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

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

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$90$

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(A) The angle $	heta$ between two vectors $\vec A$ and $\vec B$ is given by the formula: $\cos 	heta = \frac{\vec A \cdot \vec B}{|A||B|}$.First,calculate the dot product $\vec A \cdot \vec B = (3)(3) + (4)(4) + (5)(-5) = 9 + 16 - 25 = 0$.Next,calculate the magnitudes $|A| = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50}$ and $|B| = \sqrt{3^2 + 4^2 + (-5)^2} = \sqrt{9 + 16 + 25} = \sqrt{50}$.Substituting these values into the formula: $\cos 	heta = \frac{0}{\sqrt{50} \cdot \sqrt{50}} = \frac{0}{50} = 0$.Since $\cos 	heta = 0$,the angle $	heta = 90^\circ$.

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

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