A) \[-3\hat{i}+9\hat{j}+5\hat{k}\]
B) \[3\hat{i}-9\hat{j}-5\hat{k}\]
C) \[\frac{1}{2}\left( -3\hat{i}+9\hat{j}+5\hat{k} \right)\]
D) \[\frac{1}{2}\left( 3\hat{i}-9\hat{j}+5\hat{k} \right)\]
Correct Answer: C
Solution :
\[\vec{\alpha }=3\hat{i}+\hat{j}\] \[\vec{\beta }=2\hat{i}-\hat{j}+3\hat{k}\] \[\vec{\beta }={{\vec{\beta }}_{1}}-{{\vec{\beta }}_{2}}\] \[{{\vec{\beta }}_{1}}=\lambda \left( 3\hat{i}+\hat{j} \right),{{\vec{\beta }}_{2}}=\lambda \left( 3\hat{i}+\hat{j} \right)-2\hat{i}+j-3\hat{k}\] \[{{\vec{\beta }}_{2}}.\vec{\alpha }=0\] \[(3\lambda -2).3+(\lambda +1)=0\] \[9\lambda -6+\lambda +1=0\] \[\lambda =\frac{1}{2}\] \[\Rightarrow \]\[{{\vec{\beta }}_{1}}=\frac{3}{2}\hat{i}+\frac{1}{2}\hat{j}\] \[\Rightarrow \]\[{{\vec{\beta }}_{2}}=-\frac{1}{2}\hat{i}+\frac{3}{2}\hat{j}-3\hat{k}\] Now\[{{\vec{\beta }}_{1}}\times {{\vec{\beta }}_{2}}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ \frac{3}{2} & \frac{1}{2} & 0 \\ -\frac{1}{2} & \frac{3}{2} & -3 \\ \end{matrix} \right|\] \[=\hat{i}\left( -\frac{3}{2}-0 \right)-\hat{j}\left( -\frac{9}{2}-0 \right)+\hat{k}\left( \frac{9}{4}+\frac{1}{4} \right)\] \[=-\frac{3}{2}\hat{i}+\frac{9}{2}\hat{j}+\frac{5}{2}\hat{k}\] \[=\frac{1}{2}\left( -3\hat{i}+9\hat{j}+5\hat{k} \right)\] Aliter: \[\vec{\beta }={{\vec{\beta }}_{1}}-{{\vec{\beta }}_{2}}\Rightarrow \vec{\beta }.\hat{\alpha }={{\vec{\beta }}_{1}}.\hat{\alpha }=\left| {{{\vec{\beta }}}_{1}} \right|\] \[\Rightarrow \]\[{{\vec{\beta }}_{1}}=\left( \vec{\beta }.\hat{\alpha } \right)\hat{\alpha }\] \[\Rightarrow \]\[{{\vec{\beta }}_{2}}=\left( \vec{\beta }.\hat{\alpha } \right)\hat{\alpha }-\vec{\beta }\] \[\Rightarrow \]\[{{\vec{\beta }}_{1}}\times {{\vec{\beta }}_{2}}=-\left( \vec{\beta }.\hat{\alpha } \right)\hat{\alpha }\times \vec{\beta }\] \[=\frac{-5}{10}\left( 3\hat{i}+\hat{j} \right)\times \left( 2\hat{i}-\hat{j}+3\hat{k} \right)\] \[=\frac{1}{2}\left( -3\hat{i}+9\hat{j}+5\hat{k} \right)\]
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