Solution 16.10a - Revision history

Ian at 16:42, 27 March 2011

2011-03-27T16:42:09Z

←Older revision		Revision as of 16:42, 27 March 2011
Line 7:		Line 7:
	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, that is zero.	+	component. 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}

Ian at 16:23, 1 October 2010

2010-10-01T16:23:58Z

←Older revision		Revision as of 16:23, 1 October 2010
Line 1:		Line 1:
	<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>

Ian: New page: $\begin{align} & m(4\mathbf{i}+V\mathbf{j})+\lambda m(2\mathbf{i}-V\mathbf{j})=(m+\lambda m)(3\mathbf{i}+2\mathbf{j}) \\ & (4m+2\lambda m-3m)\mathbf{i}+(mV-\lambda mV-2\lambda m)\mat...$

2010-09-29T17:39:25Z

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...

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)\mathbf{j}=0\mathbf{i}+0\mathbf{j} \\
\end{align}</math>

Consider the
<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.

<math>\begin{align}
& 3m+2\lambda m-4=0 \\
& 2\lambda -1=0 \\
& \lambda =\frac{1}{2} \\
\end{align}</math>

Solution 16.10a - Revision history

Ian at 16:42, 27 March 2011

Ian at 16:23, 1 October 2010

Ian: New page: \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...

Ian: New page: $\begin{align} & m(4\mathbf{i}+V\mathbf{j})+\lambda m(2\mathbf{i}-V\mathbf{j})=(m+\lambda m)(3\mathbf{i}+2\mathbf{j}) \\ & (4m+2\lambda m-3m)\mathbf{i}+(mV-\lambda mV-2\lambda m)\mat...$