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  • Let be three vectors mutually perpendicular to each other and have same magnitude. If a vector

Let \vec{a}, \vec{b}, \vec{c} be three vectors mutually perpendicular to each other and have same magnitude. If a vector \vec{r} satisfies
\vec{a} \times\{(\vec{r}-\vec{b}) \times \vec{a}\}+\vec{b} \times\{(\vec{r}-\vec{c}) \times \vec{b}\}+\vec{c} \times\{(\vec{r}-\vec{a}) \times \vec{c}\}=\overrightarrow{0},then \vec{r} is equal to :
Option: 1 \begin{aligned} &\frac{1}{3}(\vec{a}+\vec{b}+\vec{c}) \\ \end{aligned}
Option: 2 \frac{1}{3}(2 \vec{a}+\vec{b}-\vec{c}) \\
Option: 3 \frac{1}{2}(\vec{a}+\vec{b}+\vec{c}) \\
Option: 4 \frac{1}{2}(\vec{a}+\vec{b}+2 \vec{c})

Answers (1)

best_answer

|\vec{a}|=|\vec{b}|=|\vec{c}|\: and \: \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=\vec{c} \cdot \vec{a}=0\\

Let\: \vec{r}= x\vec{a}+y \vec{b}+z \vec{c} \\

where \:\: \vec{r}\cdot \vec{a}=x|\vec{a}|^{2}, \vec{r} \cdot \vec{b}=y|\vec{b}|^{2}, \vec{r} \cdot \vec{c}=z|\vec{c}|^{2}

Give expression is

(\vec{a} \times(\vec{r} \times \vec{a}))-(\vec{a} \times(\vec{b} \times \vec{a}))+\vec{b} \times(\vec{r} \times \vec{b})-\vec{b} \times(\vec{c} \times \vec{b})+\\
\vec{c} \times(\vec{r} \times \vec{c})-(\vec{c} \times(\vec{a}\times \vec{c))=0} \\

\Rightarrow(\vec{a} \cdot \vec{r}) \vec{a}-|\vec{a}|^{2} \vec{r}-(\vec{a} \cdot \vec{b}) \vec{a}+|\vec{a}|^{2} \vec{b}+(\vec{b} \cdot \vec{r}) \vec{b}-|\vec{b}|^{2} \vec{r}-\\
(\vec{b} \cdot \vec{c}) \vec{b}+|\vec{b}|^{2}\vec{c}+(\vec{c}\cdot \vec{r})\vec{c}-|\vec{c}|^{2}\vec{r}-(\vec{c}\cdot \vec{a})\vec{a}+|\vec{c}|^{2}\vec{a}=0\\

\Rightarrow x|\vec{a}|^{2}\vec{a}+y|\vec{b}|^{2}\vec{b}+z|\vec{c}|^{2}\vec{c}-\vec{r}(|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2})+\\
|\vec{a}|^{2}\vec{b}+|\vec{b}|^{2}\vec{c}+|\vec{c}|^{2}\vec{a}=0\\

\Rightarrow |\vec{a}|^{2}(x\vec{a}+y\vec{b}+z\vec{c})-3|\vec{a}|^{2}\vec{r}+|\vec{a}|^{2}(\vec{a}+\vec{b}+\vec{c})=0\\

\Rightarrow 3\vec{r}-\vec{r}=\vec{a}+\vec{b}+\vec{c}\\

\Rightarrow \vec{r}=\frac{1}{2}\left ( \vec{a}+\vec{b}+\vec{c} \right )

Posted by

Kuldeep Maurya

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