A) \[\mathbf{r}=x\mathbf{a}+\frac{1}{\mathbf{a}\,.\,\,\mathbf{a}}(\mathbf{a}\times \mathbf{b})\]
B) \[\mathbf{r}=x\mathbf{b}-\frac{1}{\mathbf{b}\,.\,\,\mathbf{b}}(\mathbf{a}\times \mathbf{b})\]
C) \[\mathbf{r}=x\mathbf{a}\times \mathbf{b}\]
D) \[\mathbf{r}=x\mathbf{b}\times \mathbf{a}\]
Correct Answer: A
Solution :
Since \[\mathbf{a},\,\,\mathbf{b}\] and \[\mathbf{a}\times \mathbf{b}\] are non-coplanar, hence \[\mathbf{r}=x\mathbf{a}+y\mathbf{b}+z(\mathbf{a}\times \mathbf{b})\] for some scalars \[x,\,\,y\] and \[z.\] Now, \[\mathbf{b}=\mathbf{r}\times \mathbf{a}=\left\{ x\mathbf{a}+y\mathbf{b}+z(\mathbf{a}\times \mathbf{b}) \right\}\times \mathbf{a}\] \[=y(\mathbf{b}\times \mathbf{a})+z[(\mathbf{a}\times \mathbf{b})\times \mathbf{a}]\]\[=-y(\mathbf{a}\times \mathbf{b})-z\,[\mathbf{a}\times (\mathbf{a}\times \mathbf{b})]\] \[=-y(\mathbf{a}\times \mathbf{b})-z\,[(\mathbf{a}\,.\,\mathbf{b})\mathbf{a}-(\mathbf{a}\,.\,\mathbf{a})\,\mathbf{b}]\] \[=-y(\mathbf{a}\times \mathbf{b})+z(\mathbf{a}\,.\,\mathbf{a})\mathbf{b}\], \[\left\{ \because \,\mathbf{a}\,.\,\mathbf{b}=0 \right\}\] \[\Rightarrow y=0\] and \[z=\frac{1}{(\mathbf{a}\,.\,\mathbf{a})}\Rightarrow \mathbf{r}=x\mathbf{a}+\frac{1}{\mathbf{a}\,.\,\mathbf{a}}(\mathbf{a}\times \mathbf{b})\].
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