Answer:
For a part:
(x-1, y, z-3) * (-6, -3, -2) = 0 (Normal equation)
P = (1,0,3) + (1,-2,0)*s + (0,-2,3)*t (Parametric equation)
For b part:
(x-1, y, z-3) * (1, 1, 1) = 0 (Normal equation)
P = (4, 0, 0) + (-1, 1, 0)*s + (-1, 0, 1)*t (Parametric equation)
Step-by-step explanation:
a)
The parametric equation of a plane is defined by three things: a point, and two non-colinear vectors in the plane. We already have three points, so we only need to find two vectors contained in the plane (we will call them V1 and V2):
V1 = P-Q = (1,0,3)-(0,2,3) = (1,-2,0)
V2 = P-R = (1,0,3)-(1,2,0) = (0,-2,3)
Therefore, the parametric equation of our plane is given by:
P = P + V1*s + V2*t = (1,0,3) + (1,-2,0)*s + (0,-2,3)*t
The normal equation of any plane is given by [x-P]*n = 0, where x is the vector(x,y,z), P is a point contained in the plane, n is a vector normal to the plane and * stands for the dot product. Therefore, for finding the normal form of the equation, we need an orthogonal vector to the plane (n), which we find by doing the cross product of our previous vectors V1 and V2:
n = V1xV2 = (-6,-3,-2)
Substituting the required values in the formula mentioned above, we can write the normal equation of our plane as:
[(x,y,z) - (1,0,3)]*(-6,-3,-2) = 0 or (x-1, y, z-3) * (-6, -3, -2) = 0
b)
In this second exercise, it is straightforward to give the normal equation:
[(x,y,z) - (1,0,3)]*(1,1,1) = 0 or (x-1, y, z-3) * (1, 1, 1) = 0
For transforming this equation to its parametric form, first we transform it to its cartesian form:
x-1 +y +z-3 = 0, then : x+y+z = 4
Now that we have the cartesian form, we solve for variable x and get:
x = 4 -y -z
Then, we know that every point in the plane can be expressed as:
P = (4 -y -z, y, z).
Finally, rewriting the expression and converting y and z to the parameters s and t respectively we get:
P = (4, 0, 0) + (-y, y, 0) + (-z, 0, z) = (4, 0, 0) + (-1, 1, 0)*s + (-1, 0, 1)*t
