Determine (6i+2j)· (i-4j+k)
This is a dot product question and each letter has the vector symbol
in the x direction 6 * 1 = 6
int the y direction 2 * -4 = -8
in the z direction 0 * 1 = 0
6 + -8 = -2
see:
http://www.mathsisfun.com/algebra/vectors-dot-product.html
to see how to do it with
|A| * |B| * cos theta
To determine the dot product of two vectors, you need to multiply corresponding components of the vectors and then add them together. Let's calculate the dot product of (6i+2j) and (i-4j+k) step by step.
First, let's identify the corresponding components:
(6i+2j)·(i-4j+k) = (6i)·(i) + (6i)·(-4j) + (6i)·(k) + (2j)·(i) + (2j)·(-4j) + (2j)·(k)
Now, let's calculate each term:
(6i)·(i) = 6 * 1 = 6
(6i)·(-4j) = 6 * -4 * 0 = 0 (since the i-component of (6i) and the j-component of (-4j) are perpendicular)
(6i)·(k) = 6 * 0 = 0 (since the i-component of (6i) and the k-component of (k) are perpendicular)
(2j)·(i) = 2 * 0 = 0 (since the j-component of (2j) and the i-component of (i) are perpendicular)
(2j)·(-4j) = 2 * -4 * 1 = -8
(2j)·(k) = 2 * 0 = 0 (since the j-component of (2j) and the k-component of (k) are perpendicular)
Now, let's add all the calculated terms together:
(6i+2j)·(i-4j+k) = 6 + 0 + 0 + 0 + (-8) + 0
= 6 - 8
= -2
Therefore, the dot product of (6i+2j) and (i-4j+k) is -2.