easy Vector algebra

[vorcil]

p = (2,6,3)
q = (1,0,1)
r = (4,1,-1)

part a)
Find a parametric equation of the line that passes through p with direction q
x=p + t(q)
x=(2,6,3) + t(1,0,1)

part b)
find the point X on the line in part a so that RX is perpendicular to the line

I'm having trouble doing this one
please help

[Office_Shredder]

What way can you test to see if two vectors are perpendicular?

[vorcil]

What way can you test to see if two vectors are perpendicular?

if a.b = 0
or a=0
or b=0

-

It's asking me to find the point

p = (2,6,3)
q = (1,0,1)
r = (4,1,-1)

pq=p + t(q)
pq=(2,6,3) + t(1,0,1)

part b)
find the point X on the line in part a so that RX is perpendicular to the line

can you please give some more direction?

I already know the answer i just need to know how to get to it!!
(1,6,2) is the point

[vorcil]

cmon PF don't fail me now!

[HallsofIvy]

Let (x, y, z) be the point X. Then a vector from r to X is (x- 4, y- 1, z+ 1). That is the vector that must be perpendicular to (1, 0, 1) and so must have dot product with it equal to 0.

That will give you one equation for x, y, and z. Since X is "on the line in part a" it must also satisfy the equation of that line.

[vorcil]

(x-4,y-1,z+1).(1,0,1)=0
(2+t , 6 , 3+t)

(x-4).1 + (y-1).0 + (z+1).1 = 0
(2+t-4).1 + 6-1.0 + ((3+t) + 1).1 = 0
(2+t-4) + (3+t)+1 = 0
t= -1

into original equation,
2+(-1) , 6 , 3+(-1) = (1,6,2)
OH SHI-

XD =] :)