Show that the four points (2,0,1), (−1,2,3), (3,2,2) and (3,−6,−3) lie in a plane.
we know that 3 points define a plane. So, pick 3 points and determine the normal direction to their plane.
Now replace any of the points with the 4th point, and determine the new normal.
If the two normals are the same, the 4 points are coplanar.
If we label the direction vectors of the points u,v,w,x, then we need
(w-u)x(w-v) to be parallel to (x-u)x(x-v)
<1,2,1>x<4,0,-1> = <-2,5,-8>
<1,-6,-4>x<4,-8,-6> = <4,-10,16>
the two normals are parallel, so the 4 points are coplanar.
let's take the first 3 points and find the equation of the plane containing them
direction vector #1: (-3,2,2)
direction vector #2: (1,2,1)
normal to these , take the cross-product
= (-2, 5, -8)
or (2, -5, 8) in the opposite direction
equation of plane is
2x - 5y + 8z = c
but (2,0,1) lies on it
4 - 0 + 8 = c = 12
the plane equation using the first three points is
2x - 5y + 8z = 12
test for the 4th point (3,-6,-3)
LS = 2(3) -5(-6) + 8(-3) = 12 = RS
So, yes, all 4 points are coplanar.
To show that the four points (2,0,1), (-1,2,3), (3,2,2), and (3,-6,-3) lie in a plane, we need to check if they are coplanar.
First, let's consider the vectors formed by any three of the given points. Let's take the vectors formed by the points (2,0,1), (-1,2,3), and (3,2,2):
Vector AB = (B - A) = (-1, 2, 3) - (2, 0, 1) = (-3, 2, 2)
Vector AC = (C - A) = (3, 2, 2) - (2, 0, 1) = (1, 2, 1)
Next, let's calculate the cross product of vectors AB and AC:
Cross product = AB x AC = (AB2 * AC3 - AB3 * AC2, AB3 * AC1 - AB1 * AC3, AB1 * AC2 - AB2 * AC1)
= ((-3 * 2) - ( 2 * 1), ( 2 * 1) - (-3 * 1), (-3 * 2) - ( 2 * 1))
= (-8, 5, -8)
Now, let's calculate the dot product of the cross product with the vector formed by any of the four given points and the origin (0,0,0). Let's take the dot product of (-8, 5, -8) with the point (3, -6, -3):
Dot product = (-8 * 3) + (5 * -6) + (-8 * -3)
= -24 - 30 + 24
= -30
If the dot product is 0, then the point lies on the plane. However, if the dot product is not 0, then the point does not lie on the plane.
In this case, the dot product is not 0 because -30 ≠ 0. Therefore, the four points (2,0,1), (-1,2,3), (3,2,2), and (3,-6,-3) do not lie in the same plane.
To show that the four points (2,0,1), (-1,2,3), (3,2,2), and (3,-6,-3) lie in a plane, we can use the concept of coplanarity.
Two vectors in three-dimensional space are coplanar if their cross product is a zero vector. This means that the cross product of any two vectors in the plane formed by the four points will be zero.
Let's take two vectors from the given points and calculate their cross product:
Vector 1:
A = (-1, 2, 3) - (2, 0, 1) = (-3, 2, 2)
Vector 2:
B = (3, 2, 2) - (2, 0, 1) = (1, 2, 1)
To find the cross product, we can use the formula:
A x B = i(detay - detyz) - j(detax - detxz) + k(detax - detay)
Here, i, j, and k represent the unit vectors along the x, y, and z-axes, respectively.
Calculating the cross product:
A x B = i((2 x 1) - (2 x 2)) - j((-3 x 1) - (2 x 1)) + k((-3 x 2) - (-3 x 1))
= i(2 - 4) - j(-3 - 2) + k(-6 + 3)
= i(-2) - j(-5) + k(-3)
= (-2, 5, -3)
Since the result is not a zero vector, we can conclude that the four points (2,0,1), (-1,2,3), (3,2,2), and (3,-6,-3) do not lie in a plane.