Kampo tarp vektorių a ir b didumas 120. Žinoma, kad |a| = 3, |b| = 4. Apskaičiuokite skaliarinę sandaugą (3⋅a−2⋅b)⋅(a+2⋅b).
Sprendimas.
( 3* _a- 2* _b)* (_a+ 2* _b) =
( 3* _a- 2* _b)* (_a+ 2* _b) = (3⋅a−2⋅b)⋅(a+2⋅b) = 
3* 3^2+ 4* | _a |* | _b |* cos(α)- 4* _b^2 = 3⋅32+4⋅∣a∣⋅∣b∣⋅cos(α)−4⋅b2 = 
3* 3^2+ 4* | _a |* | _b |* cos(α)- 4* 4^2 = 3⋅32+4⋅∣a∣⋅∣b∣⋅cos(α)−4⋅42 = 
3* 3^2+ 4* | _a |* | _b |* cos(120)- 4* 4^2 = 3⋅32+4⋅∣a∣⋅∣b∣⋅cos(120)−4⋅42 = 
3* 3^2+ 4* 3* | _b |* cos(120)- 4* 4^2 = 3⋅32+4⋅3⋅∣b∣⋅cos(120)−4⋅42 = 
3* 3^2+ 4* 3* 4* cos(120)- 4* 4^2 = 3⋅32+4⋅3⋅4⋅cos(120)−4⋅42 = 
(3⋅a−2⋅b)⋅(a+2⋅b) = (3⋅a⋅a+3⋅a⋅2⋅b−2⋅b⋅a−2⋅b⋅2⋅b) = (3⋅a2+6⋅a⋅b−2⋅a⋅b−4⋅b2) = 3⋅a2+4⋅a⋅b−4⋅b2 = 3⋅a2+4⋅∣a∣⋅∣b∣⋅cos(α)−4⋅b2 = 3⋅32+4⋅3⋅4⋅cos(120)−4⋅42 = 27+48⋅cos(120)−4⋅42 = 27−248⋅1−4⋅42 = 27−24−64 =
(3*_a*_a+3*_a*2*_b-2*_b*_a-2*_b*2*_b) =
(3*_a^2+6*_a*_b-2*_a*_b-4*_b^2) =
3*_a^2+4*|_a|*|_b|*cos(α)-4*_b^2 =
3*3^2+4*3*4*cos(120)-4*4^2 =

Atsakymas: - 61