Pytanie # 94346

Odpowiedź:

#hat (PQR) = cos ^ (- 1) (27 / sqrt1235) #

Wyjaśnienie:

Bądź dwoma wektorami #vec (AB) # i #vec (AC) #:

#vec (AB) * vec (AC) = (AB) (AC) cos (kapelusz (BAC)) #

# = (x_ (AB) x_ (AC)) + (y_ (AB) y_ (AC)) + (z_ (AB) z_ (AC)) #

Mamy:

# P = (1; 1; 1) #

#Q = (- 2; 2; 4) #

# R = (3; -4; 2) #

w związku z tym

#vec (QP) = (x_P-x_Q; y_P-y_Q; z_P-z_Q) = (3; -1; -3) #

#vec (QR) = (x_R-x_Q; y_R-y_Q; z_R-z_Q) = (5; -6; -2) #

i

# (QP) = sqrt ((x_ (QP)) ^ 2+ (y_ (QP)) ^ 2+ (z_ (QP)) ^ 2) = sqrt (9 + 1 + 9) = sqrt (19) #

# (QR) = sqrt ((x_ (QR)) ^ 2+ (y_ (QR)) ^ 2+ (z_ (QR)) ^ 2) = sqrt (25 + 36 + 4) = sqrt (65) #

W związku z tym:

#vec (QP) * vec (QR) = sqrt19sqrt65cos (kapelusz (PQR)) #

#=(3*5+(-1)(-6)+(-3)(-2))#

#rarr cos (hat (PQR)) = (15 + 6 + 6) / (sqrt19sqrt65) = 27 / sqrt1235 #

#rarr hat (PQR) = cos ^ (- 1) (27 / sqrt1235) #