题目
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.
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.
提问时间:2020-11-19
答案
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
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
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