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.
recall that the magnitude of a×b is the area of the parallelogram formed by a and b.
So, ka×kb = k(a×kb) = k^2 a×b
think of the determinant form of the calculation. multiplying the 2nd and 3rd rows by k multiplies the determinant by k^2
To explore how the cross product behaves under scalar multiplication, let's consider the specific example of two vectors:
Vector a = (3, 2, -1)
Vector b = (-2, 5, 4)
First, we'll calculate the cross product of Vector a and Vector b:
Vector a × Vector b = ((2)(4) - (-1)(5), (-1)(-2) - (3)(4), (3)(5) - (2)(-2))
= (13, -14, 19)
Now, let's multiply this result by a scalar, k:
k(Vector a × Vector b) = k(13, -14, 19)
= (k * 13, k * (-14), k * 19)
= (13k, -14k, 19k)
Next, let's calculate the cross product of the scalar multiple of Vector a and Vector b:
(kVector a) × Vector b = k(3, 2, -1) × Vector b
= (k * 3, k * 2, k * (-1)) × (-2, 5, 4)
= (3k, 2k, -k) × (-2, 5, 4)
Expanding this cross product:
(3k, 2k, -k) × (-2, 5, 4) = ((2k)(4) - (-k)(5), (-k)(-2) - (3k)(4), (3k)(5) - (2k)(-2))
= (13k, -14k, 19k)
Similarly, let's calculate the cross product of Vector a and the scalar multiple of Vector b:
Vector a × (kVector b) = Vector a × k(-2, 5, 4)
= (3, 2, -1) × (k * (-2), k * 5, k * 4)
= (3, 2, -1) × (-2k, 5k, 4k)
Expanding this cross product:
(3, 2, -1) × (-2k, 5k, 4k) = ((2)(4k) - (-1)(5k), (-1)(-2k) - (3)(4k), (3)(5k) - (2)(4k))
= (13k, -14k, 19k)
As we can see, in both cases, the cross product of Vector a and Vector b multiplied by a scalar k is equal to the cross product of the scalar multiple of Vector a and Vector b, as well as the cross product of Vector a and the scalar multiple of Vector b.
In general, we can state that:
k(Vector a × Vector b) = (kVector a) × Vector b = Vector a × (kVector b),
where k is any scalar and Vector a and Vector b are two arbitrary vectors. This holds true for any vectors in 3-dimensional space.
To explore how the cross product behaves under scalar multiplication, let's use a specific example.
Consider two vectors, Vector a = [3, 2, 1] and Vector b = [1, -1, 2]. We will investigate whether the equation k (Vector a × Vector b) = (kVector a) × Vector b = Vector a × (kVector b) holds true.
First, we compute the cross product of Vector a and Vector b:
Vector a × Vector b = [a2 * b3 - a3 * b2, -(a1 * b3 - a3 * b1), a1 * b2 - a2 * b1]
= [(2 * 2) - (1 * -1), -((3 * 2) - (1 * 2)), (3 * -1) - (2 * 1)]
= [4 - (-1), -(6 - 2), -3 - 2]
= [5, -4, -5]
Now, let's explore the three cases:
1. k (Vector a × Vector b):
Let's choose k = 2:
2 (Vector a × Vector b) = 2 [5, -4, -5] = [10, -8, -10]
2. (kVector a) × Vector b:
We'll apply k = 2 to Vector a first:
(kVector a) = 2 [3, 2, 1] = [6, 4, 2]
Then, compute the cross product:
(kVector a) × Vector b = [b2 * (kVector a)3 - (kVector a)2 * b3, -[(b1 * (kVector a)3 - (kVector a)1 * b3)], b1 * (kVector a)2 - (kVector a)1 * b2]
= [(-4 * 2) - (2 * -5), -[(1 * 2) - (6 * -5)], 1 * 4 - 3 * 2]
= [-8 - (-10), -(2 - (-30)), 4 - 6]
= [2, -28, -2]
3. Vector a × (kVector b):
Again, we substitute k = 2 into Vector b:
(kVector b) = 2 [1, -1, 2] = [2, -2, 4]
Now, calculate the cross product:
Vector a × (kVector b) = [b2 * a3 - a2 * b3, -(b1 * a3 - a1 * b3), b1 * a2 - a1 * b2]
= [(-2 * 1) - (2 * -5), -[(-1 * 1) - (3 * -5)], (-1 * 2) - (3 * 1)]
= [-2 - (-10), -(-1 - (-15)), -2 - 3]
= [8, -14, -5]
After evaluating all three cases, we find that the result is not the same in each case. Hence, it is NOT true in general that k (Vector a × Vector b) = (kVector a) × Vector b = Vector a × (kVector b).
Therefore, we have disproven the theory in the general case.