Use the cross product to find a vector perpendicular to each of the following pairs of vectors.
(2, -1, 3) and (1, 4, -2)
Can someone please explain this step by step?
To find a vector perpendicular to two given vectors, we can use the cross product. The cross product of two vectors gives us a vector that is perpendicular to both of the original vectors.
Let's denote the first vector as A = (2, -1, 3) and the second vector as B = (1, 4, -2).
Step 1: Calculate the cross product
To find the cross product of two vectors, we use the formula:
A × B = (AyBz - AzBy, AzBx - AxBz, AxBy - AyBx)
In this case:
Ax = 2, Ay = -1, Az = 3
Bx = 1, By = 4, Bz = -2
Substituting these values into the formula, we get:
A × B = (-1 * (-2) - 3 * 4, 3 * 1 - 2 * (-2), 2 * 4 - (-1) * 1)
= (2 + 12, 3 + 4, 8 + 1)
= (14, 7, 9)
Therefore, the resultant vector (or the cross product) is (14, 7, 9).
Step 2: Verify if the cross product is perpendicular to the given vectors
To verify that the cross product is perpendicular to the original vectors, we can take the dot product of the cross product with each of the given vectors. If the dot product is zero, then the vectors are perpendicular.
Let's calculate the dot product of A × B with each of the original vectors:
(A × B) · A = 14 * 2 + 7 * (-1) + 9 * 3
= 28 - 7 + 27
= 48
(A × B) · B = 14 * 1 + 7 * 4 + 9 * (-2)
= 14 + 28 - 18
= 24
Since both dot products are non-zero, the cross product (14, 7, 9) is not perpendicular to either of the original vectors.
To find a vector perpendicular to both vectors, we need to find the cross product again with a different method.
Step 3: Calculate the alternative cross product
To find a different vector perpendicular to both vectors A and B, we can swap the order of the vectors and find the cross product again:
B × A = (ByAz - BzAy, BzAx - BxAz, BxAy - ByAx)
Using the same values as before:
B × A = (4 * 3 - (-2) * (-1), (-2) * 2 - 1 * 3, 1 * (-1) - 4 * 2)
= (12 - 2, -4 - 3, -1 - 8)
= (10, -7, -9)
Therefore, the alternative resultant vector (or the alternative cross product) is (10, -7, -9).
Step 4: Verify if the alternative cross product is perpendicular to the given vectors
Now, let's calculate the dot product of B × A with each of the original vectors:
(B × A) · A = 10 * 2 + (-7) * (-1) + (-9) * 3
= 20 + 7 - 27
= 0
(B × A) · B = 10 * 1 + (-7) * 4 + (-9) * (-2)
= 10 - 28 + 18
= 0
Since both dot products are zero, the alternative cross product (10, -7, -9) is indeed perpendicular to both vectors A and B.
Therefore, (10, -7, -9) is a vector perpendicular to the given pair of vectors (2, -1, 3) and (1, 4, -2).
The cross product is the vector that one can calculate with this determinant:
| i j k |
| 2-1 3 |
| 1 4-2 |
= (2-12) i + (3+4)j + (8+1) k
= -10i +7j +9k
or (-10, 7, 9) in your notation.
That vector, or any vector that is a scalar multiple of it, will be perpendicular to the other two vectors.