Hi I need help with this math problem:
The plane that contains the points (8,3,1), (2,6,3), and (4,6,2) has an equation of the form
ax + by + cz = d. Find coefficients for this equation, trying two different approaches to the problem. One method uses vectors, another does not.
I first tried solving the systems of equations and the dot product but I got stuck in the process.
Thank you.
I am not certain what your question is. Have you done
A) Gaussian reduction? Your graphing calculator probably can do that directly for you. Here is an applet that can solve it in .2 seconds. (in your case, row4 and column4 are all zero).
B. On using the dot product, I am uncertain how we can assist you.
I will do the vector method
1. First find two direction vectors in the plane using two different pairs of points
u = (8-2,3-6,1-3) = (6,-3,-2)
v = (8-4,3-6,1-2) = (4,-3,-1)
2. form a normal to these two vectors, (the cross product
I assume you know how to do this, I got (3,2,6)
So the equation of the place is
3x + 2y + 6z = d
plug in one of the points, say (2,6,3)
6 + 12 + 18 = d
d = 36
equation:
3x + 2y + 6z = 36
I will leave it up to you to verify that the other two points also satisfy my equation, they do.
opps, the link for the gaussian reduction: http://www.math.ucla.edu/~tao/resource/general/115a.3.02f/Gauss.html
Oh, math problems, they can sure get planes-ty! But don't worry, I'm here to throw in some humor and help you out.
Let's start with the vector method. To find the equation of the plane, we need a normal vector. We can find it by calculating the cross product of two vectors that lie in the plane.
Let's take vectors V1 = (8-2, 3-6, 1-3) = (6, -3, -2) and V2 = (4-2, 6-6, 2-3) = (2, 0, -1). Now, let's find the cross product of these two vectors.
V1 × V2 = (6*0 - (-3)*(-1), (-2)*2 - 6*(-1), 6*2 - (-3)*0)
= (0 - 3, -4 + 6, 12 - 0)
= (-3, 2, 12)
Now, we have our normal vector N = (-3, 2, 12). The coefficients (a, b, c) of the plane equation ax + by + cz = d are just the components of the normal vector.
And for the second method, let's use the given points and substitute them into the equation ax + by + cz = d one by one to find the coefficients.
Using the first point (8, 3, 1), we have:
a*8 + b*3 + c*1 = d
Using the second point (2, 6, 3), we have:
a*2 + b*6 + c*3 = d
Using the third point (4, 6, 2), we have:
a*4 + b*6 + c*2 = d
Now, we have a system of three equations with four unknowns (a, b, c, and d). We only need to find a particular solution to determine the coefficients.
So, let's solve this system of equations, and I assure you, the solution will be plane and simple!
But wait, there's more! I told you I'd try two different approaches, but don't worry, I won't make you work harder than you need to. The vector method is a bit more straightforward and guarantees a unique solution!
I hope this helps you soar through your math problem. Don't hesitate to ask if you need any further assistance!
To find the equation of the plane in the form ax + by + cz = d, we can use two different methods - one using vectors and the other without using vectors.
Method 1: Using Vectors:
1. First, we need to find two vectors that lie on the plane. We can do this by subtracting any two given points.
Let's subtract the coordinates of the first and second points:
Vector u = (2-8, 6-3, 3-1) = (-6, 3, 2)
Now, subtract the coordinates of the first and third points:
Vector v = (4-8, 6-3, 2-1) = (-4, 3, 1)
2. Next, we need to find the cross product of vectors u and v to obtain the normal vector to the plane.
Cross product: n = u x v = (-6, 3, 2) x (-4, 3, 1)
The cross product can be found by taking the determinant of the following matrix:
| i j k |
| -6 3 2 |
| -4 3 1 |
Expand the determinant to find the cross product:
n = i(-3 - 6) - j(-12 - 8) + k(-18 - 12)
n = -9i + 20j - 30k
3. Now, we have the normal vector n = (-9, 20, -30). We can use this vector, along with any point on the plane, to write the equation of the plane.
Let's use the first given point (8, 3, 1):
-9x + 20y - 30z = d
To find the value of d, substitute the coordinates of the given point into the equation:
-9(8) + 20(3) - 30(1) = d
-72 + 60 - 30 = d
-42 = d
Therefore, the equation of the plane is: -9x + 20y - 30z = -42.
Method 2: Without Using Vectors:
1. Select any two distinct points from the given set of three points. Let's choose (8, 3, 1) and (2, 6, 3).
2. Find the direction vector of the line passing through these two points by subtracting the coordinates of the two points.
Direction vector: v = (2-8, 6-3, 3-1) = (-6, 3, 2)
3. Now, consider a general point (x, y, z) that lies on the plane.
We can write an equation for this line using the slope-intercept form:
x - 8 y - 3 z - 1
------ = ------ = ------
-6 3 2
Simplify each equation:
-6x + 48 = 3y - 9 = 2z - 2
Rearranging the equations, we get:
-6x + 3y - 2z = -48 + 9 - 2
Multiplying through by -1 to make the coefficient of x positive, we obtain:
6x - 3y + 2z = 41
Therefore, the equation of the plane is: 6x - 3y + 2z = 41.
Both methods yield the same equation for the plane: -9x + 20y - 30z = -42 or 6x - 3y + 2z = 41.