Download F1.8GD1 Notes on Vectors A vector v is an ordered triple v = (x, y, z

F1.8GD1
Notes on Vectors
A vector v is an ordered triple v = (x, y, z) ∈ R3 . We also write v = xi + yj + zk,
where i, j, k denote the vectors (1, 0, 0), (0, 1, 0), (0, 0, 1) respectively. Vectors are used
in Physics to describe quantities that have direction as well as magnitude associated to
them, such as force, acceleration, velocity, etc.
[It is also possible to consider 2-dimensional vectors (x, y) ∈ R2 , or more generally
n-dimensional vectors (x1 , . . . , xn ) ∈ Rn for any n. For the purposes of these notes, we
will restrict attention to the case n = 3.]
Vectors can be added together in the obvious way:
(a, b, c)+(x, y, z) = (a+x, b+y, c+z), or (ai+bj+ck)+(xi+yj+zk) = (a+x)i+(b+y)j+(c+z)k.
They can also be multiplied by a scalar (ie real number):
λ(x, y, z) = (λx, λy, λz), or λ(xi + yj + zk) = λxi + λyj + λzk.
These algebraic operations obey a lot of obvious properties, such as commutativity
and associativity of addition:
v + w = w + v;
u + (v + w) = (u + v) + w
and the distributive law of scalar multiplication:
λ(v + w) = λv + λw.
(Formally, the operations of addition and scalar multiplication make R3 into a vector
space.)
The zero vector is 0 = (0, 0, 0) = 0i + 0j + 0k.
Dot Product
Given two vectors u = (u1 , u2 , u3 ) and v = (v1 , v2 , v3 ), we define their dot product, or
scalar product to be the real number
u · v := u1 v1 + u2 v2 + u3 v3 .
A case of particular interest is where u = v. Then we have
v · v = v12 + v22 + v32 ≥ 0,
with v · v = 0 if and only if v = 0. The non-negative real number
√
|v| := v · v
is called the length or magnitude of v.
For example, the three basic vectors i, j, k all have magnitude 1.
Example If v = 3i + 4j + 12k then |v|2 = 32 + 42 + 122 = 169, so |v| = 13.
The dot product also satisfies some obvious properties:
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• u·v =v·u
• u · (v + w) = (u · v) + (u · w)
• u · (λv) = λ(u · v)
Exercise: Check these identities using the definition of dot product.
As long as the vectors u, v are nonzero, we can also use the dot product to define the
angle θ between them (0 ≤ θ ≤ π), using the rule:
cos(θ) =
u·v
|u|.|v|
Then u, v are said to be orthogonal, or at right angles if θ = π/2, ie if u · v = 0.
The three basic vectors i, j, k are all mutually orthogonal:
i · j = i · k = j · k = 0.
Theorem (The triangle inequality) |u + v| ≤ |u| + |v|.
Proof. |u + v|2 = (u + v) · (u + v) = |u|2 + 2|u||v| cos(θ) + |v|2 ≤ (|u| + |v|)2 .
Cross product
Given two vectors u, v, we can also form their vector product or cross product, which
is another vector, written u × v. Before giving a formula for this, let me describe geometrically what it is. The vector w = u × v is to be orthogonal to both u and v.
Its magnitude is to be the area of the parallelogram of which u and v are two sides (ie
|u||v| sin(θ), where u · v = |u||v| cos(θ)). This leaves two possibilities for the direction (eg
‘upwards’ or ‘downwards’ when u and v lie in a horizontal plane). There is a ‘right-hand
rule’ convention to settle which is which.
All this can be formalised as follows:
i j k (u1 i + u2 j + u3 k) × (v1 i + v2 j + v3 k) = u1 u2 u3 .
v1 v2 v3 Here | − | denotes determinant. In particular the three basic vectors i, j, k are related by
the cross product construction in a fairly simple manner:
i × j = k, j × i = −k, j × k = i, etc.
The cross product is less easy to use than the dot product, but this is made somewhat
easier if we use some simple rules.
• u × v = −v × u. In particular u × u = −u × u, so u × u = 0.
