Let and are two vectors of magnitude 4 inclined at an angle then what is the angle between and ?

. Let be the angle between and . We know . We also find the magnitude , so . Using the dot product formula, . Thus, .

Let \vec{a} and \vec{b} are two vectors of magnitude 4 inclined at an angle \frac{\pi}{3} then what is the angle between \vec{a} and \vec{a}-\vec{b}?

A. \frac{\pi}{2}
B. \frac{\pi}{3} ✓
C. \frac{\pi}{4}
D. \frac{\pi}{6}

Correct Answer: B. \frac{\pi}{3}

Explanation

Let \alpha be the angle between \vec{a} and \vec{a}-\vec{b}. We know \vec{a} \cdot (\vec{a} - \vec{b}) = |\vec{a}|^2 - \vec{a} \cdot \vec{b} = 4^2 - 4(4)\cos\frac{\pi}{3} = 16 - 16(\frac{1}{2}) = 8. We also find the magnitude |\vec{a}-\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b} = 16 + 16 - 2(8) = 16, so |\vec{a}-\vec{b}| = 4. Using the dot product formula, \cos \alpha = \frac{\vec{a} \cdot (\vec{a}-\vec{b})}{|\vec{a}| |\vec{a}-\vec{b}|} = \frac{8}{(4)(4)} = \frac{1}{2}. Thus, \alpha = \frac{\pi}{3}.

Let \vec{a} and \vec{b} are two vectors of magnitude 4 inclined at an angle \frac{\pi}{3} then what is the angle between \vec{a} and \vec{a}-\vec{b}?

Explanation

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