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 toA. `pi/2`B. `pi/5`C. `pi/6`D. `pi/3`

If `hat(i)+hat(j), hat(j)+hat(k), hat(i)+hat(k)` are the position vect

1.	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 toA. `pi/2`B. `pi/5`C. `pi/6`D. `pi/3`
Answer» Correct Answer - D Let position vector of the vertices are `OA = hati +hatj , OB = hatj + hatk` and `OC = hati + hatk` Now, `AB = - hati + hatk` and `AC = hatk - hati` `:.cos theta = ((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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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 toA. `pi/2`B. `pi/5`C. `pi/6`D. `pi/3`

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