The area of a parallelogram whose adjacent sides are determined by the vectors $\to a = ˆ i + 2 ˆ j + 3 ˆ k, a n d \to b = - 3 ˆ i - 2 ˆ j + ˆ k$

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Solution

$\begin{matrix} \to a = ˆ i + 2 ˆ j + 3 ˆ k, a n d \to b = - 3 ˆ i - 2 ˆ j + ˆ k ∣ ∣ ∣ \to a \times \to b ∣ ∣ ∣ \to a \times \to b = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ˆ i ˆ j ˆ k 123 - 3 - 21 \end{matrix} ∣ ∣ ∣ ∣ ∣ = (2 i - 9 j - 2 k) + (6 k + 6 i - j) a \times b = 8 i - 10 j + + 4 k = \sqrt{64 + 100 + 16} = \sqrt{180} = 4 \sqrt{5} \end{matrix}$
is the Answer.

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