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The position vectors of vertices of a `DeltaABC` are `4hati - 2 hatj , hati + 4hatj - 3hatk` and `-hati + 5hatj + hatk` respectively , then `angleABC` is equal toA. `(pi)/(6)`B. `(pi)/(4)`C. `(pi)/(3)`D. `(pi)/(2)` |
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Answer» Correct Answer - D Here , `vec(AB) = - 3hati + 6hatj - 3hatk , vec(BC) = - 2hati + hatj + 4 hatk ` and `vec(AB ) * vec(BC) = 6 + 6 - 12 = 0` `implies angle ABC = (pi)/(2)` |
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