determine an equation for each plane
a) contains the point A(6,2,3), B(5,1,-3), and C(5,7,-2)
b) parallel to 2x + 6y + 4z = 1 and contains the point P(3,2,1)
vector AB = (-1,-1,-6)
vector AC = (-1,5,-5)
a normal to these two vectors is (35, 1, -6) , I assume you know how to take a cross-product
the plane has equation
35x+ y - 6z = c
plug in a point, e.g. A(6,2,3)
210 + 2 - 18 = c
c =194
plane equation: 35x + y - 6z = 194
check by subbing in points B and C, (they work)
b) Since they are parallel, they must differ only in the constant
so new equation: 2x + 6y + 4z = c
plug in your point to find c
(a)
1. Find vectors AB and AC.
2. Calculate the cross product of ABxAC, which gives the normal vector to the required plane.
3. Form the equation of the plane. The components of the normal vector correspond to the coefficients of x, y and z of the equation of the plane.
4. Find the constant term of the plane by passing the plane through one of the three points, i.e. substitute (xa,ya,za) into the equation and find the constant term.
5. Optionally, check that the other points pass through the equation of the plane.
(b)
Parallel to 2x + 6y + 4z = 1 means that the normal is <2,6,4>.
Take the left-hand side of the equation and evaluate with the point P(3,2,1), i.e. pass a plane parallel to 2x + 6y + 4z = 1 through the given point:
2(3)+6(2)+4(1)=6+12+4=16
So the required equation of the plane is
2x + 6y + 4z = 16
a) To determine the equation of a plane, we need the position vectors of three non-collinear points lying on the plane. For this scenario, we have points A(6,2,3), B(5,1,-3), and C(5,7,-2).
Step 1: Find two vectors lying on the plane.
Vector AB = B - A = (5-6, 1-2, -3-3) = (-1, -1, -6)
Vector AC = C - A = (5-6, 7-2, -2-3) = (-1, 5, -5)
Step 2: Use the two vectors to find the normal vector to the plane.
The normal vector, N, can be obtained by taking the cross product of AB and AC:
N = AB x AC = (-1, -1, -6) x (-1, 5, -5)
= (31, 1, 4)
Step 3: Plug the normal vector and a point (e.g., A) into the equation of a plane.
The equation of a plane is given by:
Ax + By + Cz = D
Substituting A=31, B=1, C=4, and x=6, y=2, and z=3 into the equation, we have:
(31)(6) + (1)(2) + (4)(3) = D
186 + 2 + 12 = D
D = 200
Therefore, the equation of the plane is:
31x + y + 4z = 200
b) To determine the equation of a plane parallel to the given equation 2x + 6y + 4z = 1 and contains the point P(3,2,1), we need to find a vector that is perpendicular to the normal vector of the given plane and passes through point P.
Step 1: Rewrite the given equation in the form Ax + By + Cz = D.
2x + 6y + 4z = 1 can be rewritten as:
2x + 6y + 4z - 1 = 0
Step 2: Identify the normal vector of the given plane.
The normal vector of the given plane is (2, 6, 4).
Step 3: Find a vector that is perpendicular to the normal vector.
One way to find a vector perpendicular to (2, 6, 4) is to take the cross product of (2, 6, 4) with any other vector not parallel to it. Let's choose the vector (1, 0, 0).
(2, 6, 4) x (1, 0, 0) = (0, -4, 6)
Step 4: Use the perpendicular vector and the point P(3,2,1) to find the equation of the desired plane.
Substituting the perpendicular vector (0, -4, 6) and the point P(3,2,1) into the equation of a plane, we have:
0(x - 3) - 4(y - 2) + 6(z - 1) = 0
-4y + 8 - 6z + 6 = 0
-4y - 6z + 14 = 0
Therefore, the equation of the plane parallel to 2x + 6y + 4z = 1 and contains the point P(3,2,1) is:
-4y - 6z + 14 = 0.