Question

# Attempt any three part of the following :If $$|\vec{a}| = 3, |\vec{b}| = 5 \space and \space |\vec{c}| = 7$$ and $$\vec{a} + \vec{b} + \vec{c} = 0$$ then prove that angle between $$\vec{a}$$ and $$\vec{b}$$ is $$\dfrac{\pi}{3}$$

Solution

## To   Prove   that  angle  between   $$\vec{a}$$    and   $$\vec{b}$$   is   $$\dfrac{\pi }{3}$$Given ,  $$\left | \vec{a} \right | = 3 ,\left | \vec{b} \right | = 5$$  and   $$\left | \vec{c} \right | = 7$$     and     $$\vec{a} + \vec{b} + \vec{c} =0$$This   can   be   written   as  $$\vec{a} + \vec{b} + \vec{c} =0$$ $$\vec{a} + \vec{b} = - \vec{c}$$ On  Squaring   both sides, we   get   $$\left | \vec{a} \right |^{2} + \left | \vec{b} \right |^{2} + 2 \vec{a} . \vec{b} = \left | \vec{c} \right |^{2}$$$$\left | \vec{a} \right |^{2} + \left | \vec{b} \right |^{2} + 2 \left | \vec{a} \right |\left | \vec{b} \right |cos\theta = \left | \vec{c} \right |^{2}$$$$(3)^{2} + (5)^{2} + 2 .3.5cos\theta = (7)^{2}$$$$9 + 25 + 30cos\theta = 49$$$$34 + 30cos\theta = 49$$$$30cos\theta = 15$$$$cos\theta = \dfrac{15}{30}$$$$cos\theta = \dfrac{1}{2}$$$$\theta = \dfrac{\pi }{3}$$Hence , angle  between   $$\vec{a}$$    and   $$\vec{b}$$   is   $$\dfrac{\pi }{3}$$ Mathematics

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