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Kampo tarp vektorių a ir b didumas 120. Žinoma, kad |a| = 3, |b| = 4. Apskaičiuokite skaliarinę sandaugą $(3\cdot \vec{a}-2\cdot \vec{b})\cdot (\vec{a}+2\cdot \vec{b})$ .

Sprendimas.

( 3* _a- 2* _b)* (_a+ 2* _b) =

( 3* _a- 2* _b)* (_a+ 2* _b) = $(3\cdot \vec{a}-2\cdot \vec{b})\cdot (\vec{a}+2\cdot \vec{b})$ =

3* 3^²+ 4* | _a |* | _b |* cos(α)- 4* _b^² = $3\cdot 3^{2}+4\cdot |\vec{a}|\cdot |\vec{b}|\cdot cos(\alpha)-4\cdot \vec{b}^{2}$ =

3* 3^²+ 4* | _a |* | _b |* cos(α)- 4* 4^² = $3\cdot 3^{2}+4\cdot |\vec{a}|\cdot |\vec{b}|\cdot cos(\alpha)-4\cdot 4^{2}$ =

3* 3^²+ 4* | _a |* | _b |* cos(120)- 4* 4^² = $3\cdot 3^{2}+4\cdot |\vec{a}|\cdot |\vec{b}|\cdot cos(120)-4\cdot 4^{2}$ =

3* 3^²+ 4* 3* | _b |* cos(120)- 4* 4^² = $3\cdot 3^{2}+4\cdot 3\cdot |\vec{b}|\cdot cos(120)-4\cdot 4^{2}$ =

3* 3^²+ 4* 3* 4* cos(120)- 4* 4^² = $3\cdot 3^{2}+4\cdot 3\cdot 4\cdot cos(120)-4\cdot 4^{2}$ =

$(3\cdot \vec{a}-2\cdot \vec{b})\cdot (\vec{a}+2\cdot \vec{b})$ =

$(3\cdot \vec{a}\cdot \vec{a}+3\cdot \vec{a}\cdot 2\cdot \vec{b}-2\cdot \vec{b}\cdot \vec{a}-2\cdot \vec{b}\cdot 2\cdot \vec{b})$ =

$(3\cdot \vec{a}^{2}+6\cdot \vec{a}\cdot \vec{b}-2\cdot \vec{a}\cdot \vec{b}-4\cdot \vec{b}^{2})$ =

$3\cdot \vec{a}^{2}+4\cdot \vec{a}\cdot \vec{b}-4\cdot \vec{b}^{2}$ =

$3\cdot \vec{a}^{2}+4\cdot |\vec{a}|\cdot |\vec{b}|\cdot cos(\alpha)-4\cdot \vec{b}^{2}$ =

$3\cdot 3^{2}+4\cdot 3\cdot 4\cdot cos(120)-4\cdot 4^{2}$ =

$27+48\cdot cos(120)-4\cdot 4^{2}$ =

$27-\frac{48\cdot 1}{2}-4\cdot 4^{2}$ =

$27-24-64$ =

$3-64$ =

$-61$

Atsakymas: - 61

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