1.

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 \( \frac{d}{d t}(\vec{u} \times \vec{v})=\vec{w} \times(\vec{u} \times \vec{v}) \).

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 \(\hat k\) = \(\vec w\)\(\vec u\)

⇒ \(\begin{vmatrix}\hat i&\hat j&\hat k\\w_1&w_2&w_3\\u_1&u_2&u_3\end{vmatrix}\) = 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 = \(\vec w\)\(\vec v\)

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 \(\hat c\) = (\(\hat a\) . \(\hat c\))\(\hat b\) - (\(\hat a\) . \(\hat b\)) \(\hat c\))

= -(\(\vec u\) . \(\vec v\)\(\vec w\) +  (\(\vec v\) . \(\vec w\))\(\vec u\) + (\(\vec u\) . \(\vec v\))\(\vec w\) - (\(\vec u\) . \(\vec w\))\(\vec v\)
(∵ \(\hat a\) . \(\hat b\)  = \(\hat b\) . \(\hat a\))

=  (\(\vec v\) . \(\vec w\))\(\vec u\)) -  (\(\vec u\) . \(\vec w\))\(\vec v\)

=  (\(\vec w\) . \(\vec v\))\(\vec u\) - (\(\vec w\) . \(\vec u\))\(\vec v\) (∵ \(\hat a\) . \(\hat b\) = \(\hat b\) . \(\hat a\))

\(\vec w\) x (\(\vec u\) x \(\vec v\))

Hence Proved.



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