OL OL i LA Therefore, ˆ a = OA OL LA . ... ( ) Similarly, ˆ b = β β ... ( ) The angle between ˆ a and ˆ b is α β and so, ˆ a b | | | cos( cos( β β ...
( ) On the other hand, from ( ) and ( ) ˆ a b ⋅ = (cos ) (cos β β β β . ... ( ) From ( ) and ( ), we get cos( β β β Example . With usual notations, in any triangle ABC, prove by vector method that B C With usual notations in triangle ABC we have BC a CA , and AB .
Then | , | BC CA and | AB Since in ABC ∆ BC CA AB , we have BC BC CA AB Simplification gives, BC CA = AB BC ... ( ) Similarly, since BC CA AB , we have CA BC CA AB = Fig. . Fig.
. Fig. . π –A π –C π –B C B B O M L ˆ a ˆ b M L β π –A π –C π –B C B Vector - - Applications of Vector Algebra On Simplification, we obtain BC CA = CA AB ...
( ) Equations ( ) and ( ), we get AB BC = CA AB BC CA So, | AB BC = | CA AB BC CA . Then, we get sin( ca B π − sin( sin( bc ab C That is, ca B = bc ab C . Dividing by abc , leads to sin A = sin B C or B C Example . Prove by vector method that sin( β β β Let ˆ a OA and b OB be the unit vectors making angles α and β respectively, with positive x -axis, where A and B are as shown in the Fig.
. . Then, we get and ˆ β β The angle between ˆ a and ˆ b is α β and, the vectors , , b a k form a right-handed