Question

Find the cross product of A = 2i + 3j + k and B = i -j + 2k. *
7i -3j -5k
7i + 3j + 5k
5i -7j +5k
5i +7j +5k​

Answer 1

Answer

Given -

$\sf \vec{A} = 2\hat{\imath} + 3\hat{\jmath} + \hat{k}$

$\sf \vec{B} = \hat{\imath} - {\jmath} + 2\hat{k}$

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To find -

Cross Product of $\sf \vec{A}$ and $\sf \vec{B}$

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Formula used -

$\sf \vec{A} \times \vec{B} = (A_2B_3 - B_2A_3 )\hat{\imath} - (A_1B_3 - B_1A_3)\hat{\jmath} + (A_1B_2 - B_1A_2)\hat{k}$

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Solution -

$\implies$$\sf \vec{A} = 2\hat{\imath} + 3\hat{\jmath} + \hat{k}$

$\implies$$\sf \vec{B} = \hat{\imath} - j\hat{\jmath} +</p><p>2\hat{k}$

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$\sf \vec{A} \times \vec{B} = [(3 \times 2) - ( - 1 \times 1)]\hat{\imath} - [(2 \times 2) - (1 \times 1)]\hat{\jmath} + [(2 \times - 1) - (3 \times 1)]\hat{k}$

$\implies$$\sf = (6 - ( - 1))\hat{\imath} - (4 - 1) \hat{\jmath} + ( - 2 - 3)\hat{k}$

$\implies$$\sf = 7\hat{\imath} - 3\hat{\jmath} - 5\hat{k}$

$\boxed{\sf \vec{A} \times \vec{B} = 7\hat{\imath} - 3\hat{\jmath} - 5\hat{k}}$

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Additional information -

$\implies$$\sf \hat{\imath} \times \hat{\jmath} = \hat{k}$

$\implies$$\sf \hat{k} \times \hat{\jmath} = \hat{\imath}$

$\implies$$\sf \hat{k} \times \hat{\imath} = \hat{\jmath}$

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$\implies$$\sf \hat{\imath} \times \hat{\imath} = \hat{\jmath} \times \hat{\jmath} = \hat{k} \times \hat{k} = 0$

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If $\sf \vec{C} = \sf \vec{A} \times \sf \vec{B}$

Then , $\sf \vec{C}$ is always perpendicular to both $\sf \vec{A}$ and $\sf \vec{B}$.

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