Solution 16.10a

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m (New page: <math>\begin{align} & m(4\mathbf{i}+V\mathbf{j})+\lambda m(2\mathbf{i}-V\mathbf{j})=(m+\lambda m)(3\mathbf{i}+2\mathbf{j}) \\ & (4m+2\lambda m-3m)\mathbf{i}+(mV-\lambda mV-2\lambda m)\mat...)
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<math>\begin{align}
<math>\begin{align}
-
& m(4\mathbf{i}+V\mathbf{j})+\lambda m(2\mathbf{i}-V\mathbf{j})=(m+\lambda m)(3\mathbf{i}+2\mathbf{j}) \\
+
& m(5\mathbf{i}+V\mathbf{j})+\lambda m(2\mathbf{i}-V\mathbf{j})=(m+\lambda m)(3\mathbf{i}-2\mathbf{j}) \\
-
& (4m+2\lambda m-3m)\mathbf{i}+(mV-\lambda mV-2\lambda m)\mathbf{j}=0\mathbf{i}+0\mathbf{j} \\
+
& (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} \\
\end{align}</math>
\end{align}</math>
 +
Consider the
Consider the
<math>\mathbf{i}</math>
<math>\mathbf{i}</math>
-
component. That is the <math>\mathbf{i}</math> component on the left hand side must be the same as the <math>\mathbf{i}</math> component on the right hand side.
+
component. That is the <math>\mathbf{i}</math> component on the left hand side must be the same as the <math>\mathbf{i}</math> component on the right hand side, that is zero.
<math>\begin{align}
<math>\begin{align}
-
& 3m+2\lambda m-4=0 \\
+
& 5m+2\lambda m-3m-3\lambda m=0 \\
-
& 2\lambda -1=0 \\
+
& 2-\lambda =0 \\
-
& \lambda =\frac{1}{2} \\
+
& \lambda =2 \\
\end{align}</math>
\end{align}</math>

Revision as of 16:23, 1 October 2010

\displaystyle \begin{align} & m(5\mathbf{i}+V\mathbf{j})+\lambda m(2\mathbf{i}-V\mathbf{j})=(m+\lambda m)(3\mathbf{i}-2\mathbf{j}) \\ & (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} \\ \end{align}


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

\displaystyle \begin{align} & 5m+2\lambda m-3m-3\lambda m=0 \\ & 2-\lambda =0 \\ & \lambda =2 \\ \end{align}

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