An equation of the plane containing the points (-8, 3, 3), (-18, 5, 1), (-8, -1, -3) is ?

Question

possible equations to choose from ....



a. 4x + 2y - 3z - 5 = 0

b. 2x - y + 4z - 12 = 0

c. x + 3y - 2z + 5 = 0

d. 2x + 3y + 2z - 2 = 0

Or None Of These?

Answer 1

C is the answer. Plug and chug, That equation works for all three points. To find it out in the other direction, you can use 2 methods. 1 method is good for answer checking, 1 method is good for showing work. The way to find it out the other direction is:

we have (-8,3,3), (-18,5,1), and (-8,-1,-3) and we try to find solution so the equation =

(0) ax + by + cz = d

or
(1) -8a + 3b +3c = d
(2) -18a + 5b + c = d
(3) -8a - 1b - 3c = d

there are many ways to go. Its a system of equations and can be solved any way you want to reduce the variables a, b, or c, into functions of d. Here is 1 example i used to verify my calculator:

Subtract Equations. (3) - (1)
(4) b = (-3/2)*c

Subtract Equations. 8*(2) - 18* (3) [ to knock out a in the equation and to get a second equation with only b,c,d]
(5) 29b+31c =-5d

put (4) into (5) to solve for c as a function of d
(6) c = (2/5)d

put (6) into (4) to solve for b as a function of d
(7) b = (-3/5)d

put (7) and (6) into (1) and solve for a as a function of d
(8) a = (-1/5)d

Now plug (6), (7), and (8) into (0). So now you have
(-1/5)d*x+(-3/5)d*y+(2/5)d*z = d

Multiply both sides by (-d/5) and you get
x+3y-2z = -5

Move the -5 to the other side

x +3y -2z +5 = 0

The other way to do it is to do equations (4) through (8) on a calculator using matrices

[[a][b][c]] = [[-8,3,3][-18,5,1][-8,-1,-3]]^ (-1)*[[1][1][1]] * [d]

Doing that gives you
[[a][b][c]] = [[-d/5][-3d/5][2d/5]]
Which is what equations (6) (7) and (8) say

Answer 2

Find the equation of the plane containing the points
P(-8, 3, 3), Q(-18, 5, 1), R(-8, -1, -3).

Define two vectors, u and v, that lie in the plane.

u = PQ = Q - P = <-18+8, 5-3, 1-3> = <-10, 2, -2>
v = PR = R - P = <-8+8, -1-3, -3-3> = <0, -4, -6>

The normal vector of the plane n, is the cross product of the two vectors u and v.

n = u X v = <-10, 2, -2> X <0, -4, -6> = <-20, -60, 40>

Any non-zero multiple of n is also a normal vector to the plane. Divide by -20.

n = <1, 3, -2>

With the normal vector and one point we can write the equation of the plane. Let's choose P(-8, 3, 3).

1(x + 8) + 3(y - 3) - 2(z - 3) = 0
x + 8 + 3y - 9 - 2z + 6 = 0
x + 3y - 2z + 5 = 0

The answer is c.

Answer 3

Label the factors, then discover the gap between them. as quickly as you're carried out with that discover the conventional vector, this allow you to be responsive to the coefficients for the airplane equation. P(0,a million,a million), Q(4,-5,8), R(3,5,-a million) PQ = <4, -6, 7>, PR = <3,4,-2> PQ x PR = N Determinant i.....j....ok 4...-6...7 3...4...-2 N = <-sixteen, 29, 34> -sixteen(x - 0) + 29(y - a million) + 34(z - a million) = 0 -16x + 29y + 34z = sixty 3 (airplane Equation)

Answer 4

The answer is C, just plug the numbers in and see if the equation holds true.