12.3:

1.<1,2,1>·<4,3,5>
#<1,2,1>·<4,3,5>=<1·4,2·3,1·5>=<4,6,5>

13.<1,1,1>,<1,-2,-2>
#cosθ=(a1b1+a2b2+a3b3)/(|A||B|)=(1-2-2)/(sqrt(3)*3)=-1/sqrt3
#θ<0
#THE TWO VECTORS ARE NOT ORTHOGONAL AND THE ANGLE BETWEEN THEM IS OBTUSE.

21.i+j,j+2k
#cosθ=((1*0)i+(1*1)j+(0*2)k)/(sqrt(i+j)sqrt(j+4k))=1/sqrt10

29.
(a)<b,3,2>,<1,b,1>
#cosθ=(b+3b+2)/sqrt((b^2+25)sqrt((b^2+4)=0
#b=-1/2
(b)<4,-2,7>,<b^2,b,0>
#cosθ=(4b^2-2b)/sqrt(69)sqrt(b^4+b^2)=0
#b=1/2

31.
#<a,x,b>
#cosθ=(2a-3b)/(|A||B|)=0
#2a-3b=0
#2a=3b
#{a=3,b=2; a=6,b=4; a=9,b=6; ... ;}
#THOSE TWO VECTORS CAN BE<3,6,2>AND<6,8,4>

57.u=5i+7j-4k, v=k
#V=u·v/|v|=(5*0)i+(7*0)j-(4*1)k/sqrt(1)k=-4

63.
#OP=u·v/|v|=((3*8)+(5*2))/(sqrt(68))=sqrt(17)

12.4  : 

1.[[1,2],[4,3]]
#det[[1,2],[4,3]]=3-6=-3

5.[[1,2,1],[4,-3,0],[1,0,1]]
#det[[1,2,1],[4,-3,0],[1,0,1]]=1(-3)-2(-4)+1(3)=8

13.(i+j)×k
#i×k+j×k

21.(u-2v)×(u+2v)
#(u-2v)×(u+2v)=u×u-2v×2v-2v×u+2v×u=<0,0,0>

25.
#-u IS EQUAL TO v×w

27.
#SINCE v IS ON THE X-AXIS, ALL OF THE LINES ON THE YZ-PLANE IS PERPENDICULAR TO v.
#SINCE w IS 45°DOWNWARD IN THE FOURTH QUADRANT ON THE YZ PLANE,
#THE LINE Y=Z IS PERPENDICULAR WITH w
#AND BY RIGHT HAND THEORY, u IS ON POSITIVE DIRECTION.
#THEREFORE, u=<0,1,1>

39.
#ON THE HANDWRITTEN PDF VERSION

41.u=<1,0,3>v=<2,1,1>
#THE AREA=|u×v|=sqrt(9+25+1)=sqrt35

43.
#ON THE HANDWRITTEN PDF VERSION

45.O(1,2)P(3,4)Q(-2,2)
#OP=<1,1>,OQ=<-3,0>
#OP×OQ=4
#THE AREA OF THE TRIANGLE=4/2=2

12.5  : 

1.
#d=x+3y+2z=4-3+2=3
#THE EQUATION IS x+3y+2z=3

5.
#d=ix=3i=3
#THE EQUATION IS x=3

9.
#x+y+z=0

11.
#(b) and (d) ARE TRUE.

13.
#9x-4y-11z=2
#ONE GROUP OF SOLUTION IS (4, 3, 2)
#THE VECTOR CAN BE <4,3,2>

15.
#3(x-4)-8(y-1)+11z=0
#ONE GROUP OF SOLUTION IS (4,1,0)
#THE VECTOR CAN BE <4,1,0>

17.
#PQ=<-1,2,-3>,PR=<1,2,-6>
#PQ×PR=<-6,-9,-4>
#-6x-9y-4z=d
#-12+9-16=-19
#THE EQUATION IS -6x-9y-4z=-19

19.
#PQ=<-1,1,1>,PR=<1,0,1>
#PQ×PR=<1,2,-1>
#x+2y-z=d
#2-1=1
#THE EQUATION IS x+2y-z=1

25.
#ix+kz=d
#-2+5=3
#THE EQUATION IS x+z=3

31.
#ON THE HANDWRITTEN PDF VERSION

53.
#FOR LINE 3x+2z=5, WHEN x=0, z=5/2; WHEN z=0, x=5/3
#A(5/3,0,0),B(0,0,5/2)
#IF C IS (0,1,0) AND IS ON THE PLANE,
#CA=<5/3,-1,0>,CB=<0,-1,5/2>
#CA×CB=<-5/2,-25/6,-5/3>
#THE EQUATION OF THE PLANE WITH POINT (1,1,0) (WHICH IS CALCULATED BY THE EQUATION OF THE LINE) IS -5/2*1-25/6*1-5/3*0=-25/6
#THE EQUATION OF THE PLANES WHOSE INTERSECTION WITH THE XZ-PLANE IS THE LINE WITH EQUATION 3x+2z=5 AND A INTERSECTION WITH THE Y AXIS ON (0,n,0) IS
#-5/2nx-25/6y-5/3nz=-25/6
#UNLESS WHEN THE PLANE HAS NO INTERSECTION WITH Y-AXIS, IN THIS CASE THE EQUATION IS 3x+2z+0y=5 

13.1  :

5.
#THE EQUATION OF THE LINE: (3,-5,7)+t<3,0,1>=<3+3t,-5,7+t>
#THE PARAMETRIZATION IS:x(t)=3+3t, y(t)=-5, z(t)=7+t

17.
#CENTER=(0,0,0) RADIUS=9 PLANE:z=0

13.2  :

3.
#THE RESULT=<e^(2t)2*i+ 1/(t+1)*j+4k>

5.
#r'(t)=<(-t)^-2,cost,0>

7.
#r'(t)=<1,2t,3t^2>

15.
#ON THE HANDWRITTEN PDF VERSION

31.
#r'(2)=<-4, 5, 24>
#THE PARAMETRIZATION OF THE TAGENT LINE IS x=-4, y=5, z=24

33.
#r'(2)=<-1, 0, 1/2>
#THE PARAMETRIZATION OF THE TAGENT LINE IS x=-1, y=0, z=1/2

41.
#-2->2(u^3*i+u^5*j)du=0

49.
#r(t)=t^3/3*i+(5t^2)/2*j+t*k+C
#r(1)=i+5/2j+k+C=j+2k
#C=-i-3/2j+k
#r(t)=i*t+5/2j*t+k*t-i-3/2j+k