• u × (v + w) = (u × v) + (u × w).
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• u × (λv) = λ(u × v).
• u · (u × v) = v · (u × v) = 0. This confirms that u × v is orthogonal to u and to v.
Exercise: Check these identities using the definition of cross product.
Equation of a plane
The standard way to write the equation of a plane in R3 is in the form
ax + by + cz = d,
where a, b, c, d ∈ R are constants and x, y, z are variables corresponding to the three
coordinates of a variable vector v = (x, y, z). Another way to write this is as
n · v = d,
where n is the constant vector (a, b, c). A plane can be determined by three points (not
in a straight line).
Example Find the equation of the plane through (1, 0, 0), (0, 1, 0) and (0, 0, −1).
Solution The equation of the plane has the form
ax + by + cz = d,
and this is satisfied when (x, y, z) is one of the three points specified. This gives us three
equations
a = d, b = d, − c = d,
in four unknowns. There are infinitely many solutions, but they are all scalar multiples
of one another. Taking for example d = 1 we get a = b = 1, c = −1, so the equation of
the plane is
x + y − z = 1.
[If we took instead d = 2, this would give rise to the equivalent equation
2x + 2y − 2z = 2
for the same plane.]
Suppose u and w are two points on the plane given by the equation n · v = d. Then
we have n · u = n · w = d. The line segment from the point represented by u to the point
represented by w is given by the vector w − u, and n · (w − u) = 0. Thus n is at right
angles to every line in the plane, and so is at right angles to the plane itself. The vector
n is called a normal vector to the plane.
Example Find the distance from the point (1, 2, 3) to the plane x + 3y + 2z = 4.
Solution Let n = (1, 3, 2), so that the plane has the equation n · v = 4. If u = (1, 2, 3)
then n · u = 13. If v is the point on the plane closest to u, then n · v = 4 and u − v is
parallel to n, say u − v = λn with λ ∈ R. Then
9 = 13 − 4 = n · u − n · v = n · (λn) = λ|n|2 = 14λ.
3
√
√
Hence λ = 9/14, and so the required distance is |λn| = (9/14) 14 = 9/ 14.
Equation(s) of a line
A line in R3 can be regarded as the intersection of two planes, and so given by a pair
of simultaneous equations
n1 · v = d1 , n2 · v = d2 ,
where n1 and n2 are fixed vectors (both normal to the line), d1 and d2 are fixed scalars,
and v is the variable vector in the line.
Alternatively, it can be given by a parametric equation
v = v0 + λu,
where v0 , u are constant vectors and λ is a variable scalar. Here v0 represents a point on
the line, and u(6= 0) is any vector parallel to the line.
Example Find a parametric equation for the line given by the simultaneous equations
x + y + z = 3, x + 2y + 3z = 6.
Solution Solving the system of equations, we get x = z, y = 3 − 2z, where z can be
any real number. To fix on v0 , take z = 0 say, to give v0 = (0, 3, 0). Taking z = 1
gives another solution v1 = (1, 1, 1) = v0 + (1, −2, 1). Hence we can put λ = z and
u = (1, −2, 1), giving the parametric equation
v = (0, 3, 0) + λ(1, −2, 1).
Distance from a point to a plane
If π is the plane given by the equation ax + by +
√cz = d, then v = (a, b, c) is a vector
normal (that is, perpendicular) to π of length |v| = a2 + b2 + c2 A point P = (x0 , y0 , z0 )
of distance ` from π can be written in the form P = P1 ± `v/|v| for some P1 ∈ π. Hence
ax0 + by0 + cz0 = d ± `|v|2 /|v| = d ± `|v|.
Rearranging this equation, we see that the distance ` from P to π is given by
`=
|ax0 + by0 + cz0 − d|
.
|v|
Example Find the distance from the point (0, −1, 1) to the plane with equation 2x + y −
2z = 9.
√
Solution
2.(0)+(−1)−2.(1)
=
−3.
The
normal
vector
(2,
1,
−2)
has
length
22 + 12 + 22 =
√
9 = 3, so the distance is (9 − (−3))/3 = 12/3 = 4.
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