Question

求过点(2,1,3)且与直线(x+1)/3=(y-1)/2=z/-1 垂直相交的直线方程

Answer 1

简单计算一下即可，答案如图所示

Answer 2

L :直线(x+1)/3=(y-1)/2=z/-1 =k
C(x,y,z) is on L
x=2k-1, y=2k+1, z=-k (1)

过点A(2,1,3)
(2k-1-2, 2k+1-1, -k-3).(3,2,-1) =0
(2k-3, 2k, -k-3).(3,2,-1) =0
3(2k-3)+2(2k)-(-k-3)=0
11k-6=0
k = 6/11

from (1)
x=2k-1 = 12/11 -1 = 1/11
y=2k+1= 12/11+1 = 23/11
z=-k = -6/11
C = (1/11, 23/11, -6/11)

equation of AC
(x-2)/(2-1/11) = (y-1)/(1- 23/11) = (z-3)/(3+6/11)
(x-2)/(-9/11) = (y-1)/(-12/11) = (z-3)/(39/11)
(x-2)/9 = (y-1)/12 = (z-3)/(-39)
(x-2)/3 = (y-1)/4 = (z-3)/(-13)

Answer 3

(x+1)/3=(y-1)/2
2x+2=3y-3
y=2x/3+5/3
过M垂直y=2x/3+5/3 y-1=-1/(2/3) (x-2)
y=4-3x/2
(x+1)/3=z/-1
3z=-x-1,z=-x/3-1/3
过M垂直z=-x/3-1/3 z-3=-1/(-1/3) (x-2)
z=3x-3
过M垂直直线L:
(4-y)+(z+3)/2=3x
8-2y+z+3=6x
x+2y-z-11=0