向量内积的定义和基本性质的解答题,1.已知向量a=(1,根号3),向量b=(-根号3,-1),求2.已知点A(X,-1),B(-2,-6),C(1,-2)且| 向量AB |=| 向量AC |,求X的值.3.已知点A(X,4),B(2,Y+3),且向量AB=(3,6),
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/27 12:27:25
![向量内积的定义和基本性质的解答题,1.已知向量a=(1,根号3),向量b=(-根号3,-1),求2.已知点A(X,-1),B(-2,-6),C(1,-2)且| 向量AB |=| 向量AC |,求X的值.3.已知点A(X,4),B(2,Y+3),且向量AB=(3,6),](/uploads/image/z/2475808-16-8.jpg?t=%E5%90%91%E9%87%8F%E5%86%85%E7%A7%AF%E7%9A%84%E5%AE%9A%E4%B9%89%E5%92%8C%E5%9F%BA%E6%9C%AC%E6%80%A7%E8%B4%A8%E7%9A%84%E8%A7%A3%E7%AD%94%E9%A2%98%2C1.%E5%B7%B2%E7%9F%A5%E5%90%91%E9%87%8Fa%3D%EF%BC%881%2C%E6%A0%B9%E5%8F%B73%EF%BC%89%2C%E5%90%91%E9%87%8Fb%3D%EF%BC%88-%E6%A0%B9%E5%8F%B73%2C-1%EF%BC%89%2C%E6%B1%822.%E5%B7%B2%E7%9F%A5%E7%82%B9A%EF%BC%88X%2C-1%EF%BC%89%2CB%28-2%2C-6%29%2CC%EF%BC%881%2C-2%EF%BC%89%E4%B8%94%7C+%E5%90%91%E9%87%8FAB+%7C%3D%7C+%E5%90%91%E9%87%8FAC+%7C%2C%E6%B1%82X%E7%9A%84%E5%80%BC.3.%E5%B7%B2%E7%9F%A5%E7%82%B9A%EF%BC%88X%2C4%EF%BC%89%2CB%EF%BC%882%2CY%2B3%EF%BC%89%2C%E4%B8%94%E5%90%91%E9%87%8FAB%3D%EF%BC%883%2C6%EF%BC%89%2C)
向量内积的定义和基本性质的解答题,1.已知向量a=(1,根号3),向量b=(-根号3,-1),求2.已知点A(X,-1),B(-2,-6),C(1,-2)且| 向量AB |=| 向量AC |,求X的值.3.已知点A(X,4),B(2,Y+3),且向量AB=(3,6),
向量内积的定义和基本性质的解答题,
1.已知向量a=(1,根号3),向量b=(-根号3,-1),求
2.已知点A(X,-1),B(-2,-6),C(1,-2)且| 向量AB |=| 向量AC |,求X的值.
3.已知点A(X,4),B(2,Y+3),且向量AB=(3,6),求X,Y的值.
向量内积的定义和基本性质的解答题,1.已知向量a=(1,根号3),向量b=(-根号3,-1),求2.已知点A(X,-1),B(-2,-6),C(1,-2)且| 向量AB |=| 向量AC |,求X的值.3.已知点A(X,4),B(2,Y+3),且向量AB=(3,6),
1.a = (1,√3),b = (-√3,-1)
cos = a•b /∣a∣∣b∣
= [1 * (-√3) + √3 * (-1)] / √[1² + (√3)²] * √[(-√3)² + (-1)²]
= -2√3 / 4
= -√3 / 2
所以 = 180° - 30° = 150°
2.AB = (-2,-6) - (x,-1)
= (-2-x,-6+1)
= (-2-x,-5)
AC = (1,-2) - (x,-1)
= (1-x,-2+1)
= (1-x,-1)
且∣AB∣=∣AC∣
√[(-2-x)² + (-5)²] = √[(1-x)² + (-1)²]
两边同时平方,x² + 4x + 4 + 25 = x² - 2x + 1 + 1
解得 x = -9/2
3.AB = (2,y+3) - (x,4)
= (2-x,y+3-4)
= (2-x,y-1)
且AB = (3,6)
所以2 - x = 3,y - 1 = 6
解得x = -1,y = 7