

12. Guided Practice u v = 0!!! Prove that the vectors u = 2, 3 and
v = –6, 4 are orthogonal
Check the dot product:
u v = 0!!!
Graphical Support???

13. First, let’s look at a brain exercise…
Page 520, #30:
Find the interior angles of the triangle with vertices (–4,1),
(1,–6), and (5,–1).
Start with a graph…
A(–4,1)
B(5,–1)
C(1,–6)

14. First, let’s look at a brain exercise…

15. First, let’s look at a brain exercise…

16. First, let’s look at a brain exercise…

17. Definition: Vector Projection
The vector projection of u = PQ onto a nonzero vector v = PS
is the vector PR determined by dropping a perpendicular from
Q to the line PS. Q u P S v R
Thus, u can be broken into components PR and RQ:
u = PR + RQ

18. Definition: Vector ProjectionQ u P S R
Notation for PR, the vector projection of u onto v:
PR = proj u v
The formula:
u v
proj u = v v
|v| 2

19. Some Practice Problems
Find the vector projection of u = 6, 2 onto v = 5, –5 .
Then write u as the sum of two orthogonal vectors, one
of which is proj u. v
Start with a graph…

20. u = proj u + u = 2, –2 + 4, 4 Some Practice Problems…
Find the vector projection of u = 6, 2 onto v = 5, –5 .
Then write u as the sum of two orthogonal vectors, one
of which is proj u. v
Start with a graph…
u = proj u + u = 2, – , 4 v 2

21. u = proj u + u = –1.8,–5.4 + 4.8,–1.6 Some Practice Problems…
Find the vector projection of u = 3, –7 onto v = –2, –6 .
Then write u as the sum of two orthogonal vectors, one
of which is proj u. v
u = proj u + u = –1.8,– ,–1.6 v 2

22. Some Practice Problems…
Find the vector v that satisfies the given conditions: 2
u = –2,5 , u v = –11, |v| = 10
A system to solve!!!

23. Some Practice Problems…
Find the vector v that satisfies the given conditions: 2
u = –2,5 , u v = –11, |v| = 10 OR

24. Some Practice Problems…
Now, let’s look at p.520: even:
What’s the plan???
If u v = 0  orthogonal!
If u = kv  parallel!
34) Neither
36) Orthogonal
38) Parallel