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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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