If with reference to the right handed system of mutually perpendi… - Vidyadip

Question

If with reference to the right handed system of mutually perpendicular unit vectors $\hat i,\hat j$ and $\hat k$, $\vec \alpha = 3\hat i - \hat j$, $\vec \beta = 2\hat i + \hat j - 3\hat k$, then express $\vec \beta $ in the form $\vec \beta = {\vec \beta _1} + {\vec \beta _2}$, where ${\vec \beta _1}$ is || to $\vec \alpha $ and ${\vec \beta _2}$ is perpendicular to $\vec \alpha $.

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Answer

Let ${\vec \beta _1} = \lambda \vec \alpha \,\,\left[ {\because \,{{\vec \beta }_1}||to\,\,\vec \alpha } \right]$
${\vec \beta _1} = \lambda \left( {3\hat i - \hat j} \right)$
$ = 3\lambda \hat i - \lambda \hat j$
${\vec \beta _2} = \vec \beta - {\vec \beta _1}$
$= \left( {2\hat i + \hat j - 3\hat k} \right) - \left( {3\lambda \hat i - \lambda \hat j} \right)$
$= \left( {2 - 3\lambda } \right)\hat i + \left( {1 + \lambda } \right)\hat j - 3\hat k$
$\vec \alpha .{\vec \beta _2} = 0\,\,\,\left[ {\because {{\vec \beta }_2} \bot \vec \alpha } \right]$
$3\left( {2 - 3\lambda } \right) - \left( {1 + \lambda } \right) = 0$
$\lambda = \frac{1}{2}$
${\vec \beta _1} = \frac{3}{2}\hat i - \frac{1}{2}\hat j$
${\vec \beta _2} = \frac{1}{2}\hat i + \frac{3}{2}\hat j - 3\hat k$

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