有向量a，b，|a|=4，|b|=5，a与b的夹角为π/3，求|a-b|及|a+b|
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|a-b|^2
= (a-b). (a-b)
=|a|^2 + |b|^2 - 2a.b
=16+25 - 2|a||b| cos(π/3)
=41-2(4)(5)1/2
=41-20
=21
so |a-b|= √21 #
Similarly
|a+b|^2 = (a+b). (a+b)
= 61
|a+b| = √61 #
= (a-b). (a-b)
=|a|^2 + |b|^2 - 2a.b
=16+25 - 2|a||b| cos(π/3)
=41-2(4)(5)1/2
=41-20
=21
so |a-b|= √21 #
Similarly
|a+b|^2 = (a+b). (a+b)
= 61
|a+b| = √61 #
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