How do you determine scalar, vector, and parametric equations for the plane that passes through the points A(1,-2,0), B(1,-2,2), and C(0,3,2)?

We seek scalar, vector, and parametric equations for the plane that passes through the coordinates:

# A(1,-2,0)#, #B(1,-2,2)#, and #C(0,3,2)#

Firstly, we have:

# bb(vec(AB)) = bb(vec(OB)) - bb(vec(OA)) = ( (1), (-2),(2) ) - ( (1), (-2),(0) ) = ( (0), (0),(2) ) #

# bb(vec(AC)) = bb(vec(OC)) - bb(vec(OA)) = ( (0), (3),(2) ) - ( (1), (-2),(0) ) = ( (-1), (5),(2) ) #

We can gain, a vector that is perpendicular to these vectors by taking their cross product. Thus we can gain a normal vector, #bb(ul n)# using:

# bb(ul n) = bb(vec(AB)) bb(xx) bb(vec(AC)) #
# \ \ \ = ( (0), (0),(2) ) bb(xx) ( (-1), (5),(2) )#

# \ \ \ = | ( bb(ul hat i), bb(ul hat j), bb(ul hat k)), (0,0,2),(-1,5,2) |#

# \ \ \ = | (0,2),(5,2) | bb(ul hat i) - | (0,2),(-1,2) | bb(ul hat j) +| (0,0),(-1,5) | bb(ul hat k) #

# \ \ \ = (0-10) bb(ul hat i) - (0+2) bb(ul hat j) + (0-0) bb(ul hat k) #

# \ \ \ = -10 bb(ul hat i) - 2 bb(ul hat j) #

Having gained the normal vector we can now form the vector equation of the plane using the vector form # bb(ul r) * bb(ul n) = bb(ul a) * bb(ul n)#, thus arbitrary using #bb(ul a)=C#, we have:

# \ \ \ \ \ bb(ul r) * ( (-10), (-2),(0) ) = ( (0), (3),(2) ) * ( (-10), (-2),(0) )#

# :. bb(ul r) * ( (-10), (-2),(0) ) = -6#, or, # ( (x), (y),(z) ) * ( (-10), (-2),(0) ) = -6#

Then if we evaluate the scalar product, we obtain the cartesian equation:

# (x)(-10) + (y)(-2) + (z)(0) = -6 #

# :. -10x -2y + 0 = -6 #

# :. 5x +y = 3 #

And we can parameterize by setting #x=t# leading to:

# 5t+y=3 => y = 3-5t #

Thus the parametric equations are

# { (x=t), (y=3-5t), (z=0) :} #

And, we can verify these results graphically: