a Determine the vector and parametric equations of the plane

a) Determine the vector and parametric equations of the plane that passes through the points Q(-3/2, 0, 0), R(0, -1, 0), and S(0, 0, 3).

b) Determine if the point P(1, 5, 6) is a point on this plane.

a) Q(-3/2, 0, 0), R(0, -1, 0), and S(0, 0, 3)

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

( xo, yo, zo ) is a point on plane.

( a. , b , c ) vector perpendicula to plane

V1 = QR = < 3/2 , -1 , 0 > and

V2 = QS = < 0 , 1 , 3>

vector product of V1 x V2 = < -3 , -9/2 , 3/2 >

So, ( a. , b , c ) = < -3 , -9/2 , 3/2 >

-3(x -xo) + (9/2)(y - yo) +(3/2)(z -zo) =0

use any of the three point for xo,yo, zo,

So, I use R : -3( x -0) - 9(y +1)/2 + 3( z-0)/2 =0

-3x -9y/2 -9/2 +3z/2 =0

-3x -9y/2 +3z/2 = 9/2

-6x -9y +6z = 9

simplify : -2x -3y +2z =3 ( cartesian Equation of plane)

vector form : we can write z = (1/2)( 3 +2x +3y)

So, ( x , y , z) = ( 0, 0, 3) + x( 1, 0 , 2) +y( 0, 1 , 3)

( x , y , z) = ( 0,0 , 3) + s( 1, 0 , 2) +t( 0, 1 , 3) vector form

b) Substitute point ( 1 , 5 , 6) to check if it lies on plane:

-2x -3y +2z =3

-2 -15 +12 =3

-5 =3

Point does not satisfy the equation of plane so, it does not lie on plane