If A is the area of the parallelogram having diagonals $3 i + j - 2 k$ and $i - 3 j + 4 k .$ Find

3 \sqrt{2} .

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6 \sqrt{2} .

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5 \sqrt{3} .

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10 \sqrt{3} .

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Solution

The correct option is B $5 \sqrt{3} .$
Area of a parallelogram of sides $a$ and $b$ is $| a \times b | .$
If $d_{1}$ and $d_{2}$ be its diagonals then $d_{1} = a + b$ and $d_{2} = a - b .$
Area of a parallelogram whose sides are $d_{1}$ and $d_{2}$ is $(d_{1} \times d_{2})$
$= [(a + b) \times (a - b)] = - 2 (a \times b)$
$= 2 \times$ Area of parallelogram of sides $a, b$
$∴$ Area of parallelogram $= \frac{1}{2} (d_{1} \times d_{2})$
Here $\frac{1}{2} (d_{1} \times d_{2}) = - i - 7 j - 5 k$
$∴$ Area $= \sqrt{(1 + 49 + 25)} = \sqrt{75} = 5 \sqrt{3} .$

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