algebraic vectors
algebraic vectors
Magnitudes of Cartesian Vectors In Two-Space
MCV4U: Calculus & Vectors
Let ~v be a position vector in the xy -plane.
Thus ~v has its tail at the origin and its head at some point
P(x, y ).
Algebraic Vectors In Two- and Three-Space
Part 2: Magnitudes and Direction Angles
Its magnitude can be calculated using the Pythagorean
Theorem, and its direction can be calculated using the
tangent ratio.
J. Garvin
J. Garvin — Algebraic Vectors In Two- and Three-Space
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algebraic vectors
algebraic vectors
Magnitudes of Cartesian Vectors In Two-Space
Magnitudes of Cartesian Vectors In Two-Space
Magnitude and Direction of a Position Vector In
Two-Space
Example
p
~ = x 2 + y 2 and the angle it
For any point P(x, y ), |OP|
makes with the positive x-axis is given by tan−1 yx .
Let ~v = (5, 12). Determine the magnitude of ~v and its
direction.
p
52 + 122
√
= 169
|~v | =
= 13
~v has a direction angle of tan−1
J. Garvin — Algebraic Vectors In Two- and Three-Space
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12
5
≈ 67◦ .
J. Garvin — Algebraic Vectors In Two- and Three-Space
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algebraic vectors
Magnitudes of Cartesian Vectors In Two-Space
algebraic vectors
Magnitudes of Cartesian Vectors In Two-Space
The magnitude of any Cartesian vector can be found using
the formula for the distance between two points P and Q.
Magnitude of a Cartesian Vector In Two-Space
For any two points P(xp , yp ) and Q(xq , yq ),
p
~ = (xq − xp )2 + (yq − yp )2 .
|PQ|
Start by expressing PQ in terms of OP and OQ.
~ = PO
~ + OQ
~
PQ
~ − OP
~
= OQ
J. Garvin — Algebraic Vectors In Two- and Three-Space
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~ has a horizontal component of xp and a vertical
OP
~ has a horizontal component of
component of yp , while OQ
xq and a vertical component of yq .
The horizontal components have a magnitude of xq − xp , and
the vertical components a magnitude of yq − yp .
Using the Pythagorean
Theorem to combine the components
p
~ = (xq − xp )2 + (yq − yp )2 as required.
yields |PQ|
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algebraic vectors
Magnitudes of Cartesian Vectors In Two-Space
Example
Given the vectors ~u = (6, −2) and ~v = (−3, 8), determine
|~u − ~v |.
algebraic vectors
Magnitudes of Cartesian Vectors In
Three-Space
The formulae for magnitudes of vectors in three-space can be
extended from those in two-space.
Magnitude of a Position Vector in Three-Space
~ =
For any point P(x, y , z), |OP|
q
|~u − ~v | = (8 − (−2))2 + (−3 − 6)2
q
= (10)2 + (−9)2
√
= 181
p
x 2 + y 2 + z 2.
Magnitude of a Cartesian Vector in Three-Space
Note that is equivalent to asking for the magnitude of the
line segment between points U(6, −2) and V (−3, 8).
For any two points P(xp , yp , zp ) and Q(xq , yq , zq ),
p
~ = (xq − xp )2 + (yq − yp )2 + (zq − zp )2 .
|PQ|
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J. Garvin — Algebraic Vectors In Two- and Three-Space
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algebraic vectors
algebraic vectors
Magnitudes of Cartesian Vectors In
Three-Space
Magnitudes of Cartesian Vectors In
Three-Space
Example
Example
Determine the magnitude of ~v = (9, 2, −1).
Determine the magnitude of ~v =
q
92 + 22 + (−1)2
√
= 81 + 4 + 1
√
= 86
|~v | =
|~v | =
=
=
=
q
q
q
√
5 2
9
25
81
+
+
16
81
5 4 2
9, 9, 9
4 2
9
+
4
81
+
.
2 2
9
45
81
5
3
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J. Garvin — Algebraic Vectors In Two- and Three-Space
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algebraic vectors
algebraic vectors
Magnitudes of Cartesian Vectors In
Three-Space
Magnitudes of Cartesian Vectors In
Three-Space
Alternate Solution
Example
Determine the magnitude of ~v =
Note that ~v = 19 (5, 4, 2).
5 4 2
9, 9, 9
.
q
(52 + 42 + 22 )
√
1
= 9 45
|~v | =
=
=
1
9
√
45
9
√
5
3
J. Garvin — Algebraic Vectors In Two- and Three-Space
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Given the vectors ~u = (8, −1, 5) and ~v = (4, 2, −6),
determine |~u − ~v |.
q
(4 − 8)2 + (2 − (−1))2 + (−6 − 5)2
q
= (−4)2 + 32 + (−11)2
√
= 146
|~u − ~v | =
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algebraic vectors
algebraic vectors
Direction Angles and Direction Cosines
Direction Angles and Direction Cosines
What about direction vectors in three-space?
To calculate direction angles, we use the direction cosines of
the vector.
Rather than introduce a new system to replace those in place
for two-space (NEWS or bearings), we can use the angles
relative to the three axes.
Direction Angles of a Vector
The direction angles, α, β and γ, are the acute or obtuse
angles formed between a vector ~v and the x-, y - and z-axes.
Direction Angles of a Vector
The direction cosines of a vector ~v are:
cos α = |~vx |
cos β = |~yv |
cos γ =
The direction cosines have the property that
cos2 α + cos2 β + cos2 γ = 1.
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algebraic vectors
Direction Angles and Direction Cosines
algebraic vectors
Questions?
Example
Determine the direction angles for ~v = (5, −2, 3).
First, determine |~v |.
q
52 + (−2)2 + 32
√
= 38
|~v | =
Next, use the direction cosines to calculate the angles.
5
cos α = √
38
α ≈ 36◦
−2
cos β = √
38
β ≈ 109◦
J. Garvin — Algebraic Vectors In Two- and Three-Space
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z
|~v |
3
cos γ = √
38
γ ≈ 61◦
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