Math 21a Vectors and the Dot Product Lecture 2

Vector vs Scalar

1. Are the following better described by vectors or scalars?

(a) The cost of a Super Bowl ticket. (b) The wind at a particular point outside. (c) The number of students at Harvard. (d) The velocity of a car. (e) The speed of a car.

~u What are ~u + ~v, ~u − ~v, 2~u, and |~u| ?

Dot Product

2. Find the following dot products. (a) h1, 2i · h3, 4i (b) h2, −1i · h1, 2i (c) h2, −1i · h2, −1i Main Property: There is an alternate definition of the dot product. Two vectors ~u and ~v determine an angle θ in their common plane, and then their dot product is simply

~u · ~v = |~u| |~v| cos(θ).

Any Other Properties?

3. Find the angle between the following two vectors. (a) h1, 2, 1i and h1, −1, 1i. (b) h1, 1, 0i and h0, 1, −1i. What does the Dot Product Compute?

4. Suppose you are pulling a sled which is going horizontally along the ground. If you are pulling at an angle of 30◦ with a force of 50 N, how much work do you do in pulling the sled 10 meters?

Vector Projection

5. This is a problem about vector projections. For each of the following sets of vectors ~u and ~v,

find (draw) proj~u ~v, the vector projection of ~v onto ~u.

~v

~u ~u ~v How to Compute Vector Projection? What is Scalar Projection?

6. Find the following vector projections proj~u ~v. You can use either the geometric meaning of ~u·~v the vector projection or the formula proj~u ~v = |~u|2 ~u. (a) ~u = h2, 0i and ~v = h3, 5i (b) ~u = h0, 1, 0i and ~v = h5, −2, 8i

(c) ~u = h1, 1i and ~v = h2, 3i (d) ~u = h1, 1, −1i and ~v = h1, 2, 1i Vectors and the Dot Product – Answers and Solutions

1. (a) Scalar – the cost is just a number. (b) Vector – the wind has both a speed and a direction. (c) Scalar. (d) Vector. The velocity is defined to be both the speed of the car (how fast it’s going) and the direction it’s going. (e) Scalar. The speed refers only to how fast the car is going; it is the magnitude of the velocity vector.

2. (a) h1, 2i · h3, 4i = (1)(3) + (2)(4) = 11 (b) h2, −1i · h1, 2i = 0 (c) h2, −1i · h2, −1i = 5

3. (a) Let ~u = h1, 2, 1i and ~v = h1, −1, 1i, and let θ be the angle between ~u and ~v. Then, we know that ~u · ~v = |~u||~v| cos θ. We calculate that ~u · ~v = (1)(1) + (2)(−1) + (1)(1) = 0, so π 0 = |~u||~v| cos θ. Since the lengths |~u| and |~v| are both positive, cos θ = 0, so θ = 2 . (b) Let ~u = h1, 1, 0i and ~v = h0, 1, −1i, and let θ be the angle between ~u and ~v. Then,

~u · ~v 1 1 cos θ = = √ √ = . |~u| |~v| 2 2 2

π So θ = 3 . 4. Let F~ denote the force and d~ the displacement vector of the sled. Then we can assume ~ ◦ ◦ ~ ~ ~ F = h50 cos(30 ), 50 sin(30√ ) and d = h10, i. The the work W is given by F · d (with the unit Nm = J). So W = 250 3 J.

F~

~ ~ projd~ F d

5.

~v ~u ~v

proj~u ~v ~u proj~u ~v

6. (a) Considering the projection of ~v onto the line spanned by ~u is equivalent to looking at

the first component of ~v = h3, 5i. So proj~u ~v = h3, 0i.

(b) The same reasoning as in (a) shows proj~u ~v = h0, −2, 0i. (c) We use the formula for the vector projection.

~u · ~v (1)(2) + (1)(3) 5 5 5 proj~u ~v = ~u = √ 2 = h1, 1i = , |~u|2 11 + 11 ~u 2 2 2

(d) ~u · ~v 2 2 2 2 proj ~v = ~u = h1, 1, −1i = , , − ~u |~u|2 3 3 3 3