A parallelogram whose sides are $2 i + 4 j - 5 k$ and $i + 2 j + 3 k .$ Find the area of the $∥^{m} .$

11 \sqrt{5}

sq.units.

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\sqrt{55}

sq.units.

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5 \sqrt{11}

sq.units.

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11 \sqrt{5} / 2

sq.units.

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Solution

The correct option is A $11 \sqrt{5}$ sq.units.
If $a$ and $b$ be the adjacent sides of a parallelogram then the diagonals are
$\to A C = a + b, \to B D = b - a$
$∴$ Unit vectors are $\frac{\to A C}{∣ ∣ \to A C ∣ ∣}$ and $\frac{\to B D}{∣ ∣ \to B D ∣ ∣} .$
Area of parallelogram $= | a \times b | .$
Ans. $\frac{1}{7} (3 i + 6 j - 2 k) \times \frac{1}{\sqrt{(69)}} (- i - 2 j + 8 k) = 11 \sqrt{5}$ sq.units.

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A parallelogram whose sides are 2i+4j−5k and i+2j+3k. Find the area of the ∥m.

The correct option is A 11√5 sq.units.If a and b be the adjacent sides of a parallelogram then the diagonals are →AC=a+b,→BD=b−a∴ Unit vectors are →AC∣∣→AC∣∣ and →BD∣∣→BD∣∣.Area of parallelogram =|a×b|. Ans.17(3i+6j−2k)×1√(69)(−i−2j+8k)=11√5 sq.units.

A parallelogram whose sides are $2 i + 4 j - 5 k$ and $i + 2 j + 3 k .$ Find the area of the $∥^{m} .$