The angle between the two vectors vec A = 3ha | vedclass

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(B) The angle $	heta$ between two vectors $\vec A$ and $\vec B$ is given by the formula: $\cos 	heta = \frac{\vec A \cdot \vec B}{|A||B|}$.Since $\v

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(B) The angle $	heta$ between two vectors $\vec A$ and $\vec B$ is given by the formula: $\cos 	heta = \frac{\vec A \cdot \vec B}{|A||B|}$.Since $\vec A = 3\hat i + 4\hat j + 5\hat k$ and $\vec B = 3\hat i + 4\hat j + 5\hat k$,we see that $\vec A = \vec B$.The dot product is $\vec A \cdot \vec B = (3)(3) + (4)(4) + (5)(5) = 9 + 16 + 25 = 50$.The magnitudes are $|A| = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50}$ and $|B| = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{50}$.Substituting these values into the formula: $\cos 	heta = \frac{50}{\sqrt{50} \cdot \sqrt{50}} = \frac{50}{50} = 1$.Since $\cos 	heta = 1$,the angle $	heta = \cos^{-1}(1) = 0^o$.

The angle between the two vectors $\vec A = 3\hat i + 4\hat j + 5\hat k$ and $\vec B = 3\hat i + 4\hat j + 5\hat k$ is ....... $^o$

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