Find the area of the parallelogram whose adjacent sides are determined by the vector $\to a =^i -^j + 3^k$ and $\to b = 2^i - 7^j +^k$ .

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Solution

The area of the parallelogram whose adjacent sides are $\to a$ and $\to b$ is $∣ ∣ \to a \times \to b ∣ ∣$ .
Adjacent sides are given as
$\to a =^i -^j + 3^k$ and $\to b = 2^i - 7^j +^k$
$∴ \to a \times \to b = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 1 & - 1 & 3 2 & - 7 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ =^i (- 1 + 21) -^j (1 - 6) +^k (- 7 + 2) = 20^i + 5^j - 5^k$
$∣ ∣ \to a \times \to b ∣ ∣ = \sqrt{20^{2} + 5^{2} + 5^{2}} = \sqrt{400 + 25 + 25} = 15 \sqrt{2}$
Hence, the area of the given parallelogram is $15 \sqrt{2}$ square units.

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Find the area of the parallelogram whose adjacent sides are determined by the vectors $\to a =^i -^j + 3^k and \to b = 2^i - 7^j +^k$ .

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