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
you will need two direction vectors on the plane
how about vectors RS and RQ ?
vector RS = (0, 1, 3)
vector RQ = (-3/2, 0, -3) or more civil-like : (1, 0, 2)
vector equation:
(x,y,z) = (0, -1, 0) + s(0, 1, 3) + t(1, 0 , 2)
parametric form:
x = 0 + t
y = -1 + s
z = 3s + 2t
b) is (1,5,6) on it?
1 = 0+t , t = 1
5 = -1 + s, s = 6
then 6 = 3(6) + 2(1), which is false, so (1,5,6) is NOT on the plane
other method:
it is easy to find the cross-product of our two direction vectors to be (2,3,-1)
so the plane equation is 2x + 3y - z = c
but (0,0,3) lies on it, so 0 + 0 - 3 = c
plane equation is 2x + 3y - z = -3
plug (1,5,6) into it ...
LS = 2(1) + 3(5) - 6 ≠ -3
so the point is NOT on the plane
To determine the vector equation of the plane, we need to find two non-parallel vectors that lie in the plane. We can achieve this by taking the difference between the given points.
Vector QR = R - Q = (0, -1, 0) - (-3/2, 0, 0) = (3/2, -1, 0)
Vector QS = S - Q = (0, 0, 3) - (-3/2, 0, 0) = (3/2, 0, 3)
Now, we can use these two vectors to find the cross-product, which will give us the normal vector of the plane.
Normal vector N = QR x QS
= (3/2, -1, 0) x (3/2, 0, 3)
To calculate the cross product, we can use the formula:
(i) (j) (k)
(3/2) (-1) (0)
(3/2) (0) (3)
N = [(-1)(3)(3/2), (3/2)(3)(3/2) - (0)(0), (3/2)(0) - (-1)(3/2)]
N = [-9/2, 27/4, 3/2]
Therefore, the normal vector of the plane is N = (-9/2, 27/4, 3/2).
Next, we can write the vector equation of the plane using the point Q(-3/2, 0, 0) and the normal vector N:
(X, Y, Z) = Q + tN
where (X, Y, Z) represents any point on the plane, Q is the given point (-3/2, 0, 0), t is a parameter, and N is the normal vector (-9/2, 27/4, 3/2).
So, the vector equation of the plane is:
(X, Y, Z) = (-3/2, 0, 0) + t(-9/2, 27/4, 3/2)
To find the parametric equations of the plane, we can express each component (X, Y, Z) as a function of another parameter u and v:
X = -3/2 - (9/2)t
Y = (27/4)t
Z = (3/2)t
Finally, to determine whether the point P(1, 5, 6) lies on this plane, we substitute the coordinates of P into the parametric equations of the plane:
1 = -3/2 - (9/2)t
5 = (27/4)t
6 = (3/2)t
Solving these equations simultaneously, we can determine the value of t. If a solution exists, then the point P lies on the plane.
To determine the vector equation of a plane, we need a normal vector and a point on the plane.
To find the normal vector, we can use the cross product of two vectors formed by the given points.
Let's start by finding two vectors on the plane:
- V1 = QR = R - Q = (0, -1, 0) - (-3/2, 0, 0) = (3/2, -1, 0)
- V2 = QS = S - Q = (0, 0, 3) - (-3/2, 0, 0) = (3/2, 0, 3)
Now, we can find the normal vector N by taking the cross product of V1 and V2:
N = V1 x V2 = (3/2, -1, 0) x (3/2, 0, 3)
To find the cross product, we can use the determinant method:
N = ((-1)(3), (0)(3/2) - (0)(0), (3/2)(0) - (-1)(3/2))
= (-3, 0 - 0, 0 - (-3/2))
= (-3, 0, 3/2)
So, the normal vector of the plane is N = (-3, 0, 3/2).
Now, let's find the point P on the plane. We can choose any of the given points Q, R, or S since they all lie on the plane. Let's choose Q(-3/2, 0, 0).
Therefore, the vector equation of the plane is:
P = Q + aN
= Q + a(-3, 0, 3/2) (where 'a' is a scalar parameter)
To find the parametric equations of the plane, we can write the vector equation in terms of x, y, and z:
x = -3/2 - 3a
y = 0
z = 3/2 + (3/2)a
b) To determine if the point P(1, 5, 6) is on this plane, we substitute the values of x, y, and z into the parametric equations:
1 = -3/2 - 3a
5 = 0
6 = 3/2 + (3/2)a
From the second equation, we can see that y ≠ 0, so P(1, 5, 6) is not a point on the plane determined by the vectors Q, R, and S.