Prove: vectors u•v=u1•v1+u2+v2 from a• b=|a|•|b|•cos D
u•v=u1•v1+u2+v2 from a• b=|a|•|b|•cos D
|u| = sqrt(u1^2 + u2^2)
|v| = sqrt (v1^2 + v2^2)
so
|u| |v| = sqrt(u1^2 + u2^2) * sqrt (v1^2 + v2^2)
then for the angle
cos (alpha-beta) = cos alpha cos beta + sin alpha sin beta
= v1/sqrt(v1^2+v2^2)*u1/sqrt(u1^2+u2^2)
+ v2/sqrt(v1^2+v2^2)*u1/sqrt(u1^2+u2^2)
so
v1* u1 - v2*u1
whew
are u1,u2,v1,v2 the components of u and v?
If so, then I think you must mean u•v=u1•v1+u2•v2
But that is poor notation, since one • means vector dot product, and the other • means scalar multiplication.
First, it is each to show that
i•i = 1 and j•j=1
since |i|=|j|=1 and cosθ=1 since θ=0
also, i•j=0 since cosθ=0
Now you can see that
u•v = (u1 i + u2 j)•(v1 i + v2 j)
Now just expand that out. All the i•i and j•j are 1, and i•j = j•i = 0
I mean:
so
v1* u1 + v2*u1
whew again
To prove the equation u•v = u1•v1 + u2•v2, we need to show that the dot product of two vectors u and v can be calculated by taking the product of their corresponding components and then summing them.
First, let's define the vectors u and v as follows:
- u = u1i + u2j
- v = v1i + v2j
Here, i and j represent the unit vectors in the x and y directions, respectively.
Now, let's calculate the dot product of u and v:
u•v = (u1i + u2j) • (v1i + v2j)
= (u1•v1)i•i + (u1•v2)i•j + (u2•v1)j•i + (u2•v2)j•j
Since i•i = j•j = 1 (the dot product of any unit vector with itself is 1) and i•j = j•i = 0 (the dot product of orthogonal vectors is 0), the equation simplifies to:
u•v = u1•v1 + u2•v2
Therefore, we have proved that the dot product of two vectors u and v can be calculated by taking the product of their corresponding components and summing them, which is u•v = u1•v1 + u2•v2.
Now, let's move on to proving the equation a•b = |a|•|b|•cos θ, where a and b are vectors and θ is the angle between them.
The magnitude (or length) of a vector a, denoted by |a|, can be calculated using the Pythagorean theorem:
|a| = sqrt(a1^2 + a2^2)
Similarly, the magnitude of vector b, denoted by |b|, can be calculated as:
|b| = sqrt(b1^2 + b2^2)
Now, let's rewrite vectors a and b in terms of their magnitudes and unit vectors:
a = |a|•cos α•i + |a|•sin α•j
b = |b|•cos β•i + |b|•sin β•j
Here, α and β represent the angles that vectors a and b make with the positive x-axis, respectively.
Now, let's calculate the dot product of a and b:
a•b = (|a|•cos α•i + |a|•sin α•j) • (|b|•cos β•i + |b|•sin β•j)
= (|a|•cos α•i)•(|b|•cos β•i) + (|a|•cos α•i)•(|b|•sin β•j)
+ (|a|•sin α•j)•(|b|•cos β•i) + (|a|•sin α•j)•(|b|•sin β•j)
Using the properties of the dot product we discussed earlier (i•i = j•j = 1, i•j = j•i = 0), this equation simplifies to:
a•b = |a|•|b|•cos α•cos β + |a|•|b|•sin α•sin β
= |a|•|b|•(cos α•cos β + sin α•sin β)
Using the trigonometric identity cos (α-β) = cos α•cos β + sin α•sin β, we can rewrite the equation as:
a•b = |a|•|b|•cos (α-β)
Since α-β is the angle (let's call it θ) between vectors a and b, we have:
a•b = |a|•|b|•cos θ
Therefore, we have proved that the dot product of two vectors a and b is equal to the product of their magnitudes and the cosine of the angle between them, which is a•b = |a|•|b|•cos θ.