N where X = [x,y,z] and A = [1,1,2] as given,gives cartesian eqn of plane:0x + 21y +0z = 1(0) + 1(21)+2(0) =21so 21y = 21 or y =1 is eqn of plane,......
we then cross-multiply to get N the normal to the plane, N
N = AB x AC (vector product = vector) = [0, 21, 0] I assume you know how to do this!
since A is on plane then (X -A).N where X = [x,y,z] and A = [1,1,2] as given,
gives cartesian eqn of plane:
0x + 21y +0z = 1(0) + 1(21)+2(0) =21
so 21y = 21 or
y =1 is eqn of plane,
this is a plane that lies parallel to (x,z) -plane but through the ordinate y =1!
This is easy as you already got the perpendicular vector.
The equation of plane is:
r dot n = d
where r is just simply r (leave it as it is: to get the first vector form)
n is A which is perpendicular to the line on the plane.
and d is the dot product of A to either B or C. You can use any of the two.
thus
r dot A = A dot B
OR
r dot A = A dot C
To get the second form:
you can turn r into (x, y, z) to get its cartesian form. Just do (x, y, z) dot (A) = A dot B (or C) whichever you used on the first form.
The explanation is messy, but I hope that helps.
