Re: [中學] 向量一題
: 第六題 感謝了
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: Sent from JPTT on my Sony F8332.
a^表示a向量，|a^| = a = |b^|
a, b之夾角 = 2k
a^ + b^ = 2acos(k)
a^ - b^ = 2asin(k)
2a[cosk - sink] = sqrt(2)a
=> sqrt(2)cos(Pi/4 + k) = 1/sqrt(2)
=> cos(k + Pi/4) = 1/2
=> k = 15
a, b之間的夾角為30度
另解
|a^ + b^|^2 + |a^ - b^|^2 - 2|a^ + b^||a^ - b^| = 2a^2
=> 2a^2 + 2b^2 - 2|a^ + b^||a^ - b^| = 2a^2
=> b^4 = [a^2 + b^2 + 2abcos(2k)][a^2 + b^2 - 2abcos(2k)]
=> 0 = a^4 + 2(ab)^2 - 4[abcos(2k)]^2
= a^4 + 2(ab)^2 [1 - 2(cos(2k))^2]
=> cos(2k) = sqrt[a^2[a^2 + 2b^2]/(2ab)^2]
= sqrt[3/4]
=> a, b之間的夾角為30度
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