oblicz babilon: 3. Obliczyć wysokość i jeden z kątów trójkąta o wierzchołkach A(−3,−2,4), B(3,1,−2) i C(0,1,2).

janek191: A = ( −3, −2, 4) B = ( 3, 1, −2) C = ( 0,1, 2) → AB = [ 6,3,− 6] ⇒ I AB I = √81 = 9 → AC = [ 3, 3, −2 ] ⇒ I AC I = √22 → →

	6*3+ 3*3 + 6*2		39		13√22
cos ∡ (AB , AC ) =		=		=
	9*√22		9√22		66

janek191:

	13√22
cos α =
	66

	13√22		169		29
sin2α = 1 − (		)2 = 1 −		=
	66		198		198

sin α = √²⁹₁₉₈

	√29
PΔ = 0,5*9*√22*		= 1,5 √29
	√198

0,5 IABI *h = 1,5 √29 / *2 9*h = 3 √29

	√29
h =
	3

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