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Question Number 126778 by Ar Brandon last updated on 24/Dec/20

	Answered by Dwaipayan Shikari last updated on 24/Dec/20

	$\sqrt{{({\vec{A}}_{1})}^{2} + {({\vec{A}}_{2})}^{2} + 2 ∣ {\vec{A}}_{1} {\vec{A}}_{2} ∣ c o s φ} = 3$ $4 + 9 + 12 c o s φ = 9 \Rightarrow φ = {c o s}^{- 1} (- \frac{1}{3})$ $({\vec{A}}_{1} + 2 {\vec{A}}_{2}) (3 {\vec{A}}_{1} - 4 {\vec{A}}_{2}) = 3 ∣ {\vec{A}}_{1} ∣^{2} + 2 ∣ {\vec{A}}_{1} {\vec{A}}_{2} ∣ c o s φ - 8 ∣ {\vec{A}}_{2} ∣^{2}$ $= 12 - 4 - 8.9 = - 64$
	Answered by mr W last updated on 24/Dec/20

	$∣ A_{1} + A_{2} ∣^{2} = {(a_{1 x} + a_{2 x})}^{2} + {(a_{1 y} + a_{2 y})}^{2}$ $= (a_{1 x}^{2} + a_{1 y}^{2}) + (a_{2 x}^{2} + a_{2 y}^{2}) + 2 (a_{1 x} a_{2 x} + a_{1 y} a_{2 y})$ $=∣ A_{1} ∣^{2} + ∣ A_{2} ∣^{2} + 2 A_{1} \cdot A_{2}$ $3^{2} = 2^{2} + 3^{2} + 2 A_{1} \cdot A_{2}$ $\Rightarrow A_{1} \cdot A_{2} = - 2$ $(A_{1} + 2 A_{2}) \cdot (3 A_{1} - 4 A_{2})$ $= 3 A_{1} \cdot A_{1} + 2 A_{1} \cdot A_{2} - 8 A_{2} \cdot A_{2}$ $= 3 ∣ A_{1} ∣^{2} + 2 A_{1} \cdot A_{2} - 8 ∣ A_{2} ∣^{2}$ $= 3 \times 2^{2} + 2 \times (- 2) - 8 \times 3^{2}$ $= - 64$
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