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parallelogram law of vel Two vectors \( \operatorname{are} \vec{A}=3 \vec{\imath}-3 \vec{\jmath}+\vec{k} \) and \( \vec{B}=4 \vec{\imath}+9 \vec{\jmath}+2 \vec{k} \). Solve \( \vec{A} X \vec{B} \) |
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Answer» We have \(\vec A\) = \(3\hat i\) - \(3\hat j\) + \(\hat k\) \(\vec B\) = \(4\hat i\) + \(9\hat j\) + \(2\hat k\) \(\vec A \times \vec B\) = (\(3\hat i\) - \(3\hat j\) + \(\hat k\)) x ( \(4\hat i\) + \(9\hat j\) + \(2\hat k\)) = \(\begin{vmatrix} \hat i &\hat j &\hat k \\[0.3em] 3 &-3 &1 \\[0.3em] 4 & 9 & 2 \end{vmatrix}\) = \(\hat i\begin{vmatrix} -3 &1 \\[0.3em] 9 &2 \end{vmatrix}\) - \(\hat j\begin{vmatrix} 3 &1 \\[0.3em] 4 &2 \end{vmatrix}\) + \(\hat k\begin{vmatrix} 3 &-3 \\[0.3em] 4 &9 \end{vmatrix}\) = \(\hat i\)(-3 x 2 - 9 x 1) - \(\hat j\)(3 x 2 - 4 x 1) + \(\hat k\)(3 x 9 - 4 x -3) = \(\hat i\)(- 6 - 9) - \(\hat j\)(6 - 4) + \(\hat k\) (27 + 12) = \(-15\hat i\) - \(2\hat j\) + \(39\hat k\). |
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