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If \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) and \(\mathbf{r}\) are vectors such that \(\mathbf{a}\) is not perpendicular to \(\mathbf{b} \cdot \mathbf{r} \times \mathbf{b}=\mathbf{c} \times \mathbf{b}\) and \(\mathbf{r} \cdot \mathbf{a}=\mathbf{0}\) then \(\mathbf{r}=\)
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Verified Answer
The correct answer is:
\(\mathbf{c}-\frac{(\mathbf{c} \cdot \mathbf{a})}{(\mathbf{b} \cdot \mathbf{a})} \mathbf{b}\)
Given,
\(\begin{aligned}
& \mathbf{r} \times \mathbf{b}=\mathbf{c} \times \mathbf{b} \\
& \Rightarrow \quad(\mathbf{r}-\mathbf{c}) \times \mathbf{b}=\mathbf{0} \\
& \Rightarrow \quad \mathbf{r}-\mathbf{c} \text { is parallel to } \mathbf{b} \text {. } \\
& \Rightarrow \quad(\mathbf{r}-\mathbf{c})=\lambda \mathbf{b} \\
& \mathbf{r}=\mathbf{c}+\lambda \mathbf{b} \\
\end{aligned}\)
or \(\mathbf{r}=\mathbf{c}+\lambda \mathbf{b}\) ...(i)
Also, \(\mathbf{r} \cdot \mathbf{a}=0 \Rightarrow(\mathbf{c}+\lambda \mathbf{b}) \cdot \mathbf{a}=0\)
\(\Rightarrow \quad \mathbf{c} \cdot \mathbf{a}+\lambda \mathbf{b} \cdot \mathbf{a}=\mathbf{0}\)
or \(\lambda=-\left(\frac{\mathbf{c} \cdot \mathbf{a}}{\mathbf{b} \cdot \mathbf{a}}\right)\) ...(ii)
So, from Eq. (i) we substitute ' \(\lambda\) ' from Eq. (ii), we get
\(\mathbf{r}=\mathbf{c}-\left(\frac{\mathbf{c} \cdot \mathbf{a}}{\mathbf{b} \cdot \mathbf{a}}\right) \mathbf{b}\)
\(\begin{aligned}
& \mathbf{r} \times \mathbf{b}=\mathbf{c} \times \mathbf{b} \\
& \Rightarrow \quad(\mathbf{r}-\mathbf{c}) \times \mathbf{b}=\mathbf{0} \\
& \Rightarrow \quad \mathbf{r}-\mathbf{c} \text { is parallel to } \mathbf{b} \text {. } \\
& \Rightarrow \quad(\mathbf{r}-\mathbf{c})=\lambda \mathbf{b} \\
& \mathbf{r}=\mathbf{c}+\lambda \mathbf{b} \\
\end{aligned}\)
or \(\mathbf{r}=\mathbf{c}+\lambda \mathbf{b}\) ...(i)
Also, \(\mathbf{r} \cdot \mathbf{a}=0 \Rightarrow(\mathbf{c}+\lambda \mathbf{b}) \cdot \mathbf{a}=0\)
\(\Rightarrow \quad \mathbf{c} \cdot \mathbf{a}+\lambda \mathbf{b} \cdot \mathbf{a}=\mathbf{0}\)
or \(\lambda=-\left(\frac{\mathbf{c} \cdot \mathbf{a}}{\mathbf{b} \cdot \mathbf{a}}\right)\) ...(ii)
So, from Eq. (i) we substitute ' \(\lambda\) ' from Eq. (ii), we get
\(\mathbf{r}=\mathbf{c}-\left(\frac{\mathbf{c} \cdot \mathbf{a}}{\mathbf{b} \cdot \mathbf{a}}\right) \mathbf{b}\)
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