Finding a plane given 3 points

In summary, the conversation discusses finding a plane through three given points and involves calculating the cross product and determinant. There is also a discussion about a possible mistake in the calculation and a correction is provided. The conversation then continues with a similar problem and the correct solution is given.
  • #1
the7joker7
113
0

Homework Statement



Find a plane through the points (6,-7,2), (-4,7,-4), (4,8,-2)

The Attempt at a Solution



A = <-10, 14, -6>

B = <-2, 15, -4>

i j k
-10 14 -6
-2 15 -4

i(-146) - j(28) + k(-178)

-146(x - 6) - 28(y + 7) - 178(z + 2)

-146x - 28y - 178z = 324

Actual answer: `34*x -28*y -122*z = 156`

ugh
 
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  • #2
u evaluated the determinant of the cross product matrix incorrectly, it should be 34i-28y-122z, did u use cramer's rule?
 
  • #3
Okay, here's my work for a similar problem with the same concept. Does this look right?

P = (-8, -3, 7)

Q = (-8, 7, 4)

R = (-2, -4, -8)

PQ = (0, 10, -3)

PR = (6, -1, -15)

i j k
0 10 -3
6 -1 -15

i(-150 + 3) - j(-18) + k(60) = 0

147(x + 8) + 18(y - 3) + 60(z - 7) = 0

147x + 18y + 60z = -1176 + 54 + 420 = -702
 
  • #4
Ugh! It's wrong...what am I doing wrong?!
 
  • #5
You have the right idea, but perhaps you are calculating the determinant too fast, it should go like this:
i((10*-15)-(-3*-1))-j((0*-15)-(-3*6))+k((0*-1)-(10*6)=
i(-150-3)-j(0-(-18))+k(0-60)=
-153i-18j-60k
Now, as you are having the right idea, take the dot product of our newfound normal vector with P and set it equal to zero since, by definition, it is normal:
(-153i-18j-60k).(i(x-(-8))+j(y-(-3))+k(z-7))=
153(x+8)-18(y+3)-60(z-7)=0
You should be able to solve it from there. Let me know if this answer does not coincide with the book. There are lots of double negatives that you seemed to have misplaced in your calculations...be careful!
 

1. What is the importance of finding a plane given 3 points?

Finding a plane given 3 points is important in various fields such as engineering, mathematics, and computer graphics. It allows for the determination of the orientation and position of a plane in a 3-dimensional space, which is crucial in many practical applications.

2. What information is needed to find a plane given 3 points?

To find a plane given 3 points, we need the coordinates of these points in a 3-dimensional space. This means knowing the x, y, and z coordinates of each point. Without this information, it is impossible to determine the equation of the plane.

3. What is the equation used to find a plane given 3 points?

The equation used to find a plane given 3 points is called the point-normal form of a plane. It is given by Ax + By + Cz = D, where A, B, and C are the components of the normal vector of the plane and D is a constant value.

4. Can a plane be uniquely determined by 3 points?

Yes, a plane can be uniquely determined by 3 non-collinear points. This means that if the 3 points are not in a straight line, there will be only one plane that passes through all 3 points. However, if the points are collinear, there are infinite planes that can be drawn through them.

5. What are some practical applications of finding a plane given 3 points?

Finding a plane given 3 points has many practical applications, such as in computer graphics for rendering 3D objects, in engineering for designing structures and calculating forces, and in mathematics for solving problems involving planes and 3-dimensional geometry.

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