(4)$A$矢量的方向余弦(与三个坐标轴的夹角余弦)的大小是：
A: $cos\alpha=3/\sqrt{14},cos\beta=-1/\sqrt{14},cos\gamma=3/\sqrt{14}$
B: $cos\alpha=4/\sqrt{14},cos\beta=-1/\sqrt{14},cos\gamma=3/\sqrt{14}$
C: $cos\alpha=2/\sqrt{14},cos\beta=-1/\sqrt{14},cos\gamma=3/\sqrt{14}$
D: $cos\alpha=3/\sqrt{14},cos\beta=9/\sqrt{14},cos\gamma=3/\sqrt{14}$
A: $cos\alpha=3/\sqrt{14},cos\beta=-1/\sqrt{14},cos\gamma=3/\sqrt{14}$
B: $cos\alpha=4/\sqrt{14},cos\beta=-1/\sqrt{14},cos\gamma=3/\sqrt{14}$
C: $cos\alpha=2/\sqrt{14},cos\beta=-1/\sqrt{14},cos\gamma=3/\sqrt{14}$
D: $cos\alpha=3/\sqrt{14},cos\beta=9/\sqrt{14},cos\gamma=3/\sqrt{14}$
举一反三
- 以\( (2,2,1) \)为起点，以\( (1,3,0) \)为终点的向量的方向余弦为（ ）. A: \( \cos \alpha = { { - 1} \over {\sqrt 3 }},\cos \beta = {1 \over {\sqrt 3 }},\cos \gamma = { { - 1} \over {\sqrt 3 }} \) B: \( \cos \alpha = {1 \over {\sqrt 3 }},\cos \beta = { { - 1} \over {\sqrt 3 }},\cos \gamma = { { - 1} \over {\sqrt 3 }} \) C: \( \cos \alpha = { { - 1} \over {\sqrt 3 }},\cos \beta = { { - 1} \over {\sqrt 3 }},\cos \gamma = { { - 1} \over {\sqrt 3 }} \) D: \( \cos \alpha = { { - 1} \over {\sqrt 3 }},\cos \beta = { { - 1} \over {\sqrt 3 }},\cos \gamma = {1 \over {\sqrt 3 }} \)
- \((\cos\alpha \cos\beta, \cos\alpha \sin\beta, \sin\alpha),(1,1,0),(1,2,1)\)张成六面体体积最大为___. A: \(\sqrt{3}\) B: \(2\sqrt{3}\) C: \(\sqrt{6}\)
- 计算[img=127x40]1803177bfc8c3e2.png[/img]的程序表达为 A: sqrt(sin alpha ^ 2 + cos beta ^ 2) B: sqrt(sin^2(alpha) + cos^2(beta)) C: sqrt(pow(sin(alpha), 2) + pow(cos(beta), 2)) D: sqrt pow(sin(alpha), 2) + pow(cos(beta), 2)
- 14、求平方根函数的函数名为（ ）。 A: cos B: abs C: pow D: sqrt
- 曲线$\left\{ \matrix{ {x^2} + {y^2} + {z^2} = 9 \cr y = x \cr} \right.$的参数方程为( ). A: $$\left\{ \matrix{ x = \sqrt 3 \cos t \cr y = \sqrt 3 \cos t \cr z = \sqrt 3 \sin t \cr} \right.(0 \le t \le 2\pi )$$ B: $$\left\{ \matrix{ x = {3 \over {\sqrt 2 }}\cos t\cr y = {3 \over {\sqrt 2 }}\cos t \cr z = 3\sin t \cr} \right.(0 \le t \le 2\pi )$$ C: $$\left\{ \matrix{ x = \cos t\cr y = \cos t\cr z = \sin t \cr} \right.(0 \le t \le 2\pi )$$ D: $$\left\{ \matrix{ x = {{\sqrt 3 } \over 3}\cos t\cr y = {{\sqrt 3 } \over 3}\cos t \cr z = {{\sqrt 3 } \over 3}\sin t\cr} \right.(0 \le t \le 2\pi )$$