Use the triple scalar product to determine whether the given points lie in the same plane.
A(1, 2, 3), B(3, -4, 8),
C(7, -1, 1), D(5, 5, -4)
The scalar triple product P among three vectors U,V and W is defined as follows:
P=U.(VxW)
|P| represents the volume of the parallelepiped formed by the vectors U, V and W.
Consequently, if U, V and W are coplanar, P=0.
Since 4 points (A,B,C,D) are given, three vectors can be formed where
U=AB, V=AC, and W=AD.
Calculate the triple scalar product as described above and if the result is zero, then the four points are coplanar.
yes
To determine whether the given points A(1, 2, 3), B(3, -4, 8), C(7, -1, 1), and D(5, 5, -4) lie in the same plane or not, we can use the triple scalar product.
The triple scalar product is defined as the dot product of the first vector with the cross product of the second and third vectors. In other words, it can be written as:
(V1 dot (V2 x V3))
Let's designate AB as vector V1, AC as vector V2, and AD as vector V3.
V1 = B - A = (3, -4, 8) - (1, 2, 3) = (2, -6, 5)
V2 = C - A = (7, -1, 1) - (1, 2, 3) = (6, -3, -2)
V3 = D - A = (5, 5, -4) - (1, 2, 3) = (4, 3, -7)
Now we can calculate the triple scalar product:
(V1 dot (V2 x V3)) = (2, -6, 5) dot ((6, -3, -2) x (4, 3, -7))
To find the cross product of (6, -3, -2) and (4, 3, -7), we can use the formula:
((6, -3, -2) x (4, 3, -7)) = ((-9) - (-6), (-42) - (-14), (-12) - (-12)) = (-3, -28, 0)
Now we can calculate the dot product:
(2, -6, 5) dot (-3, -28, 0) = 2*(-3) + (-6)*(-28) + 5*0 = -6 + 168 + 0 = 162
If the result is zero, then the points lie in the same plane. However, if the result is nonzero, then the points do not lie in the same plane.
In this case, the result is 162, which is nonzero. Therefore, the given points A(1, 2, 3), B(3, -4, 8), C(7, -1, 1), and D(5, 5, -4) do not lie in the same plane.
To determine whether the given points A(1, 2, 3), B(3, -4, 8), C(7, -1, 1), and D(5, 5, -4) lie in the same plane, we can use the triple scalar product.
The triple scalar product of three vectors is calculated as follows:
[Triple Scalar Product] = (A ⨯ B) ⋅ C
where A and B are vectors in the plane and C is a vector perpendicular to the plane. If the triple scalar product is zero, then the points lie in the same plane.
Let's calculate the value of the triple scalar product for the given points:
Vector AB = (3 - 1, -4 - 2, 8 - 3) = (2, -6, 5)
Vector AC = (7 - 1, -1 - 2, 1 - 3) = (6, -3, -2)
Now, calculate the cross product of AB and AC:
AB ⨯ AC = (2, -6, 5) ⨯ (6, -3, -2)
Using the formula for calculating the cross product:
AB ⨯ AC = (6(-2) - (-3)(5), (5)(6) - 2(-2), 2(-3) - (-6)(-6)) = (-27, 34, -30)
Finally, calculate the dot product of the obtained cross product with vector AD (or any other vector not parallel to AB and AC):
Triple Scalar Product = (-27, 34, -30) ⋅ (5 - 1, 5 - 2, -4 - 3) = (-27, 34, -30) ⋅ (4, 3, -7)
Using the dot product formula:
Triple Scalar Product = (-27)(4) + (34)(3) + (-30)(-7) = -108 + 102 + 210 = 204
Since the triple scalar product is not zero (it is 204), the given points A, B, C, and D do not lie in the same plane.