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12. If 4i - 5j + k, 21i+ 3j + 3k and xi +yj + zk are
respectively three sides AB, BC and AC of traingle ABC
then the values of x, y, z are :
(A) x = 6, y = -2, z = 4
(B) x = 6, y = 2, z = 4
(C) x = 6, y = -1, z = 4
(D) x = 2, y = -8,2= -2
Answers
Given info : 4i - 5j + k, 2i + 3j + 3k and xi +yj + zk are respectively three sides AB, BC and AC of traingle ABC.
To find : The values of x , y and z are ..
solution : AB = 4i - 5j + k
BC = 2i + 3j + 3k
AC = xi + yj + zk ⇒CA = -(xi + yj + zk)
AB , BC and CA form a triangle so AB + BC + CA = 0
⇒(4i - 5j + k) + (2i + 3j + 3k) + {-(xi + yj + zk)} = 0
⇒(4 + 2 - x)i + (-5 + 3 - y)j + (1 + 3 - z)k = 0
⇒4 + 2 - x = 0 ⇒x = 6
⇒-5 + 3 - y = 0 ⇒y = -2
⇒1 + 3 - z = 0 ⇒z = 4
Therefore the values of x , y and z are 6 , -2 and 4 respectively.
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