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Given A=3i-j+3k and 2i+j-2k find as A.B by the angle between A and B​

User Tholle
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Final answer:

To find the dot product of vectors A and B, multiply the corresponding components and add them together, resulting in A.B = -1. To find the angle between the vectors, use the formula that involves the dot product and the magnitudes of the vectors, applied by taking the inverse cosine.

Step-by-step explanation:

To find the dot product of the vectors A and B, we use the formula:

A.B = Ax Bx + Ay By + Az Bz

Assuming the second vector mentioned in the question is B = 2i + j - 2k, we can calculate the dot product as follows:

A.B = (3)(2) + (-1)(1) + (3)(-2) = 6 - 1 - 6 = -1

To find the angle between vectors A and B, we use the inverse cosine of the dot product divided by the product of the magnitudes of A and B:

cos(theta) = (A.B) / (|A| |B|)

We can calculate the magnitudes (|A| and |B|), then use them to find theta (the angle).

The provided equations are not directly relevant to the question, as they represent other vector operations.

User Michael Hale
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