Given A→1=2,A→2=3 and A→1+A→2=3. Find the value of A→1+2A→2⋅3A→1−4A→2

A1=2,A2=3,A→1+A→2=3⇒ A→1+A→22=9⇒ A12+A22+2A→1⋅A→2=9⇒ 22+32+2A→1⋅A→2=9⇒A→1⋅A→2=−2 Now, A→1+2A→2⋅3A→1−4A→2=3A12−8A22+2A→1⋅A→2 =3(2)2−8(3)2+2(−2)=−64

Given A→1=2,A→2=3 and A→1+A→2=3. Find the value of A→1+2A→2⋅3A→1−4A→2

A1=2,A2=3,A→1+A→2=3⇒ A→1+A→22=9⇒ A12+A22+2A→1⋅A→2=9⇒ 22+32+2A→1⋅A→2=9⇒A→1⋅A→2=−2 Now, A→1+2A→2⋅3A→1−4A→2=3A12−8A22+2A→1⋅A→2 =3(2)2−8(3)2+2(−2)=−64

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Given $|{\vec{A}}_{1}| = 2, |{\vec{A}}_{2}| = 3 and |{\vec{A}}_{1} + {\vec{A}}_{2}| = 3$ . Find the value of $({\vec{A}}_{1} + 2 {\vec{A}}_{2}) \cdot (3 {\vec{A}}_{1} - 4 {\vec{A}}_{2})$

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-60

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Detailed Solution

$\begin{array}{l} A_{1} = 2, A_{2} = 3, |{\vec{A}}_{1} + {\vec{A}}_{2}| = 3 \\ \Rightarrow {|{\vec{A}}_{1} + {\vec{A}}_{2}|}^{2} = 9 \\ \Rightarrow A_{1}^{2} + A_{2}^{2} + 2 {\vec{A}}_{1} \cdot {\vec{A}}_{2} = 9 \\ \Rightarrow 2^{2} + 3^{2} + 2 {\vec{A}}_{1} \cdot {\vec{A}}_{2} = 9 \Rightarrow {\vec{A}}_{1} \cdot {\vec{A}}_{2} = - 2 \\ \begin{aligned} Now, ({\vec{A}}_{1} + 2 {\vec{A}}_{2}) \cdot (3 {\vec{A}}_{1} - 4 {\vec{A}}_{2}) & = 3 A_{1}^{2} - 8 A_{2}^{2} + 2 {\vec{A}}_{1} \cdot {\vec{A}}_{2} \\ = 3 (2)^{2} - 8 (3)^{2} + 2 (- 2) = - 64 \end{aligned} \end{array}$