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Revision as of 16:29, 1 October 2010 (edit) Ian (Talk | contribs) (New page: From part a) <math>(5m+2\lambda m-3m-3\lambda m)\mathbf{i}+(mV-\lambda mV+2m+2\lambda m)\mathbf{j}=0\mathbf{i}+0\mathbf{j}</math> Consider the <math>\mathbf{j}</math> component. That is...) ← Previous diff		Current revision (16:43, 27 March 2011) (edit) (undo) Ian (Talk | contribs)
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	& V=6 \\		& V=6 \\
	\end{align}</math>		\end{align}</math>
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-	Note that <math>V</math> is a dimensionless number.

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Current revision

From part a)

\displaystyle (5m+2\lambda m-3m-3\lambda m)\mathbf{i}+(mV-\lambda mV+2m+2\lambda m)\mathbf{j}=0\mathbf{i}+0\mathbf{j}

Consider the \displaystyle \mathbf{j} component. That is the \displaystyle \mathbf{j} component on the left hand side must be the same as the \displaystyle \mathbf{j} component on the right hand side, that is zero.

\displaystyle \begin{align} & mV-\lambda mV+2m+2\lambda m=0 \\ & V-2V+2+4=0 \\ & -V+6=0 \\ & V=6 \\ \end{align}