解4題高二數學題

2.行列式|a d g|
|b e h|=6
|c f i|求行列式的值|a+d d+g g+a|
|b+e e+h h+b| = ?
|c+f f+i i+c|
*|a d g|*|d g a|
=|b e h|+|e h b|
*|c f i|*|f i c|
*|a d g|*|d a g|
=|b e h|-|e b h|
*|c f i|*|f c i|
*|a d g|*|a d g|
=|b e h|+|b e h|
*|c f i|*|c f i|= 2*6= 12
3.過A(1，-1，1)，B(2，1，1)，C(3，5，0)三點平面方程式AB = B - A = (1,2,0)AC = C - A = (2,6,-1)N = Normal= AB x AC=|2 *0 1 2|
*|6 -1 2 6|=(-2-0;0+1;6-4)=(-2,1,2)0 = -2(x-1) + (y+1) + 2(z-1)= -2x + y + 2z + 2 + 1 - 2= -2x + y + 2z + 1
4.x-y+2z=3與2x+y+cz=4夾角=60度，c=?N1 = (1,-1,2)N2 = (2,1,c)N1 dot N2 = 2 - 1 + 2c= (2c + 1)= √[6*(c^2+5)]*cos(60) or cos(120)= √[6*(c^2+5)]*(+-1/2)4(2c + 1)^2 = 6*(c^2 + 5)2(4c^2 + 4c + 1) = 3c^2 + 150 = 5c^2 + 8c - 13 = (c - 1)(5c + 13) c = 1 or -13/5

2015-05-03 12:52:46 補充：
1.i,j,k = Unit Vectors

u = (x-2)i + yj + 2z*k

v = 2i - j + 2k

Cauchy Inequality:

u dot v

= 2(x-2) - y + 4z

<= √[(x-2)^2+y^2+4z^2]*√(4+1+4)

=> [2(x-2) - y + 4z]^2 <= [(x-2)^2+y^2+4z^2]*9 = 81

-9 <= (2x - y + 4z - 4) <= 9

-5 <= (2x - y + 4z - 4) <= 13

=> min = -5

=> max = 13

2015-05-03 14:02:36 補充：
1-2. m,n = max,min; Pm=?, Pn=?


(1) Ellipsoid: s=sin, c=cos


A = 0~2pi

B = 0~pi


x = 2 + 3*cosA*sinB

y = 3*sinA*sinB

z = 1.5*cosB



(2) F(A,B) = ?

F(A,B) = 2x - y + 4z

= 4 + 6*cA*sB - 3*sA*sB + 6cB

2015-05-03 14:04:13 補充：
(3) Partial Differences

Fa = -6*sA*sB - 3*cA*sB

= -3*sB*(2sA + cA)

= 0

=> tanA = -1/2, sA = +-1/√5, cA = -+2/√5



Fb = 3[cB*(2cA - sA) - 2sB] = 0

=> tanB = (2cA - sA)/2 = -+√5/2

=> sinB = +-√5/3, cosB= -+2/3 or +-2/3



(4) Pm, Pn = ?

x = 2 + 3*cA*sB

= 2 +- 3*(2/√5)*(√5/3)

= 2 +- 2

2015-05-03 14:05:05 補充：
= 4 or 0

y = 3*sA*sB

= -+ 3*(1/√5)*(√5/3)

= -+ 1

z = +- 1.5*cB

= +- 1.5*2/3

= +- 1


Pm = (4, -1, 1)

Pn = (0, 1, -1)

第一題螞蟻知識長用柯西不等式的方法算精簡
有平方項求最大或最小值可考慮使用

至於第四題的話
兩平面的夾角為60°和120°
跟兩法向量的夾角一樣~

1.設實數x，y，z滿足(x－2)^2+y^2+4z^2=9，求2x─y+4z的最大值與最小值，並分別求有最大值與最小值時x，y，z的值
Sol
[(x－2)^2+y^2+(2z)^2]*[2^2+(－1)^2+2^2]>=[2(x－2)－y+2(2z)]^2
[(x－2)^2+y^2+4z^2]*9>=(2x－y+4z－4)^2
81>=(2x－y+4z－4)^2
－9<=2x－y+4z－4<=9
－5<=2x－y+4z<=13