InterviewSolution
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Express cross product of two vectors in cartesian co-ordinate. |
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Answer» Let \(\vec A\) and \(\vec B\) be the two vectors. \(\vec A\) =\(\vec{(A_x}\hat{i}+\vec{A_y}\hat{j}+\vec{A_z}\hat{k})\) and \(\vec R\) = \(\vec{(B_x}\hat{i}+\vec{B_y}\hat{j}+\vec{B_z}\hat{k})\) \(\vec A\) × \(\vec B\) =\(\vec{(A_x}\hat{i}+\vec{B_y}\hat{j}+\vec{B_z}\hat{k})\) AxBx(\(\hat{i}\times\hat{i}\)) + AxBy(\(\hat{i}\times\hat{j}\)) + AxBy(\(\hat{i}\times\hat{k}\)) + AyBx(\(\hat{j}\times\hat{i}\)) + AyBy(\(\hat{j}\times\hat{j}\)) + AyBz (\(\hat{j}\times\hat{k}\)) + AzBx(\(\hat{k}\times\hat{i}\)) + AzBx(\(\hat{k}\times\hat{j}\)) + AzBy (\(\hat{k}\times\hat{k}\)) or\(\vec{A}\) × \(\vec{B}\) = (AyBz − AzBy)\(\hat{i}\)– (AxBy − AzBx)\(\hat{j}\)+ (AxBy − AyBx)\(\hat{k}\) or \(\vec{A}\) × \(\vec{B}\) = AxBx(0) + AxBy(\(\hat{k}\)) + AxBz (\(-\hat{j}\)) + ̂ AyBx(\(-\hat{k}\)) + AyBy(0) + AyBz (\(\hat{i}\)) + AzBx(\(\hat{j}\)) + AzBy(\(-\hat{i}\)) + AzBz(0) It can be written in determinant from as \(\vec{A}\) ×\(\vec{B}\) = \(\begin {vmatrix} \hat{i} & \hat{j} &\hat{k} \\ A_x &A_y&A_z \\ A_x&B_y&B_z \end{vmatrix}\) |
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