ů = uiltų 5 are non zero vector
V = V, 2 tu, ] avey parallel it and
UV - U₂ Y = 0
than they
only for
Answers
Answer:
Step-by-step explanation:
2. Dot product
Definition 2.1. Let ~v = (v1, v2, v3) and ~w = (w1, w2, w3) be two vectors
in R
3
. The dot product of ~v and ~w, denoted ~v · ~w, is the scalar
v1w1 + v2w2 + v3w3.
Example 2.2. The dot product of ~v = (1, −2, −1) and ~w = (2, 1, −3)
is
1 · 2 + (−2) · 1 + (−1) · (−3) = 2 − 2 + 3 = 3.
Lemma 2.3. Let ~u, ~v and ~w be three vectors in R
3 and let λ be a
scalar.
(1) (~u + ~v) · ~w = ~u · ~w + ~v · ~w.
(2) ~v · ~w = ~w · ~v.
(3) (λ~v) · ~w = λ(~v · ~w).
(4) ~v · ~v = 0 if and only if ~v = ~0.
Proof. (1–3) are straightforward.
To see (4), first note that one direction is clear. If ~v = ~0, then
~v · ~v = 0. For the other direction, suppose that ~v · ~v = 0. Then
v
2
1 + v
2
2 + v
2
3 = 0. Now the square of a real number is non-negative and
if a sum of non-negative numbers is zero, then each term must be zero.
It follows that v1 = v2 = v3 = 0 and so ~v = ~0.
Definition 2.4. If ~v ∈ R
3
, then the norm or length of ~v = (v1, v2, v3)
is the scalar
kvk =
√
~v · ~v = (v
2
1 + v
2
2 + v
2
3
)
1/2
.
It is interesting to note that if you know the norm, then you can
calculate the dot product:
(~v + ~w) · (~v + ~w) = ~v · ~v + 2~v · ~w + ~w · ~w
(~v − ~w) · (~v − ~w) = ~v · ~v − 2~v · ~w + ~w · ~w.
Subtracting and dividing by 4 we get
~v · ~w =
1
4
((~v + ~w) · (~v + ~w) − (~v − ~w) · (~v − ~w))
=
1
4
(k~v + ~wk
2 − k~v − ~wk
2
).
Given two non-zero vectors ~v and ~w in space, note that we can define
the angle θ between ~v and ~w. ~v and ~w lie in at least one plane (which
is in fact unique, unless ~v and ~w are parallel). Now just measure the
angle θ between the ~v and ~w in this plane. By convention we always
take 0 ≤ θ ≤ π.
1
Theorem 2.5. If ~v and ~w are any two vectors in R
3
, then
~v · ~w = k~vk k~wk cos θ.
Proof. If ~v is the zero vector, then both sides are equal to zero, so that
they are equal to each other and the formula holds (note though, that
in this case the angle θ is not determined).
By symmetry, we may assume that ~v and ~w are both non-zero. Let
~u = ~w−~v and apply the law of cosines to the triangle with sides parallel
to ~u, ~v and ~w:
k~uk
2 = k~vk
2 + k~wk
2 − 2k~vkk~wk cos θ.
We have already seen that the LHS of this equation expands to
~v · ~v − 2~v · ~w + ~w · ~w = k~vk
2 − 2~v · ~w + k~wk.
Cancelling the common terms k~vk
2 and k~wk
2
from both sides, and
dividing by 2, we get the desired formula.
We can use (2.5) to find the angle between two vectors:
Example 2.6. Let ~v = −ˆı + ˆk and ~w = ˆı + ˆ. Then
−1 = ~v · ~w = k~vkk~wk cos θ = 2 cos θ.
Therefore cos θ = −1/2 and so θ = 2π/3.
Definition 2.7. We say that two vectors ~v and ~w in R
3 are orthogonal if ~v · ~w = 0.
Remark 2.8. If neither ~v nor ~w are the zero vector, and ~v · ~w = 0 then
the angle between ~v and ~w is a right angle. Our convention is that the
zero vector is orthogonal to every vector.
Example 2.9. ˆı, ˆ and ˆk are pairwise orthogonal.
Given two vectors ~v and ~w, we can project ~v onto ~w. The resulting
vector is called the projection of ~v onto ~w and is denoted proj ~w ~v. For
example, if F~ is a force and ~w is a direction, then the projection of F~
onto ~w is the force in the direction of ~w.
As proj ~w ~v is parallel to ~w, we have
proj ~w ~v = λ ~w,
for some scalar λ. Let’s determine λ. Let’s deal with the case that
λ ≥ 0 (so that the angle θ between ~v and ~w is between 0 and π/2). If
we take the norm of both sides, we get
k proj ~w ~vk = kλ ~wk = λk~wk,
2
(note that λ ≥ 0), so that
λ =
k proj ~w ~vk
k~wk
.
But
cos θ =
k proj ~w ~vk
k~vk
,
so that
k proj ~w ~vk = k~vk cos θ.
Putting all of this together we get
λ =
k~vk cos θ
k~wk
=
k~vkk~wk cos θ
k~wk
2
=
~v · ~w
k~wk
2
.
There are a number of ways to deal with the case when λ < 0 (so
that θ > π/2). One can carry out a similar analysis to the one given
above. Here is another way. Note that the angle φ between ~w and
~u = −~v is equal to π − θ < π/2. By what we already proved
proj ~w ~u =
~u · ~w
k~wk
2
~w.
But proj ~w ~u = − proj ~w ~v and ~u· ~w = −~v · ~w, so we get the same formula
in the end. To summarise:
Theorem 2.10. If ~v and ~w are two vectors in R
3
, where ~w is not zero,
then
proj ~w ~v =
~v · ~w
k~wk
2
~w.
Example 2.11. Find the distance d between the line l containing the
points P1 = (1, −1, 2) and P2 = (4, 1, 0) and the point Q = (3, 2, 4).
Suppose that R is the closest point on the line l to the point Q. Note
that −→QR is orthogonal to the direction −−→P1P2 of the line. So we want the
length of the vector −−→P1Q − proj−−−→ P1P2
−−→P1Q, that is, we want
Answer:
tvjjggfddvnvdeerfgvwdhnddkol
gghjkkoolu y
hhhnmh