Če vec (a) = 2i + 2j + 2k, vec (b) = - i + 2j + k, vec (c) = 3i + j so takšni, da je vec (a) + jvec (b) pravokotno na vec (c) ), poiščite vrednost j?

Odgovor:

# j = 8 #

Pojasnilo:

#costheta = ((a + jb).c) / (abs (a + jb) abs (c)) #

Vendar pa # theta = 90 #, Torej # cos90 = 0 #

# (a + jb).c = 0 #

# a + jb = ((2), (2), (2)) + j ((- 1), (2), (1)) = ((2-j), (2 + 2j), (2) + j)) #

#c = ((3), (1), (0)) #

# (a + jb).c = 3 (2-j) + 2 + 2j = 6-3j + 2 + 2j = 8-j = 0 #

# j = 8 #