Arturo O. answered • 09/16/17
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It is sufficient to show that
a·b > 0
(4b -a) ⊥ (2a - b) ⇒ (4b -a) · (2a - b) = 0
(4b -a) · (2a - b) = 8b·a - 4b2 - 2a2 + a·b = 0
9a·b = 2a2 + 4b2
a·b = (2a2 + 4b2)/9 > 0
Therefore,
a·b > 0
The angle between a and b is acute.
Arturo O.
Mark,
The distributive property applies here too, as long as you use the dot product in the multiplications. Also, note that
a·a = |a|2 = a2
b·b = |b|2 = b2
a·b = b·a
For vectors a and b,
(4b)·(2a) = (4)(2)(b·a) = 8b·a = 8a·b
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09/16/17
Arturo O.
I should add that (<4b>)(<2a>) is not defined. The only products between vectors defined here are the dot and cross products.
(4b)·(2a) is defined
(4b)X(2a) is defined
(4b)(2a) is not defined
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09/16/17
Mark M.
09/16/17