Find the volume of the parallelepiped with the vectors A = 2i + j +3k , B = i -2j + k , C = 3i + 4j+ 2k
as coterminal edges.
using A as a vertex, that would be
|(B-A)x(C-A)|
That would be the scalar triple product
cross-product of <1, -2, 1> with <3, 4, 2> = < -8, 1, 10>
(I assume you know how to find that)
volume = <2, 1, 3> dot < -8, 1, 10>
= -16 + 1 + 30
= 15 cubic units
rats. Reiny is correct. I was thinking of the area of the base.
To find the volume of a parallelepiped with the given vectors as coterminal edges, you can use the vector triple product or the scalar triple product.
The vector triple product states that the volume of a parallelepiped formed by three vectors A, B, and C is given by the magnitude of the dot product of A with the cross product of B and C:
V = |A · (B × C)|
Now, let's calculate the volume step by step:
1. Calculate the cross product of vectors B and C:
B × C = (i - 2j + k) × (3i + 4j + 2k)
To find the cross product, we can use the determinant method:
(i - 2j + k)
(3i + 4j + 2k)
= (8i - k) - (2i - 6j)
= 6i + 6j - k
2. Take the dot product of the resulting vector with vector A:
A · (B × C) = (2i + j + 3k) · (6i + 6j - k)
Distribute and simplify:
= 12(i · i) + 12(i · j) - 2(i · k) + 6(j · i) + 6(j · j) - (j · k) + 18(k · i) + 18(k · j) - 3(k · k)
Since the dot product of two orthogonal vectors is zero, we can simplify further:
= 12(1) + 12(0) - 2(0) + 6(0) + 6(1) - (0) + 18(0) + 18(0) - 3(1)
= 12 + 6 - 3
= 15
3. Take the magnitude of the dot product:
V = |A · (B × C)| = |15|
The volume of the parallelepiped formed by the given vectors is 15 cubic units.