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If $i + 2 j + 3 k, 3 i + 2 j + k$ are sides of a parallelogram then a unit vector parallel to one of the diagonals of the parallelogram is

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$\frac{i + j - k}{\sqrt{3}}$

$\frac{i + j + k}{\sqrt{3}}$

$\frac{- i + j + k}{\sqrt{3}}$

$\frac{i - j + k}{\sqrt{3}}$

answer is A.

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Detailed Solution

$a = i + 2 j - 3 k b = 3 i + 2 j + k AC = a + b = 4 i + 4 j + 4 k \overset{}{\vec{AC}} = \frac{4 (i + j + k)}{4 \sqrt{3}} = \frac{i + j + k}{\sqrt{3}}$

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