Posted by Soroban on September 09, 2002 at 11:54:07:
In Reply to: Vector Calc posted by Ccp511 on September 09, 2002 at 02:37:45:
: if v X w= 2i-3j+5k, and v.w=3, find tan(theta) where theta is the angle between vectors v and w. What is the general formula that relates dot product and cross product and how can I use that to find the angle between the two vectors?
What a devious problem!
I don't think there is a "general formula" that relates Dot product
to Cross Product, but we can invent one!
I had to dig through the cobwebs in my brain to come up with this.
See what you think.
We have two vectors V and W.
Recall that the area of the parallelogram determined by V and W
is given by: A = |V x W|.
Also, the area of that parallelogram is: A = |V||W|sin @
where @ is the angle between the vectors.
So, |V||W|sin @ = |VxW|. Hence, sin @ = |VxW|/|V||W|.
Now, the angle between V and W is: cos @ = V*W/|V||W|.
Divide the last two equations:
(sin @)/(cos @) = |VxW|/V*W
We got it! We find that: |VxW| = sqrt(38),
so: tan @ = sqrt(38)/3
I got 64.05 degrees.
[Before I shout "It's Miller time!", is this a
well-known derivation? Is it a centuries-old
formula? Did I just "reinvent the wheel"?]
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