First find a vector perpendicular to both P1=i+j(1,1,0) and P2=j+k (0,1,1) by taking the cross product of P1 and P2:
|i j k|
|1 1 0|
|0 1 1|
By evaluating the above determinant which represents the cross product of P1 and P2, we get
P3 = i(1)-j(1)+k(1) = i-j+k =(1,-1,1)
Take any point P(x,y,z) on the plane perpendicular to P3 and passing through P0, the vector P-P0 should be normal to P3, so we have
P-P0 = (x,y,z)-(4,3,0)
P3 = (1,-1,1)
For perpendicularity:
P3.(P3-P0)=0 (dot product)
(1,-1,1).(P-P0) = 0
or
P.(1,-1,1) = P0.(1,-1,1)
using properties of distributivity and commutativity.
(x,y,z).(1,-1,1) = (4,3,0).(1,-1,1)
=>
x-y+z = 1
The required plane is therefore
x -y +z -1 = 0
Checks:
4 -3 +0 -1 =0 => P0 is on the plane
(1,1,0).(1,-1,0)=0 => perp. to i+j
(0,1,1).(1,-1,0)=0 => perp. to j+k
Find parametric equations and symmetric equations for the line through P0 and perpendicular to both given vectors. (P0 corresponds to t = 0.)
P0 = (4, 3, 0)
i + j and j + k
Not exactly sure what to do, how do I find what is perpendicular to both vectors?
1 answer
1 answer