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If \( \frac{d}{d t} \vec{u}=\vec{w} \times \vec{u} \) and \( \frac{d}{d t} \vec{v}=\vec{w} \times \vec{v} \), then show that \( \frac{d}{d t}(\vec{u} \times \vec{v})=\vec{w} \times(\vec{u} \times \vec{v}) \). |
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Answer» d/dt \(\vec u\) = \(\vec w\) x \(\vec u\) and d/dt \(\vec v\) = \(\vec w\) x \(\vec v\) Let \(\vec u\) = u1\(\hat i\) + u2\(\hat j\) + u3\(\hat k\) and \(\vec v\) = v1\(\hat i\) + v2\(\hat j\) + v3\(\hat k\) \(\vec w\) = w1\(\hat i\) + w2\(\hat j\) + w3\(\hat k\) d\(\vec u\)/dt = du1/dt \(\hat i\) + du2/dt \(\hat j\) + du3/dt \(\hat k\) = \(\vec w\) x \(\vec u\) ⇒ \(\begin{vmatrix}\hat i&\hat j&\hat k\\w_1&w_2&w_3\\u_1&u_2&u_3\end{vmatrix}\) = du1/dt \(\hat i\) + du2/dt \(\hat j\) + du3/dt \(\hat k\) ⇒ \(\hat i\)(w2 u3 - u2 w3) + \(\hat j\)(w3 u1 - u3 w1) + \(\hat k\)(w1 u2 - u1 w2) = du1/dt \(\hat i\) + du2/dt \(\hat j\) + du3/dt \(\hat k\) ⇒ du1/dt = w2 u3 - u2 w3, du2/dt = w3 u1 - u3 w1 and du3/dt = w1 u2 - u1 w2 Similarly given d\(\vec v\)/dt = \(\vec w\) x \(\vec v\) So we get dv1/dt = w2 v3 - v2 w3, dv2/dt = w3 v1 - v3 w1 and dv3/dt = w1 v2 - v1 w2 L.H.S. = d/dt (\(\vec u\) x \(\vec v\)) = d/dt \(\begin{vmatrix}\hat i&\hat j&\hat k\\u_1&u_2&u_3\\v_1&v_2&v_3\end{vmatrix}\) = d/dt (\(\hat i\)(u2 v3 - v2u3) + \(\hat j\)(u3 v1 - v3 u1) + \(\hat k\)(u1 v2 - v1 u2)) = \(\hat i\)(du2/dt v3 + u2 dv3/dt - dv2/dt u3 - v2 du3/dt) + \(\hat j\)(du3/dt v1 + u3 dv1/dt - dv3/dt u1 - v3 du1/dt) + \(\hat k\)(du1/dt v2 + u1 dv2/dt - dv1/dt u2 - v1 du2/dt) = [\(\hat i\)(du2/dt u3 - v2 du3/dt) + \(\hat j\)(du3/dt v1 - v3 du1/dt) + \(\hat k\)(du1/dt v2 - v1 du2/dt)] + [\(\hat i\)(u2 dv3/dt - dv2/dt u3) + \(\hat j\)(u3 dv1/dt - dv3/dt u1) + \(\hat k\)(u1 dv2/dt - dv1/dt u2)] = (d\(\vec u\)/dt x \(\vec v\)) + (\(\vec u\) x d\(\vec v\)/dt) = (\(\vec w\) x \(\vec u\)) x \(\vec v\)+ (\(\vec u\) x (\(\vec w\) x \(\vec v\))) = -\(\vec v\)x (\(\vec w\) x \(\vec u\)) + \(\vec u\) x (\(\vec w\) x \(\vec v\)) (∵ \(\hat a\) x \(\hat b\) = -\(\hat b\) x \(\hat a\)) = -((\(\vec v\) . \(\vec u\))\(\vec w\) - (\(\vec v\) . \(\vec w\)) . \(\vec u\)) + (\(\vec u\) . \(\vec v\))\(\vec w\) - (\(\vec u\) . \(\vec w\))\(\vec v\) (∵ \(\hat a\) x \(\hat b\) x \(\hat c\) = (\(\hat a\) . \(\hat c\))\(\hat b\) - (\(\hat a\) . \(\hat b\)) \(\hat c\)) = -(\(\vec u\) . \(\vec v\)) \(\vec w\) + (\(\vec v\) . \(\vec w\))\(\vec u\) + (\(\vec u\) . \(\vec v\))\(\vec w\) - (\(\vec u\) . \(\vec w\))\(\vec v\) = (\(\vec v\) . \(\vec w\))\(\vec u\)) - (\(\vec u\) . \(\vec w\))\(\vec v\) = (\(\vec w\) . \(\vec v\))\(\vec u\) - (\(\vec w\) . \(\vec u\))\(\vec v\) (∵ \(\hat a\) . \(\hat b\) = \(\hat b\) . \(\hat a\)) = \(\vec w\) x (\(\vec u\) x \(\vec v\)) Hence Proved. |
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