Lecture 2 (Sep. 10)
Dot Product (reading: 12.3)
Definition: The dot product of two vectors a = ha1 , a2 , a3 i and b = hb1 , b2 , b3 i is
the number
a · b = a1 b 1 + a2 b 2 + a3 b 3 .
Properties:
1. a · a = |a|2
2. a · b = b · a
3. a · (b + v) = a · b + a · v
4. (ca) · b = c(a · b) = a · (cb)
Geometric picture:
a · b = |a||b| cos(✓)
Proof of this:
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Note: if a and b are non-zero vectors,
cos(✓) =
a·b
.
|a||b|
Definition: We say two non-zero vectors are orthogonal (or perpendicular) if the
angle between them is ⇡/2.
So we see a and b are orthogonal , a · b = 0.
Note: if a and b are parallel, say b = ca, then cos(✓) = ca · a/(|a||c||a|) = ±1. So
✓ = 0 or ✓ = ⇡.
Projections
Definition: The scalar projection of b onto a is
compa b :=
a·b
.
|a|
The projection of b onto a is
proja b := (compa b)
Example: Let a = ha1 , a2 , a3 i. Find projî a.
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a
(a · b)a
=
.
|a|
|a|2
Cross Product (reading: 12.4)
Definition: The cross product (or vector product) of two vectors a = ha1 , a2 , a3 i,
and b = hb1 , b2 , b3 i is the vector
a ⇥ b := ha2 b3
a3 b2 , a3 b1
a1 b3 , a1 b2
a2 b1 i.
Determinant notation for remembering this formula:
Geometric picture of cross product. Check that a ⇥ b is orthogonal to both a and b:
We also have |a ⇥ b| = |a||b| sin(✓).
In particular, a and b are parallel , a ⇥ b = 0.
A special case of this fact: a ⇥ a = 0.
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A geometric interpretation of the cross product:
Example: Find the area of the triangle shown:
Properties of the cross-product:
1. a ⇥ b =
b⇥a
2. (ca) ⇥ b = c(a ⇥ b) = a ⇥ (cb)
3. a ⇥ (b + v) = a ⇥ b + a ⇥ v
4. a · (b ⇥ v) = (a ⇥ b) · v
5. a ⇥ (b ⇥ v) = (a · v)b
(a · b)v
Some remarks:
• (1) is important – cross products do not “commute”!
• The quantity in (4) is the triple product. Geometric interpretation:
• (5) is not equal to (a ⇥ b) ⇥ v in general – the cross product is not “associative”!
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Equations of Lines and Planes (reading: 12.5)
The line L through a point r0 2 R3 in the direction v is
L = {r0 + tv 2 R3 | t 2 R}.
That is, L is described by the parametric equation r0 + tv as the parameter t runs
over all values 1 < t < 1.
This is equivalent to the three scalar parametric equations
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< x = x0 + ta
y = y0 + tb
: z = z + tc
0
where r0 = hx0 , y0 , z0 i, and v = ha, b, ci.
Example: Find a vector equation for the line passing through the points r0 and r1 .
Another way to describe L is to eliminate the parameter t from the scalar parametric
equations above, to get the so-called symmetric equations for L:
x
x0
a
=
y
y0
b
=
z
z0
c
.
This makes sense if a, b, and c are not all 0 (if, for example, a = 0, we have x = x0
and (y y0 )/b = (z z0 )/c).
Example: Find parametric and symmetric equations for the line through (1, 0, 0) and
(1, 1, 1).
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Planes: a plane in R3 is determined by a point, r0 , on it, and a vector, n, orthogonal
to it. The vector n is called a normal vector to the plane.
Thus a point r = hx, y, zi is on the plane if and only if
n · (r
r0 ) = 0.
This is an equation of the plane.
If n = ha, b, ci and r0 = hx0 , y0 , z0 i, the corresponding scalar equation is
a(x
x0 ) + b(y
y0 ) + c(z
z0 ) = 0.
Another way to write this is
ax + by + cz = d
(here we have d = ax0 + by0 + cz0 ).
Example: Find an equation for the plane passing through (1, 2, 2), (1, 1, 1), and
(0, 0, 1). Try and sketch this plane.
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Example: Find a formula for the distance from the point (x0 , y0 , z0 ) to the plane
ax + by + cz = d.
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