Find the area of the parallelogram whose adjacent sides are given by the vectors $a = 3 ˆ i + ˆ j + 4 ˆ k$ and $b = ˆ i - ˆ j + ˆ k$ .

\sqrt{42} s q . u n i t s

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\sqrt{40} s q . u n i t s

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\sqrt{45} s q . u n i t s

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\sqrt{35} s q . u n i t s

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Solution

The correct option is B $\sqrt{42} s q . u n i t s$
$a = 3 ˆ i + ˆ j + 4 ˆ k$
$b = ˆ i - ˆ j + ˆ k$
Area of parallelogram = $∣ ∣ ∣ \to a \times \to b ∣ ∣ ∣$
$\to a \times \to b$ = $∣ ∣ ∣ ∣ ∣ \begin{matrix} ˆ i & ˆ j & ˆ k 3 & 1 & 4 1 & - 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$
= $5 ˆ i + ˆ j - 4 ˆ k$
Hence,
Area of parallelogram = $∣ ∣ ∣ \to a \times \to b ∣ ∣ ∣$ = $\sqrt{5^{2} + 1^{2} + (- 4)^{2}}$ = $\sqrt{42}$ sq. units.

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Similar questions

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

\to a = 3^i +^j + 4^k

\to b =^i -^j +^k

\to a = ˆ i + 2 ˆ j + 3 ˆ k, a n d \to b = - 3 ˆ i - 2 ˆ j + ˆ k

a = -^i - 2^j - 3^k

b = -^i + 2^j - 3^k

\to a =^i -^j + 3^k

\to b = 2^i - 7^j +^k

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