<多選題>坐標空間中，有\(\overset{\rightharpoonup}{a}\)、\(\overset{\rightharpoonup}{b}\)、\(\overset{\rightharpoonup}{c}\)、\(\overset{\rightharpoonup}{d}\)四個向量，滿足外積\(\overset{\rightharpoonup}{a}×\overset{\rightharpoonup}{b}=\overset{\rightharpoonup}{c}\)，\(\overset{\rightharpoonup}{a}×\overset{\rightharpoonup}{c}=\overset{\rightharpoonup}{d}\)，且\(\overset{\rightharpoonup}{a}\)、\(\overset{\rightharpoonup}{b}\)、\(\overset{\rightharpoonup}{c}\)的向量長度均為4。設向量\(\overset{\rightharpoonup}{a}\)與\(\overset{\rightharpoonup}{b}\)的夾角為\(\theta\)（其中\(0\leq\theta\leq\pi\)），試選出正確的選項。
(1)\(\cos\theta=\frac{1}{4}\)
(2)\(\overset{\rightharpoonup}{a}\)、\(\overset{\rightharpoonup}{b}\)、\(\overset{\rightharpoonup}{c}\)所張出的平行六面體的體積為16
(3)\(\overset{\rightharpoonup}{a}\)、\(\overset{\rightharpoonup}{c}\)、\(\overset{\rightharpoonup}{d}\)兩兩互相垂直
(4)\(\overset{\rightharpoonup}{d}\)的長度等於4
(5)\(\overset{\rightharpoonup}{b}\)與\(\overset{\rightharpoonup}{d}\)的夾角等於\(\theta\)
(1) 根据向量外积公式\(\vert\overset{\rightharpoonup}{a}×\overset{\rightharpoonup}{b}\vert=\vert\overset{\rightharpoonup}{a}\vert\vert\overset{\rightharpoonup}{b}\vert\sin\theta\)，已知\(\vert\overset{\rightharpoonup}{a}\vert = \vert\overset{\rightharpoonup}{b}\vert = \vert\overset{\rightharpoonup}{c}\vert = 4\)，且\(\overset{\rightharpoonup}{a}×\overset{\rightharpoonup}{b}=\overset{\rightharpoonup}{c}\)，则\(\vert\overset{\rightharpoonup}{c}\vert=\vert\overset{\rightharpoonup}{a}\vert\vert\overset{\rightharpoonup}{b}\vert\sin\theta\)，即\(4 = 4×4×\sin\theta\)，可得\(\sin\theta=\frac{1}{4}\)。又因为\(\sin^{2}\theta+\cos^{2}\theta = 1\)，\(0\leq\theta\leq\pi\)，所以\(\cos\theta=\pm\sqrt{1 - (\frac{1}{4})^{2}}=\pm\frac{\sqrt{15}}{4}\)，(1)错误。
(2) \(\overset{\rightharpoonup}{a}\)、\(\overset{\rightharpoonup}{b}\)、\(\overset{\rightharpoonup}{c}\)所张出的平行六面体的体积\(V=\vert(\overset{\rightharpoonup}{a}×\overset{\rightharpoonup}{b})\cdot\overset{\rightharpoonup}{c}\vert\)，因为\(\overset{\rightharpoonup}{a}×\overset{\rightharpoonup}{b}=\overset{\rightharpoonup}{c}\)，所以\(V=\vert\overset{\rightharpoonup}{c}\cdot\overset{\rightharpoonup}{c}\vert=\vert\overset{\rightharpoonup}{c}\vert^{2}=16\)，(2)正确。
(3) 因为\(\overset{\rightharpoonup}{a}×\overset{\rightharpoonup}{c}=\overset{\rightharpoonup}{d}\)，所以\(\overset{\rightharpoonup}{d}\)与\(\overset{\rightharpoonup}{a}\)、\(\overset{\rightharpoonup}{c}\)都垂直；又\(\overset{\rightharpoonup}{a}×\overset{\rightharpoonup}{b}=\overset{\rightharpoonup}{c}\)，所以\(\overset{\rightharpoonup}{a}\perp\overset{\rightharpoonup}{c}\)，因此\(\overset{\rightharpoonup}{a}\)、\(\overset{\rightharpoonup}{c}\)、\(\overset{\rightharpoonup}{d}\)两两互相垂直，(3)正确。
