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If hat(i)+hat(j), hat(j)+hat(k), hat(i)+hat(k) are the position vectors of the vertices of a Delta ABC taken in order, then angleA is equal to |
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Answer» `pi/2` `OA = HATI +hatj , OB = hatj + hatk` and `OC = hati + hatk` Now, `AB = - hati+ hatk` and `AC = hatk -hati` `:.costheta = ((AB).(AC))/(|AB||AC|) = ((-hati + hatk).(hatk -hatj))/(SQRT(1^(2) + 1^(2))sqrt(1^(2) +1^(2))) = (1)/(sqrt(2)sqrt(2)) = 1/2` `rArr theta= (pi)/(3)` |
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