Calculate the cross product between $a = (3, - 3, 1)$ and $b = (4, 9, 2)$ .

- 15 ˆ i - 2 ˆ j + 39 ˆ k

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5 ˆ i - 2 ˆ j + 9 ˆ k

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- ˆ i - 8 ˆ j + 3 ˆ k

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none of these

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Solution

The correct option is A $- 15 ˆ i - 2 ˆ j + 39 ˆ k$
$\to a \times \to b$ = $∣ ∣ ∣ ∣ ∣ \begin{matrix} ˆ i & ˆ j & ˆ k 3 & - 3 & 1 4 & 9 & 2 \end{matrix} ∣ ∣ ∣ ∣ ∣$
= $ˆ i (- 6 - 9) - ˆ j (6 - 4) + ˆ k (27 + 12)$
= $- 15 ˆ i - 2 ˆ j + 39 ˆ k$

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Similar questions

\vec{r} = \hat{i} - \hat{j} + λ (\hat{i} + \hat{j} + \hat{k}) + μ (\hat{i} - 2 \hat{j} + 3 \hat{k})

\vec{r} \cdot (5 \hat{i} - 2 \hat{j} - 3 \hat{k}) = 7

\vec{r} \cdot (5 \hat{i} + 2 \hat{j} - 3 \hat{k}) = 7

\vec{r} \cdot (5 \hat{i} - 2 \hat{j} + 3 \hat{k}) = 7

a = i - 2 j + 3 k, b = 3 i - 6 j + 9 k

a = - 2 i - 4 k, b = i - 2 j - k, c = i - 4 j + 3 k

\vec{a} = \hat{i} + 2 \hat{j} - 3 \hat{k} and \vec{b} = 2 \hat{i} + 4 \hat{j} + 9 \hat{k},

\vec{a} + \vec{b}

\vec{a} = \hat{i} + 2 \hat{j} - \hat{k}, \vec{b} = 3 \hat{i} + 2 \hat{j} + 7 \hat{k}, \vec{c} = 5 \hat{i} + 6 \hat{j} + 5 \hat{k}

\vec{a} = - 4 \hat{i} - 6 \hat{j} - 2 \hat{k}, \vec{b} = - \hat{i} + 4 \hat{j} + 3 \hat{k}, \vec{c} = - 8 \hat{i} - \hat{j} + 3 \hat{k}

\hat{a} = \hat{i} - 2 \hat{j} + 3 \hat{k}, \hat{b} = - 2 \hat{i} + 3 \hat{j} - 4 \hat{k}, \hat{c} = \hat{i} - 3 \hat{j} + 5 \hat{k}

\vec{a} = 3 \hat{i} + 2 \hat{j} + 9 \hat{k} and \vec{b} = \hat{i} + p \hat{j} + 3 \hat{k}

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