Let A B C be a triangle such that B C → = a → , C A → = b → , A B → = c → , a → = 6 2 , b → = 2 3 and b → · c → = 12 Consider the statements : S 1 : a → × b → + c → × b → - c → = 6 2 2 - 1 S 2 : ∠ A B

Question

Let A B C be a triangle such that B C → = a → , C A → = b → , A B → = c → , a → = 6 2 , b → = 2 3 and b → · c → = 12 Consider the statements : S 1 : a → × b → + c → × b → - c → = 6 2 2 - 1 S 2 : ∠ A B C = cos - 1 2 3 . Then

TestHub · Accepted Answer

We know that by triangle law of vector addition,
$\vec{a} + \vec{b} + \vec{c} = 0$
$\Rightarrow \vec{b} + \vec{c} = - \vec{a}$
Squaring both side we get,
${|\vec{b}|}^{2} + {|\vec{c}|}^{2} + 2 \vec{b} \cdot \vec{c} = {|\vec{a}|}^{2}$
Now putting the given value $|\vec{a}| = 6 \sqrt{2}, |\vec{b}| = 2 \sqrt{3} & \vec{b} . \vec{c} = 12$
We get, ${|\vec{c}|}^{2} = 36$
So, $|\vec{c}| = 6$
Now, $S 1 : |\vec{a} \times \vec{b} + \vec{c} \times \vec{b}| - |\vec{c}|$
$\Rightarrow |(\vec{a} + \vec{c}) \times \vec{b}| - |\vec{c}|$
$\Rightarrow 0 - 6 = - 6$
And $S 2 : \vec{a} + \vec{b} + \vec{c} = 0$
$\vec{a} + \vec{b} = - \vec{c}$
Squaring both side we get,
${|\vec{a}|}^{2} + {|\vec{b}|}^{2} - 2 |\vec{a}| |\vec{b}| \cos (∠ A C B) = {|\vec{c}|}^{2}$
$\Rightarrow 72 + 12 - 24 \sqrt{6} \cos (∠ A C B) = 36$
$\Rightarrow \cos (∠ A C B) = \frac{48}{24 \sqrt{6}} = \frac{2}{\sqrt{6}} = \sqrt{\frac{2}{3}}$

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