3D geometry question ( Vectors)

[aaazureee]

I recently got this HW and there is only 1 part i cant do and i want to seek some help from u
We are given 3 points A(2,0,1), B(1,-1,2), C(3,-3,0)
a) Find EQN of plane ABC in form r.n = p
So this q is quite straightforward and i got this result r(1,0,1) = 3
b) Given that line L1 is the perpendicular bisector of line segment BC on the plane ABC, explain why the vector EQN of L1 can be expressed as
r= (2,-2,1) + t(1,2,-1), t= real parameter
So i can only find K, the mid point of BC which this line passes which is (2,-2,1) but i cant figure the way to find the directional vector of this line? Am i missing some important information related to this line and the plane?

 

[Ishuda]

I recently got this HW and there is only 1 part i cant do and i want to seek some help from u
We are given 3 points A(2,0,1), B(1,-1,2), C(3,-3,0)
a) Find EQN of plane ABC in form r.n = p
So this q is quite straightforward and i got this result r(1,0,1) = 3
b) Given that line L1 is the perpendicular bisector of line segment BC on the plane ABC, explain why the vector EQN of L1 can be expressed as
r= (2,-2,1) + t(1,2,-1), t= real parameter
So i can only find K, the mid point of BC which this line passes which is (2,-2,1) but i cant figure the way to find the directional vector of this line? Am i missing some important information related to this line and the plane?

In general, the normal vector of the plane
a x + d y + c z + d = 0
is (a, b, c). Note that the scalar form above can be written in vector form as
(a, b, c)\(\displaystyle \cdotp\)(x, y, z) = -d

 

[aaazureee]

Yeah i know that already its basically switched between cartesian form and vector form. But i am asking the part of the directional vector of the per. bisector of BC. How is this helping me