Find the vector, not with determinants, but by using properties of cross products.
k × (i − 5j)
what a waste of time. Well, you know that
|axb| = |a| |b| sinθ
and is ⊥ to both a and b
|k| = 1
|i-5j| = √26
Now, we know that k ⊥ i-5j, so θ=90°
The cross product will also lie in the x-y plane, and it is easy to show that it will be parallel to -5i+j (think of rotating i-5j). It looks like our magnitude is already ok,
So, k × (i − 5j) = -5i+j
You can verify using determinants if you like.
To find the cross product of two vectors without using determinants, you can use the properties of cross products.
Given the vectors k and (i - 5j), we can express them in terms of their components:
k = (0, 0, 1)
(i - 5j) = (1, -5, 0)
To find the cross product of these two vectors, you can use the following formula:
k × (i - 5j) = (k2 * (i - 5j) - k3 * (i - 5j), k3 * (i - 5j) - k1 * (i - 5j), k1 * (i - 5j) - k2 * (i - 5j))
Let's substitute the values:
= (1 * (1, -5, 0) - 0 * (i - 5j), 0 * (i - 5j) - 1 * (1, -5, 0), 0 * (1, -5, 0) - 0 * (1, -5, 0))
Simplifying each component, we have:
= (1, -5, 0) - (0, 0, 0), (0, 0, 0) - (1, -5, 0), (0, 0, 0) - (0, 0, 0)
The final result is:
k × (i - 5j) = (1, -5, 0) - (0, 0, 0) = (1, -5, 0)
To find the cross product between two vectors, we can use the properties of cross products. The cross product of two vectors, A and B, denoted as A × B, is a vector that is perpendicular to both A and B.
Let's calculate the cross product between k and (i - 5j) using the properties of cross products:
k × (i - 5j)
To calculate the cross product, we can expand it using the determinant notation:
k × (i - 5j) = (0)i - (0)(-5j) - (1)k
Calculating this expression, we get:
k × (i - 5j) = - k
Therefore, the vector is -k.