The diagonals of a parallelogram are given by the vector $i - 3 j + 4 k$ & $3 i + j + 2 k$ . Find the area of parallelogram.

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Solution

Given that,

The diagonals

$\to A = i - 3 j + 4 k$

$\to B = 3 i + j + 2 k$

Now, the area of parallelogram

$= \frac{1}{4} | (\to A + \to B) \times (\to A - \to B) |$

$= \frac{1}{4} | ((^i - 3^j + 4^k) + (3^i +^j + 2^k) \times (^i - 3^j + 4^k) - (3^i +^j + 2^k))$

$= \frac{1}{4} | (4^i - 2^j + 6^k) \times (- 2^i - 4^j + 2^k) |$

$= \frac{1}{4} | (20^i - 20^j - 20^k)$

$= \frac{1}{4} \sqrt{{(20)}^{2} + {(- 20)}^{2} + {(- 20)}^{2}}$

$= \frac{1}{4} \times 20 \sqrt{3}$

$= 5 \sqrt{3} m^{2}$

Hence, the area of parallelogram is $5 \sqrt{3} m^{2}$

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