✔ 最佳答案
Let a, b be 2 vectors.
|a∙b|
=|a||b|cosθ|
∵ 1≥|cosθ|≥0
|a||b|≥|a||b||cosθ|≥|a||b|X0
∴ |a||b|≥|a||b||cosθ|≥0
|a∙b|≤|a||b|
2012-07-07 00:01:16 補充:
So WHAT'S WRONG with your process?
∵ |cosθ| ≥ 0 correct
∴ |a| |b| |cosθ| ≥ |a| |b| WRONG !
∵ |cosθ| ≥ 0
Both sides multiplied by |a||b|, we have
( |a| |b| ) |cosθ| ≥ ( |a| |b| ) X 0
∴ |a| |b| |cosθ| ≥ 0
2012-07-07 00:02:06 補充:
If |a| |b| |cosθ| ≥ |a| |b|
Then |cosθ| ≥ 1 which is impossible
In fact, 1 ≥ |cosθ|
Both sides multiplied by |a| |b|, we have
|a| |b| ≥ |a| |b| |cosθ|