Question

how do you write the scalar equation of the plane:
[x,y,z] = [2,3,5] + s[-1,3,4] + t[6,1,-2] Thanks.

Answer 1

is it supposed to be [23,5,0] ?

Answer 2

The answer's supposed to be 10x-22y+19z-49=0

Answer 3

oops wait one sec the point is supposed to be [2,3, 5]

Answer 4

cross [-1,3,4] and [6,1,-2] to get the normal vector.
and the plane passes through [2,3, 5]

equation of a plane is
a(x- xo) + b(y - yo) + c(z - zo) = 0

where (a,b,c) is the normal vector and (xo, yo, zo) = [2,3, 5]

Answer 5

Oh it works! Thanks a lot! It really helped :)

Answer 6

no prob