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).

 

This is what I got...

 

AB

AB=OA-OB

=(1,-2,2)-(1,-2,0)

=(0,0,2)

 

AC=OA-OC

=(0,3,2)-(1,-2,0)

=(-1,5,2)

 

AB x AC

 

(0)(2)-(5)(2)

=0-10

=-10

 

(2)(-1)-(2)(0)

=-2-0

=-2

 

(0)(5)-(-1)(0)

=0

 

The normal vector is (-10,-2,0).

 

-10x-2y+0z+d=0

-10(1)-2(-2)+0(0)+d=0

-10+4+d=0

-6+d=0

d=6

 

The scalar equation is -10x-2y+6=0

 

The vector equation is (1,-2,0) + s(0,0,2) + t(-1,5,2)

 

Parametric Equations:

x=1-t

y=-2+5t

z=2s+2t

Did I do this right?

Thanks for the help in advance.

Edited April 11, 201510 yr by pravin19

The scalar equation is -10x-2y+6=0

 

Is the coefficient of z truly zero?

Yeah, are my answers right?

Thar's a bit abrupt, I was wondering if you realised the implications of that statement.

 

However, according to WolframAlpha yes that is the correct plane.

 

To check, input the determinant as follows

 

matrix {{x,y,z,1},{1,-2,0,1},{1,-2,2,1},{0,3,2,1}}