Calculate the area of the parallelogram when adjacent sides are given by the vectors $\to A$ = $^i + 2^j + 3^k$ and $\to B$ = $2^i - 3^j +^k$

195 sq unit

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$\sqrt[3]{195}$ sq unit

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$\sqrt{195}$ sq unit

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Solution

The correct option is D
$\sqrt{195}$ sq unit

We know that the area of the parallelogram is equal to the magnitude of the cross product of given vectors.

Now,
$\to A \times \to B = ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 1 & 2 & 3 2 & - 3 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ =^i (2 + 9) +^j (6 - 1) +^k (- 3 - 4) = 11^i + 5^j - 7^k S o a r e a o f p a r a l l e l o g r a m : | \to A \times \to B | = \sqrt{11^{2} + 5^{2} + (- 7)^{2}} = \sqrt{195} s q . u n i t$

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Similar questions

\to A

^i + 2^j + 3^k

\to B

2^i - 3^j +^k

\to A

^i + 2^j + 3^k

\to B

2^i - 3^j +^k

Calculate the area of the parallelogram when adjacent sides are given by the vectors $\to A$ = $^i + 2^j + 3^k$ and $\to B$ = $2^i - 3^j +^k$

\to A =^i -^j + 2^k

\to B = 2^i - 3^j + 4^k

^i + 2^j + 3^k

3^i - 2^j +^k

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