Use a specific example to explore how the cross product behaves under scalar multiplication. Is it true that k(Vector a × Vector b) = (kVector a) × Vector b = Vector a × (kVector b)? Expand to the general case to prove your theory.

To explore how the cross product behaves under scalar multiplication, let's consider a specific example using vectors in three-dimensional space.

Let Vector a = (2, -3, 4) and Vector b = (-1, 2, 0). We will now calculate the cross product of Vector a and Vector b:

Vector a × Vector b = (2, -3, 4) × (-1, 2, 0)

To find the cross product, we can use the determinant-based formula:

Vector a × Vector b = (aybz - azby, azbx - axbz, axby - aybx)

Calculating further:

Vector a × Vector b = ( (-3 * 0) - (4 * 2), (4 * (-1)) - (2 * 0), (2 * 2) - (-3 * (-1)) )

= (-8, -4, 7)

Now, let's apply scalar multiplication to the cross product. Let k be a scalar, so we have:

k(Vector a × Vector b) = k(-8, -4, 7) = (-8k, -4k, 7k)

Next, let's consider the cross product between the scalar multiple of Vector a, kVector a, and Vector b:

(kVector a) × Vector b = (k(2, -3, 4)) × (-1, 2, 0)

Performing the cross product:

(kVector a) × Vector b = (2k, -3k, 4k) × (-1, 2, 0)

Using the determinant-based formula again:

(kVector a) × Vector b = ( (-3k * 0) - (4k * 2), (4k * (-1)) - (2k * 0), (2k * 2) - (-3k * (-1)) )

= (-8k, -4k, 7k)

The result is the same as k(Vector a × Vector b).

Now, let's consider the cross product of Vector a and the scalar multiple of Vector b, Vector a × (kVector b):

Vector a × (kVector b) = (2, -3, 4) × k(-1, 2, 0)

Calculating the cross product:

Vector a × (kVector b) = (2, -3, 4) × ( -k, 2k, 0)

Using the determinant-based formula:

Vector a × (kVector b) = ( (-3 * 0) - (4 * 2k), (4 * (-k)) - (2 * 0), (2 * 2k) - (-3 * (-k)) )

= (-8k, -4k, 7k)

Again, the result is the same as k(Vector a × Vector b).

From the specific example, we can conclude that k(Vector a × Vector b) = (kVector a) × Vector b = Vector a × (kVector b).

Expanding this to the general case, let Vector a = (a1, a2, a3) and Vector b = (b1, b2, b3). The cross product of Vector a and Vector b is:

Vector a × Vector b = (a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1)

Now, when we apply scalar multiplication to the cross product, we get:

k(Vector a × Vector b) = (ka2b3 - ka3b2, ka3b1 - ka1b3, ka1b2 - ka2b1)

Using the associative property of scalar multiplication, we can rewrite the above equation as:

k(Vector a × Vector b) = (k(a2b3 - a3b2), k(a3b1 - a1b3), k(a1b2 - a2b1))

This is equal to multiplying each component of the cross product by k, resulting in:

k(Vector a × Vector b) = (ka2b3 - ka3b2, ka3b1 - ka1b3, ka1b2 - ka2b1)

Expanding further, we can rewrite k(Vector a × Vector b) as:

k(Vector a × Vector b) = (ka2b3 - ka3b2, ka3b1 - ka1b3, ka1b2 - ka2b1)

which is equal to (kVector a) × Vector b and Vector a × (kVector b).

Therefore, it is indeed true that k(Vector a × Vector b) = (kVector a) × Vector b = Vector a × (kVector b) for any vectors Vector a and Vector b, and any scalar k.