Kaj je navzkrižni produkt [-3, 5, -3] in [4, -11, 11]?

Odgovor:

#vec C = 22i + 21j + 13 k #

Pojasnilo:

# "navzkrižni produkt dveh vektorjev je podan kot:" #

#vec A = (a, b, c) #

#vec B = (d, e, f) #

#vec C = vec A X vec B #

#vec C = i (b * f-c * e) -j (a * f-c * d) + k (a * e-b * d) #

# "Tako:" #

#vec C = i (5 * 11-11 * 3) -j (-3 * 11 - (- 3 * 4)) + k ((- 3) * (- 11) -5 * 4) #

#vec C = i (55-33) -j (-33 + 12) + k (33-20) #

#vec C = 22i + 21j + 13 k #

Kaj je navzkrižni produkt <0,8,5> in <-1, -1,2>?

<21,-5,8> We know that vecA xx vecB = ||vecA|| * ||vecB|| * sin(theta) hatn, where hatn is a unit vector given by the right hand rule. So for of the unit vectors hati, hatj and hatk in the direction of x, y and z respectively, we can arrive at the following results. color(white)( (color(black){hati xx hati = vec0}, color(black){qquad hati xx hatj = hatk}, color(black){qquad hati xx hatk = -hatj}), (color(black){hatj xx hati = -hatk}, color(black){qquad hatj xx hatj = vec0}, color(black){qquad hatj xx hatk = hati}), (color(black){hatk xx hati = hatj}, color(black){qquad hatk xx hatj = -hati}, color(black){qquad hatk xx hatk

Kaj je navzkrižni produkt [0,8,5] in [1,2, -4]?

[0,8,5] xx [1,2, -4] = [-42,5, -8] Rezultat izdelka vecA in vecB je vecA xx vecB = || vecA || * || vecB || * sin (theta) hatn, kjer je theta pozitiven kot med vecA in vecB, in hatn je enota vektor s smerjo, podano z desnim pravilom. Za enote vektorje hati, hatj in hatk v smereh x, y in z, barva (bela) ((barva (črna) {hati xx hati = vec0}, barva (črna) {qquad hati xx hatj = hatk} , barva (črna) {qquad hati xx hatk = -hatj}), (barva (črna) {hatj xx hati = -hatk}, barva (črna) {qquad hatj xx hatj = vec0}, barva (črna) {qquad hatj xx hatk = hati}), (barva (črna) {hatk xx hati = hatj}, barva (črna) {qquad hatk xx hatj = -hati},

Kaj je navzkrižni produkt [-1,0,1] in [0,1,2]?

Prečni produkt je = 1,2 - 1,2, -1〉 Prečni produkt se izračuna z determinanto (veci, vecj, veck), (d, e, f), (g, h, i) | kjer sta, d, e, f〉 in, g, h, i〉 2 vektorja Tukaj imamo veca = 1,0 - 1,0,1〉 in vecb = ,2 0,1,2〉 Zato, | (veci, vecj, veck), (-1,0,1), (0,1,2) | = veci | (0,1), (1,2) | -vecj | (-1,1), (0,2) | + veck | (-1,0), (0,1) | = veci (-1) -vecj (-2) + veck (-1) = 1,2 - 1,2, -1〉 = vecc Preverjanje z 2 točkovnimi izdelki ,2 -1,2, -1〉. 〈- 1, 0,1〉 = 1 + 0-1 = 0 ,2 -1,2, -1〉. ,2 0,1,2〉 = 0 + 2-2 = 0 Torej je vecc pravokoten na veca in vecb