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Calculus 3
Lia Vas
Review of vectors. The dot and cross products
Review of vectors in two and three dimensions. A two-dimensional vector is an ordered
pair ~a = 〈a 1
, a 2
〉 of real numbers. The coordinate representation of the vector ~a corresponds to the
arrow from the origin (0, 0) to the point (a 1 , a 2 ). Thus, the length of ~a is |~a| =
√
a
2
1
2
2
. Analogously,
we have the following.
A three-dimensional vector is an ordered
triple
~a = 〈a 1
, a 2
, a 3
of real numbers. The coordinate representation
of the vector ~a corresponds to the arrow from the
origin (0, 0 , 0) to the point (a 1
, a 2
, a 3
The length of ~a is
|~a| =
√
a
2
1
2
2
2
3
Using the coordinate representation the vector addition and scalar multiplication can be realized
as follows.
Vector Addition - by coordinates 〈a 1
, a 2
, a 3
〉 + 〈b 1
, b 2
, b 3
〉 = 〈a 1
, a 2
, a 3
Scalar multiplication - by coordinates k〈a 1
, a 2
, a 3
〉 = 〈ka 1
, ka 2
, ka 3
This corresponds to the geometrical representation illustrated in the figure below.
Using its coordinates, a vector ~a = 〈a 1
, a 2
〉 in
xy-plane can be represented as a linear combina-
tion of vectors
i = 〈 1 , 0 〉 and
j = 〈 0 , 1 〉 as follows.
~a = a 1
i + a 2
j
The coordinates of a vector and geometrical
representation have analogous relation in three
dimensional space.
If
i = 〈 1 , 0 , 0 〉, ~j = 〈 0 , 1 , 0 〉, and
k = 〈 0 , 0 , 1 〉 and a vector ~a can be represented as ~a = 〈a 1
, a 2
, a 3
then
~a = a 1
i + a 2
j + a 3
k.
In the next section, it will be relevant to determine the coordinates of the vector from one point
to the other. Let P = (a 1
, a 2
, a 3
) and Q = (b 1
, b 2
, b 3
), be two points in space. If O denotes the origin
(0, 0 , 0), then the vector
OP can be represented as 〈a 1 , a 2 , a 3 〉, and the vector
OQ as 〈b 1 , b 2 , b 3 〉.
Since
OP +
P Q =
OQ we have that
P Q =
OQ −
OP = 〈b 1 , b 2 , b 3 〉 − 〈a 1 , a 2 , a 3 〉 = 〈b 1 − a 1 , b 2 − a 2 , b 3 − a 3 〉.
In some cases, we may need to find the vector
with same direction and sense as a nonzero vec-
tor ~a but of length 1. Such vector is called the
normalization of ~a.
The normalization of ~a is
the vector of length 1 in the direction of ~a,
ˆa =
~a
|~a|
Practice problems.
- Let P be the point (2, −1) and Q be the point (1, 3). Determine and sketch the vector
P Q.
- Let ~a = 〈 2 , − 1 〉 and
b = 〈 1 , 3 〉. Sketch ~a +
b, ~a −
b, 2 ~a, 2 ~a − 3
b.
- Let ~a = 〈 3 , 4 , 0 〉 and
b = 〈− 1 , 4 , 2 〉. Determine |~a|, 2 ~a + 3
b, 3 ~a − 2
b.
- Let ~a =
i + 4
j − 8
k and
b = − 2
i +
j + 2
k. Determine |~a|, ~a +
b, 2 ~a − 3
b.
- Find the normalization of the vector ~a =
i + 4
j + 8
k.
- Find the normalization of the vector ~a = 〈 3 , 4 , 0 〉.
Solutions. 1.
P Q = 〈− 1 , 4 〉 3. |~a| = 5, 2 ~a + 3
b = 〈 3 , 20 , 6 〉, 3 ~a − 2
b = 〈 11 , 4 , − 4 〉
- |~a| = 9, ~a +
b = 〈− 1 , 5 , − 6 〉, 2 ~a − 3
b = 〈 8 , 5 , − 22 〉 5. The length of ~a is |~a| =
2
2
√
81 = 9 so ˆa =
1
9
i +
4
9
j +
8
9
k 6. The length of ~a is |~a| =
2
2
2
25 = 5 so ˆa = 〈
3
5
4
5
The Dot Product
If one is to define a meaningful product of two vectors, ~a = 〈a 1
, a 2
, a 3
〉 and
b = 〈b 1
, b 2
, b 3
the first idea that comes to mind would probably be to consider coordinate-wise multiplication
〈a 1 b 1 , a 2 b 2 , a 3 b 3 〉. However, since this type of product is geometrically not very meaningful nor appli-
cable, one consider two other types of multiplication, the dot and the cross product.
