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If the vectors AB=3hati+4hatk and AC=(5hati-2hatj+4hatk) are the sides of a DeltaABC, then the length of the median through A is |
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Answer» `sqrt(18)` `therefore AB+BC=CA=0` `rArr BC=AC-AB` `rArr BM=(AC-AB)/(2) `[SINCE, M is mid-point of BC] ALSO, `AB+BM+MA=0 ` [ property of triangle] `rArr AB+(AC-AB)/(2) =AM` `rArr AM=(AB+AC)/(2) = AM` `rArr AM=(AB+AC)/(2) = (3 hati+4hatk+5hati-2hatj+4hatk)/(2)` `=4hati - hatj +4 hatk` `rArr |AM| = sqrt(4^(2)+1^(2)+4^(2)) = sqrt(33)` |
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