The sides of a parallelogram are $2 i + 4 j - 5 k$ and $i + 2 j + 3 k .$ Find the unit vectors parallel to the diagonals.

\frac{3 i + 6 j - 2 k}{\sqrt{49}} .

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\frac{- 3 i - 6 j + 2 k}{\sqrt{49}} .

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\frac{- i - 2 j + 8 k}{\sqrt{69}} .

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\frac{i + 2 j - 8 k}{\sqrt{69}} .

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Solution

The correct option is B $\frac{- i - 2 j + 8 k}{\sqrt{69}} .$
$\to A B = 2 i + 4 j - 5 k = 2, 4 - 5 \to A D = i + 2 j + 3 k = 1, 2, 3$
$∴ \to A C = \to A B + \to B C = 3, 6, - 2$
$∴$ Unit vector along AC
$\frac{\to A C}{| A C |} = \frac{3, 6, - 2}{\sqrt{(9 + 36 + 4)}} = \frac{3 i + 6 j - 2 k}{7}$

Similarly $\to B D = \to A D - \to A B = - 1, - 2, 8.$
$∴$ Unit vector along BD
$\frac{\to B D}{∣ ∣ \to B D ∣ ∣} = \frac{- 1, - 2, 8}{\sqrt{(1 + 4 + 64)}} = \frac{- i - 2 j + 8 k}{\sqrt{69}} .$

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