Let #vecv = # the vector from #(2, 0, 1)# to #(0,4,-3)#:
#vecv = (0-2)hati+(4-0)hatj+(-3-1)hatk#
#vecv = -2hati+4hatj-4hatk#
Let #vecu = # the vector from #(2, 0, 1)# to #(-2, 5, 0)#:
#vecu = (-2-2)hati+(5-0)hatj+(0-1)hatk#
#vecu = -4hati+5hatj-1hatk#
Compute the dot-product by multiplying the respective components:
#vecv*vecu = (-2)(-4) + (4)(5)+(-4)(-1)#
#vecv*vecu = 32#
Compute the magnitudes:
#|vecv| = sqrt((-2)^2+4^2+ (-4)^2)#
#|vecv| = 6#
#|vecu| = sqrt((-4)^2+5^2+(-1)^2)#
#|vecu| = sqrt(42)#
Use the equation, #vecv*vecu = |vecv||vecu|cos(theta)#, to find the cosine of the angle between the two vectors:
#32= 6sqrt42cos(theta)#
#cos(theta) = 32/(6sqrt42)#
If the points were co-linear, then the cosine of the angle between their vectors would have been #+-1#; it is not, therefore, the points are not co-linear.
Use the cross-product to find a vector that is perpendicular to both vectors. I use a 5 column determinant to compute the cross-product.
First the major diagonals:
#|(color(red)(hati),color(green)(hatj),color(blue)(hatk),hati,hatj),
(-2,color(red)(4),color(green)(-4),color(blue)(-2),4),
(-4,5,color(red)(-1),color(green)(-4),color(blue)(5))
| = #
#color(red)((4)(-1)hati)+color(green)((-4)(-4)hatj) + color(blue)((-2)(5)hatk)#
Subtract from that the minor diagonals:
#|(hati,hatj,color(blue)(hatk),color(red)(hati),color(green)(hatj)),
(-2,color(blue)(4),color(red)(-4),color(green)(-2),4),
(color(blue)(-4),color(red)(5),color(green)(-1),-4,5)
| = #
#color(red)({(4)(-1)-(-4)(5)}hati)+color(green)({(-4)(-4)-(-2)(-1)}hatj) + color(blue)({(-2)(5)-(4)(-4)}hatk)=#
#16hati+14hatj+6hatk#
Divide the vector by 2:
#8hati+7hatj+3hatk#
These coefficients match the coefficients for #x, y, and z#, respectively in the scalar equation for a plane:
#8x+7y+3z = c#
To determine the value of c, substitute in one of the points. I shall use the first one:
#8(2)+7(0)+3(1) = c#
#c = 19#
The scalar equation of the plane that contains all 3 points is:
#8x+7y+3z = 19#
