If A=4i+3j-2k, B=8i+6j-4k then the angle would be?
A.B=ABcos@
@= Cos-(A.B/AB)
To find the angle between two vectors, we can use the dot product formula:
A · B = |A| * |B| * cosθ
Where A · B is the dot product of vectors A and B, |A| and |B| are the magnitudes (or lengths) of vectors A and B, and θ is the angle between the two vectors.
First, let's find the magnitudes of vectors A and B:
|A| = √(4^2 + 3^2 + (-2)^2) = √(16 + 9 + 4) = √29
|B| = √(8^2 + 6^2 + (-4)^2) = √(64 + 36 + 16) = √116 = 2√29
Next, let's calculate the dot product of A and B:
A · B = (4 * 8) + (3 * 6) + (-2 * -4) = 32 + 18 + 8 = 58
Now we can use the formula to find the angle (θ):
58 = √29 * 2√29 * cosθ
58 = 58 * cosθ
Dividing both sides of the equation by 58:
1 = cosθ
Since the cosine of the angle is 1, we know that the angle is 0 degrees.
Therefore, the angle between vectors A and B is 0 degrees.