The vectors $A = 3 i - k, B = i + 2 j$ are adjacent sides of a parallelogram. Find the area.

\sqrt{17}

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\sqrt{29}

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\sqrt{31}

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\sqrt{41}

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Solution

The correct option is C $\sqrt{41}$
Area of parallelogram is $= \to A \times \to B$
$= (3 i + 0 j - k) \times (i + 2 j + 0 k)$
$= 2 i - j + 6 k$
Hence $\to A r e a = 2 i - j + 6 k$
$| \to A r e a | = | 2 i - 6 j + k |$
$= \sqrt{41}$ sq units.

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