InterviewSolution
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InterviewSolution
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If the vertices A,B,C of a triangle ABC are (1,2,3),(-1,0,0) ,(0,1,2) , respectively, then find `angleABC`. |
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Answer» The vertices of `triangleABC` are given as A(1,2,3) , B(-1,0,0), and C ( 0,1,2) Also, it is given that `angle ABC` is the angle between the vectors `vec(BA) and vec(BC)` thus `vec(BA)={1-(-1)}hati+(2-0)hatj+ (3-0)hatk` `2hati+ 2hatj + 3hatk` `vec(BC)={0-(-1)}hati+(1-0)hatj + (2-0)hatk` `= hati + hatj + 2hatk` `vec(BA).vec(BC)=(2hati+ 2hatj + 3hatk). (hati + hatj + 2hatk)` `2xx1+2xx1+3xx2` `= 2 + 2 +6 =10` `|vec(BA)|=sqrt(2^(2)+2^(2)+3^(2))=sqrt(4+4+9)=sqrt17` `|vec(BA)|=sqrt(1+1+2^(2))=sqrt6` Now, it is known that : `vec(BA). vec(BC)= |vec(BA)||vec(BC)|cos(angleABC)` `10 =sqrt17 xx sqrt6 cos(angleABC)` `or cos(angleABC) = 10/(sqrt17xxsqrt6)` `or angleABC = cos^(-1)(10/sqrt102)` |
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