The area of a parallelogram whose adjacent sides are represented by the vectors
$a = -^i - 2^j - 3^k$ and $b = -^i + 2^j - 3^k$ is

\sqrt{14}

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\sqrt{6}

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\frac{49}{36}

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4 \sqrt{10}

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Solution

The correct option is D $4 \sqrt{10}$
$∴$ Area of parallelogram $= | a \times b |$
$= ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k - 1 & - 2 & - 3 - 1 & 2 & - 3 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$= ∣ ∣^i (6 + 6) -^j (3 - 3) +^k (- 2 - 2) ∣ ∣ = ∣ ∣ 12^i - 4^k ∣ ∣$
$= \sqrt{12^{2} + 4^{2}} = 4 \sqrt{10}$

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Similar questions

^i + 2^j + 3^k

3^i - 2^j +^k

\to A

^i + 2^j + 3^k

\to B

2^i - 3^j +^k

^i +^j + 3^k

-^i - 2^j +^k

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

3^i -^k

^i + 2^j

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