Hey jakobjakob.
The first one is spot on.
I'll take a look at the second one in a sec.
I was hoping if someone could confirm if those 2 exercises are solved correctly.
1. we have a plane: (5x-y+z-2=0) and vector line in form: ((t)=x/-2=(y+1)/-3=(z-2)/1). What is the intersection between the two of them?
What i got is:
x=-2t
y=-3t-1
z=t+2
5(-2t)-(-3t-1)+t+2-2=0 t=1/6
x=-1/3
y=-3/2
z=13/6
2.find the perpendicular projection T' of point T(2,-1,5) on vector line : (x-2)/3=(1-z)/2 , y=-3. Also find a point what is a reflection of point T based on the projection T'.
What i did is:
s (3,0,-2) = n ( normal of plane that contains point T)
x=3t + 2
y=-3
z=1-2t
plane: 3x + 0y -2z = d ---> 3*2 -2*5=-4=d
3x-2z+4=0
3(3t+2)- 2(1-2t) +4 =0 ---> t = -8/13
T' (intersection between plane and vector line):
x= 3t+2=2/13
y=-3
z=1-2t=29/13
to get the point T'' i just calculated the difference between x,y,z coordinates of point T and T' and did the same for coordinates of T' and T'' so i got :
T''
x= 2/13 - 24/13 = -22/13
y= -3 -2 = -5
z = 29/13 - 36/13 = -7/13
so i got T'' (-24/13,-5,,-7/13)
Could this be correct? thank you