Given A1→=2, A2→=3 and A1→+A2→=3 . Find the value of A1→+2A2→.3A1→-4A2→ :-

If θ is the angle between A1→ and A2→ , thenA2=A12+A22+2A1A2cosθor 32=22+32+2×2×3cosθ∴cosθ=-13Now, A1→+2A2→3A1→-4A2→=3A1→.A1→+6A2→.A1→-4A1→.A2→-8A2→.A2→ =3×22+2×2×3×-13-8×32=-64⇒A1→+2A2→.3A1→-4A2→=64

Given A1→=2, A2→=3 and A1→+A2→=3 . Find the value of A1→+2A2→.3A1→-4A2→ :-

If θ is the angle between A1→ and A2→ , thenA2=A12+A22+2A1A2cosθor 32=22+32+2×2×3cosθ∴cosθ=-13Now, A1→+2A2→3A1→-4A2→=3A1→.A1→+6A2→.A1→-4A1→.A2→-8A2→.A2→ =3×22+2×2×3×-13-8×32=-64⇒A1→+2A2→.3A1→-4A2→=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 $|(\vec{A_{1}} + 2 \vec{A_{2}}) . (3 \vec{A_{1}} - 4 \vec{A_{2}})|$ :-

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Detailed Solution

If $θ$ is the angle between $\vec{A_{1}}$ and $\vec{A_{2}}$ , then

$A^{2} = {A_{1}}^{2} + {A_{2}}^{2} + 2 A_{1} A_{2} \cos θ$

or $3^{2} = 2^{2} + 3^{2} + 2 \times 2 \times 3 \cos θ$

$∴ \cos θ = - \frac{1}{3}$

Now, $(\vec{A_{1}} + 2 \vec{A_{2}}) (3 \vec{A_{1}} - 4 \vec{A_{2}})$

$= 3 \vec{A_{1}} . \vec{A_{1}} + 6 \vec{A_{2}} . \vec{A_{1}} - 4 \vec{A_{1}} . \vec{A_{2}} - 8 \vec{A_{2}} . \vec{A_{2}} = 3 \times 2^{2} + 2 \times 2 \times 3 \times (- \frac{1}{3}) - 8 \times 3^{2} = - 64$