Question

plane: 2x+3y-6z=8
→
OS= 3i + 5j − 6k
The point N is the foot of the perpendicular from S to this plane. Find the position vector of N
and show that the length of SN is 7.

Answer 1

I think, SN is one of normal vector of the plane. And we know that the normal vector of the plane is \[\vec n\]= <2,3,-6>

Answer 2

and let N <a,b,c> is the foot of SN . we have \[\vec SN\]=<a-3,b-5,c+6>
so, a-3 =2---> a=5
b-5=3---> b=8
c=6= -6 ---> c=0
therefore N = < 5,8,0>
what do you think?

Answer 3

and then the norm of \[\vec n\] =sqrt (2^2 + 3^2 + (-6)^2 = sqrt (49) =7

Answer 4

hey, I am working and you disappear? is it fair?

Answer 5

i am there

Answer 6

got it?

Answer 7

c would be -12
then the dist SN sqrt{(5-3)^2+(8-5)^2+(-12+6)^2}

Answer 8

yes , you are right, my mistake, good guy.

Answer 9

sorry for my bad.