- Find the angle between the vectors ~a = 〈 3 , − 1 , 2 〉 and
b = 〈 2 , 4 , − 1 〉.
- Find the projection of ~a onto
b if ~a = 〈 1 , − 1 , 0 〉 and
b = 〈 1 , 0 , 1 〉.
Solutions. 1. ~a ·
b = −2 + 9 − 4 = 3. 2. If θ denotes the angle between the vectors, then
cos θ =
~a·
~ b
|~a||
~ b|
15+
(5)(13)
63
65
= 969. θ = cos
− 1 (.969) = .249 radians or 14.25 degrees.
- cos θ =
~a·
~ b
|~a||
~ b|
0
|~a||
~ b|
= 0. So, θ = 90 degrees and the vectors are perpendicular.
- proj ~ b
~a =
~a·
~ b
|
~ b|
2
b =
1
(
√
2
1
2
1
2
1
2
The Cross Product
As opposed to the dot product which results in a scalar, the cross product of two vectors is again
a vector. If ~a and
b are two vectors, their cross product is denoted by ~a ×
b.
The vector ~a ×
b is perpendicular to the plane determined by ~a and
b. This determines the
direction of ~a ×
b. The sense of ~a ×
b is determined by the right hand rule: if ~a and is the thumb
and
b the middle finger, the index finger has the same sense as ~a ×
b. Using the right hand rule, you
can see that the cross product is not not commutative, ~a ×
b 6 =
b × ~a in general, and that
b × ~a = −~a ×
b.
The length of ~a ×
b is the same as the area of the parallelogram determined by ~a and
b. If
θ is the angle between ~a and
b, then
|~a ×
b| = |~a||
b| sin θ.
The cross product of ~a = 〈a 1 , a 2 , a 3 〉 and
b = 〈b 1 , b 2 , b 3 〉 can be computed using the coordinates
as follows.
~a ×
b = 〈a 2
b 3
− a 3
b 2
, a 3
b 1
− a 1
b 3
, a 1
b 2
− a 2
b 1
∣
∣
∣
∣
∣
∣
∣
i
j
k
a 1
a 2
a 3
b 1 b 2 b 3
∣
∣
∣
∣
∣
∣
∣
Since |~a ×
b| = |~a||
b| sin θ, ~a and
b are parallel exactly when sin θ = 0 which happens exactly when
~a ×
b =
- Thus,
~a and
b are parallel if and only if ~a ×
b =
Another way to check if the two vectors are parallel is to check if one is a scalar multiple of the other
(i.e. if ~a = k
b for some k). In this case, for
b 6 =
0 , the coordinates are such that
a 1
b 1
a 2
b 2
a 3
b 3
Practice Problems.
- Let ~a = 〈 1 , 2 , 0 〉 and
b = 〈 0 , 3 , 1 〉. Find ~a ×
b.
- Do the same for ~a = 2
i +
j −
k and
b =
j + 2
k.
- Let ~a = 〈− 5 , 3 , 7 〉 and
b = 〈 6 , − 8 , 2 〉. Determine if the vectors are parallel, perpendicular or
neither.
- Do the same for ~a = −
i + 2
j + 5
k and
b = 3
i + 4
j −
k.
- Find a vector perpendicular to the plane through the points P (1, 0 , 0), Q(0, 2 , 0) and R(0, 0 , 3)
and find the area of the triangle P QR.
- Do the same for P (0, 0 , 0), Q(1, − 1 , 1) and R(4, 3 , 7).
Solutions. 1. 〈 2 , − 1 , 3 〉 2. 3
i − 4
j + 2
k
- ~a·
b = − 40 6 = 0 so the vectors are not perpendicular. Also, the coordinates are not proportional
− 5
6
3
− 8
7
2
) so the vectors are not parallel either. Alternatively, find that the cross product is
a ×
b = 〈 62 , 52 , 22 〉 6 = 〈 0 , 0 , 0 〉 so the vectors are not parallel.
- ~a ·
b = 0, thus the vectors are perpendicular.
- Since vectors
P Q and
P R are in the plane, their cross product
P Q ×
P R is perpendicular to
the plane. Calculate
P Q = 〈− 1 , 2 , 0 〉 and
P R = 〈− 1 , 0 , 3 〉,
P Q ×
P R = 〈 6 , 3 , 2 〉. The area of the
triangle determined by P, Q, and R is half of the area of the parallelogram determined by the vectors
P Q and
P R which is the magnitude of
P Q ×
P R. Thus the triangle area is
1
2
7
2
- Similarly to previous problem, find a vector perpendicular to the plane to be 〈− 10 , − 3 , 7 〉 and
the area of the triangle to be