(4) 由\(\vert\overset{\rightharpoonup}{d}\vert=\vert\overset{\rightharpoonup}{a}×\overset{\rightharpoonup}{c}\vert=\vert\overset{\rightharpoonup}{a}\vert\vert\overset{\rightharpoonup}{c}\vert\sin\angle(\overset{\rightharpoonup}{a},\overset{\rightharpoonup}{c})\)，因为\(\overset{\rightharpoonup}{a}\perp\overset{\rightharpoonup}{c}\)，\(\sin\angle(\overset{\rightharpoonup}{a},\overset{\rightharpoonup}{c}) = 1\)，\(\vert\overset{\rightharpoonup}{a}\vert = \vert\overset{\rightharpoonup}{c}\vert = 4\)，所以\(\vert\overset{\rightharpoonup}{d}\vert = 16\)，(4)错误。
(5) 计算\(\overset{\rightharpoonup}{b}\cdot\overset{\rightharpoonup}{d}=\overset{\rightharpoonup}{b}\cdot(\overset{\rightharpoonup}{a}×\overset{\rightharpoonup}{c})\)，根据向量混合积的性质\(\overset{\rightharpoonup}{b}\cdot(\overset{\rightharpoonup}{a}×\overset{\rightharpoonup}{c})=\overset{\rightharpoonup}{a}\cdot(\overset{\rightharpoonup}{b}×\overset{\rightharpoonup}{c})\)，又\(\overset{\rightharpoonup}{a}×\overset{\rightharpoonup}{b}=\overset{\rightharpoonup}{c}\)，则\(\overset{\rightharpoonup}{b}×\overset{\rightharpoonup}{c}=\overset{\rightharpoonup}{b}×(\overset{\rightharpoonup}{a}×\overset{\rightharpoonup}{b})\)，根据向量积的运算规则\(\overset{\rightharpoonup}{b}×(\overset{\rightharpoonup}{a}×\overset{\rightharpoonup}{b})=(\overset{\rightharpoonup}{b}\cdot\overset{\rightharpoonup}{b})\overset{\rightharpoonup}{a}-(\overset{\rightharpoonup}{b}\cdot\overset{\rightharpoonup}{a})\overset{\rightharpoonup}{b}\)。
\(\vert\overset{\rightharpoonup}{b}\vert = 4\)，设\(\overset{\rightharpoonup}{a}\cdot\overset{\rightharpoonup}{b}=\vert\overset{\rightharpoonup}{a}\vert\vert\overset{\rightharpoonup}{b}\vert\cos\theta = 16\cos\theta\)，则\(\overset{\rightharpoonup}{b}×(\overset{\rightharpoonup}{a}×\overset{\rightharpoonup}{b}) = 16\overset{\rightharpoonup}{a}-16\cos\theta\overset{\rightharpoonup}{b}\)，\(\overset{\rightharpoonup}{a}\cdot(\overset{\rightharpoonup}{b}×\overset{\rightharpoonup}{c})=\overset{\rightharpoonup}{a}\cdot(16\overset{\rightharpoonup}{a}-16\cos\theta\overset{\rightharpoonup}{b}) = 16\vert\overset{\rightharpoonup}{a}\vert^{2}-16\cos\theta\overset{\rightharpoonup}{a}\cdot\overset{\rightharpoonup}{b}=256 - 256\cos^{2}\theta\)。
\(\vert\overset{\rightharpoonup}{b}\vert = 4\)，\(\vert\overset{\rightharpoonup}{d}\vert = 16\)，设\(\overset{\rightharpoonup}{b}\)与\(\overset{\rightharpoonup}{d}\)的夹角为\(\alpha\)，\(\cos\alpha=\frac{\overset{\rightharpoonup}{b}\cdot\overset{\rightharpoonup}{d}}{\vert\overset{\rightharpoonup}{b}\vert\vert\overset{\rightharpoonup}{d}\vert}=\frac{256 - 256\cos^{2}\theta}{4×16}=4 - 4\cos^{2}\theta\neq\cos\theta\)（一般情况下），所以\(\overset{\rightharpoonup}{b}\)与\(\overset{\rightharpoonup}{d}\)的夹角不等于\(\theta\)，(5)错误。
答案为(2)(3)。 報錯
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