已知向量m(根号3,1),向量n是与向量m夹角为60°的单位向量求(!)向量n,(2)若向量n与向量Q=(-根号3,1)共线,与向量p=(根号3x^2,x-y^2)垂直.求t=y^2+5x+4的最大值
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![已知向量m(根号3,1),向量n是与向量m夹角为60°的单位向量求(!)向量n,(2)若向量n与向量Q=(-根号3,1)共线,与向量p=(根号3x^2,x-y^2)垂直.求t=y^2+5x+4的最大值](/uploads/image/z/4521518-62-8.jpg?t=%E5%B7%B2%E7%9F%A5%E5%90%91%E9%87%8Fm%28%E6%A0%B9%E5%8F%B73%2C1%EF%BC%89%2C%E5%90%91%E9%87%8Fn%E6%98%AF%E4%B8%8E%E5%90%91%E9%87%8Fm%E5%A4%B9%E8%A7%92%E4%B8%BA60%C2%B0%E7%9A%84%E5%8D%95%E4%BD%8D%E5%90%91%E9%87%8F%E6%B1%82%28%21%29%E5%90%91%E9%87%8Fn%2C%282%29%E8%8B%A5%E5%90%91%E9%87%8Fn%E4%B8%8E%E5%90%91%E9%87%8FQ%3D%28-%E6%A0%B9%E5%8F%B73%2C1%29%E5%85%B1%E7%BA%BF%2C%E4%B8%8E%E5%90%91%E9%87%8Fp%3D%EF%BC%88%E6%A0%B9%E5%8F%B73x%5E2%2Cx-y%5E2%EF%BC%89%E5%9E%82%E7%9B%B4.%E6%B1%82t%3Dy%5E2%2B5x%2B4%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC)
已知向量m(根号3,1),向量n是与向量m夹角为60°的单位向量求(!)向量n,(2)若向量n与向量Q=(-根号3,1)共线,与向量p=(根号3x^2,x-y^2)垂直.求t=y^2+5x+4的最大值
已知向量m(根号3,1),向量n是与向量m夹角为60°的单位向量
求(!)向量n,(2)若向量n与向量Q=(-根号3,1)共线,与向量p=(根号3x^2,x-y^2)垂直.求t=y^2+5x+4的最大值
已知向量m(根号3,1),向量n是与向量m夹角为60°的单位向量求(!)向量n,(2)若向量n与向量Q=(-根号3,1)共线,与向量p=(根号3x^2,x-y^2)垂直.求t=y^2+5x+4的最大值
(1)
m =(√3,1)
let n be (x,y )
|n| =1
=> x^2+y^2=1
m.n = |m||n|cos60°
(√3,1).(x,y) = 2(1/2)
√3x+ y = 1
y = 1-√3x
x^2+y^2=1
x^2 + 3x^2 -2√3x +1 = 1
2x(2x-√3) =0
x=0 or x = √3/2
when x= 0 y= 1
when x=√3/2,y= -1/2
n = (0,1) or (√3/2,-1/2)
(2)
for n=(0,1),n与Q不能共线
ie n= (√3/2,-1/2)
p=(√3x^2,x-y^2)
n与p垂直 =>
n.p =0
(√3/2,-1/2).(√3x^2,x-y^2)=0
3x^2 -x + y^2 =0
t = y^2 + 5x +4
= (-3x^2+x) +5x +4
= -3x^2+6x +4
t' = -6x+6 =0
x= 1
t''=-6
1
n=(0,1) OR n=(√3/2,-1/2)
2
由1,n=(√3/2,-1/2)
因为 n与p垂直,所以 n*p=(3x^2-x+y^2)/2=0,
因此,t=x-3x^2+5x+4=-3(x-1)^2+7,
当x=1时,t取最大值7.