Expand to the general case to explore how the cross product behaves under scalar multiplication k (a x b) = (ka) x b = a x (kb).
Just set up your determinants, and you will see how it works.
If you multiply any row of a determinant by k, the value also is multiplied by k.
To explore how the cross product behaves under scalar multiplication, let's consider the general case with two vectors, a and b, and a scalar, k.
We start by computing the cross product of a and b, denoted as a x b. The cross product between two vectors in three-dimensional space yields another vector that is perpendicular to both a and b. The magnitude of the resulting vector is given by ||a x b|| = ||a|| ||b|| sin(θ), where θ is the angle between a and b.
Now, let's evaluate the first case, (k * a) x b. We multiply vector a by the scalar k, resulting in a new vector, ka. Then, we compute the cross product of ka and b, which gives another vector perpendicular to both ka and b. The magnitude of this resulting vector is ||ka x b|| = ||ka|| ||b|| sin(θ). Since k is a scalar, ||ka|| = |k| ||a||. Hence, ||ka x b|| = |k| ||a|| ||b|| sin(θ). This implies that (k * a) x b is equal to the product of the magnitude of a x b and |k|.
Next, we evaluate the second case, a x (k * b). Similarly, we multiply vector b by the scalar k, yielding kb. Then, we compute the cross product of a and kb, giving us another vector perpendicular to both a and kb. The magnitude of this resulting vector is ||a x kb|| = ||a|| ||kb|| sin(θ). Again, since k is a scalar, ||kb|| = |k| ||b||. Hence, ||a x kb|| = ||a|| |k| ||b|| sin(θ). Therefore, a x (k * b) is also equal to the product of the magnitude of a x b and |k|.
From the above analysis, we can conclude that (k * a) x b = a x (k * b) = |k| (a x b). This means that the cross product behaves under scalar multiplication by a factor of |k|, and the resulting vector will be parallel to the original a x b vector, with its magnitude scaled by |k|.