1. Tardigrade
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  4. The vectors AB = 3i + 5j + 4k and AC = 5i − 5j + 2k are side of a Δ ABC. The length of the median through A is

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Q. The vectors $AB = 3i + 5j + 4k$ and $AC = 5i − 5j + 2k$ are side of a $\Delta \,ABC$. The length of the median through $A$ is

UPSEEUPSEE 2012

Solution:

Position vector of AD
$=\frac{(3+5) i +(5-5) j +(4+2) k }{2}$
image
$A D =\frac{8 i +6 k }{2}=4 i +3 k$
$\therefore $ Length of median $=| A D |$
$=\sqrt{16+9}=5$ units