Several math problems in senior two! Urgent! Steps ~

1, let D(x, y, z), vector db = (-x, 1-y, -z) AC = (- 1, 0, 1) DC = (-x, -y, 650).

Parallel availability

-x=-k

y= 1

z=-k

Answer a.

2. it is a necessary and sufficient condition that x belongs to b, that is to say, b is a subset of a.

a = { x |-2 & lt； = x & lt=5}

So m+1>; =-2

2m- 1 & lt； =5

2m- 1 & gt； m+ 1

solve

2 & ltm & lt=3

3. The maximum point to the right focus is: a+c, and the minimum point to the focus is a-c. If the tolerance is greater than1100 and n is the maximum value, then PnF=a+c, P 1F=a-c, PNF-P/.

So the maximum value of n is 200.

4、

Because the angle formed by AD and BF, that is, the angle formed by BC and BF connects CF, the triangle BFC is an isosceles triangle (because the dihedral angle is 60 degrees, it connects CE, that is, CE=BC, so that the triangle BEF and the triangle CEF are congruent, FB=FC).

Therefore, let FG be the midline (also the high line) on the bottom BC of BFC, that is, find cos (angle FBC)=BG/FB.

BG=BC/2，FB=√2BC，

Cos (angle FBC) = BG/FB = (1/2)/√ 2 = √ 2/4.

5、

Solution: The sum of the point (1, 3/2) and the focal point (-1, 0) is.

2a=√( 1+ 1)？ +(3/2-0)? +√( 1- 1)? +(3/2-0)? =5/2+3/2=4

a=2

c= 1

b？ =a？ -c？ =4- 1=3

Ellipse: x? /4+y？ /3= 1

Let AE slope be k and AF be-K.

Straight AE: y = k (x- 1)+3/2.