Lecture 2 - Wednesday, April 2
VECTORS (§12.2)
Given two points P = (x1, y1) and Q = (x2, y2), their differ−→
ence PQ = (x2 − x1, y2 − y1) is called a vector.
Example:
Q = (5, 4)
4
3
2
B = (1, 2)
P = (4, 2)
1
A = (0, 0)
−1
−1
1
2
3
4
5
−→
PQ = (5 − 4, 4 − 2) = (1, 2)
−→
AB = (1 − 0, 2 − 0) = (1, 2)
−→
−→
The vectors PQ and AB are equivalent.
7
Adding vectors
~ = (b1, b2) define
~ = (a1, a2) and b
For a
~ = (a1 + b1, a2 + b2).
~+b
a
−
→
−
→
a + b
−
→
b
b2
a2
−
→
a
a1
b1
a1 + b1
8
a2 + b2
Subtracting vectors
~ = (b1, b2) define
~ = (a1, a2) and b
For a
~ = (a1 − b1, a2 − b2).
~−b
a
Example:
→
−
−
→
a − b
−
→
a
−
→
b
~ = (3, 4) we have
~ = (−2, 6), b
For a
~ = (−2 − 3, 6 − 4) = (−5, 2)
~−b
a
9
Scaling:
~ = (a1, a2) and a scalar c ∈ R, the
For a vector a
scalar product is defined as c~
a = (ca1, ca2).
Example:
→
2−
a
−
→
a
→
−2−
a
~ = (2, 3), we have 2~
For a
a = (4, 6) and −2~
a = (−4, −6)
10
~ = (x, y) is |~
The magnitude of a vector a
a| =
p
x 2 + y2
The magnitude is the distance between the start point and
the end point of the vector.
~ is
The direction of a vector a
1
~.
a
|~
a|
~ = (0, 0).
In R2 the zero vector is 0
The zero vector has magnitude 0, but no direction.
Some observations:
1
~| = 1
•| a
|~
a|
a|
• |c~
a| = |c| · |~
|{z} |{z}
|{z}
magnitude
abs.val magnitude
~ have same direction
• If c > 0, then c~
a and a
~
• If c < 0, then c~
a has the opposite direction to a
11
Now everything in 3D
• For P = (x1, y1, z1) and Q = (x2, y2, z2), the vector from
−→
P to Q is PQ = (x2 − x1, y2 − y1, z2 − z1).
p
~ = (x, y, z) has magnitude |~
• A vector a
a| = x 2 + y2 + z 2
~ = (0, 0, 0)
• The zero vector is 0
~ = (b1, b2, b3) gives
~ = (a1, a2, a3) and b
• Adding a
(a1 + b1, a2 + b2, a3 + b3)
• etc..
12
THE STANDARD BASIS
In R3, the vectors ~i = (1, 0, 0), ~j = (0, 1, 0), k~ = (0, 0, 1) are
called the standard basis vectors.
Every vector in R3 can be written as a linear combination
of the standard basis vectors.
Example:
(4, −2, 9) = (4, 0, 0) + (0, −2, 0) + (0, 0, 9)
= 4 (1, 0, 0) − 2 (0, 1, 0) + 9 (0, 0, 1)
= 4~i − 2~j + 9k~
~ How large
~ = ~i − ~j + 2k~ and b = 3~i − k.
Example: Given a
~
is |~
a − 2b|?
~ = (1−2·3)~i+(−1)~j+(2+2·1)k~ = (−5, −1, 4)
~ −2b
We have a
p
√
2
2
2
~
and |~
a − 2b| = (−5) + (−1) + 4 = 42.
13
Exercise (Spring 2013 Midterm 1, ex 4)
Let ℓ be the line in R3 that passes though the points (1, 2, 3)
and (4, 1, −1). Find the coordinates of the point where ℓ
intersects the xz-plane.
Solution: Let P = (1, 2, 3), Q = (4, 1, −1).
−→
PQ = (4 − 1, 1 − 2, −1 − 3) = (3, −1, −4)
The line ℓ is
−→
{P + t PQ : t ∈ R} = {(1, 2, 3) + t(3, −1, −4) : t ∈ R}
Which point on it has the form (x, 0, z)?
Calculate 2 − t = 0 ⇔ t = 2.
The point is (1, 2, 3) + 2(3, −1, −4) = (7, 0, −5)
14