MCQ
જો $\vec{a}= \hat{i} + \hat{j} + \hat{k} ,\vec{b} = \hat{i} + \hat{j} , \vec{c} = \hat{i}$ અને $(\vec{a} \times \vec{b}) \times \vec{c} = \lambda \vec{a} + \mu \vec{b}$ હોય તો $\lambda + \mu$ ની કિમત મેળવો.
- A$0$
- B$1$
- C$2$
- D$3$
$ = - \widehat i + \widehat j$
$ \Rightarrow (\vec a \times \vec b) \times \vec c = \left| {\begin{array}{*{20}{c}}
{\widehat i}&{\widehat j}&{\widehat k}\\
{ - 1}&1&0\\
1&0&0
\end{array}} \right| = - \widehat k{\rm{ }}$
Now, $\lambda \vec a + \mu \vec b = \lambda \left( {\begin{array}{*{20}{c}}
{\hat i + \widehat j + \widehat k}
\end{array}} \right) + \mu (\widehat i + \widehat j)$
$=(\lambda+\mu) \hat{1}+(\lambda+\mu) \hat{\jmath}+\lambda \hat{k} $
$\because $ $\lambda \vec a + \mu \vec b = (\vec a \times \vec b) \times \vec c$
$\Rightarrow$ $(\lambda+\mu) \hat{i}+(\lambda+\mu) \hat{j}+\lambda \hat{k}=-\hat{k}$
On equating the coefficient of $\hat{i},$ we get $\lambda+\mu=0$
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