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 $\vec{\alpha}$.

Answer


 Let $\vec{\beta}_1=\lambda \vec{\alpha}\left[\because \vec{\beta}_1 \| t o \vec{\alpha}\right]$
$\vec{\beta}_1=\lambda(3 \hat{i}-\hat{j})$
$=3 \lambda \hat{i}-\lambda \hat{j}$
$\vec{\beta}_2=\vec{\beta}-\vec{\beta}_1$
$=(2-3 \lambda) \hat{i}+(1+\lambda) \hat{j}-3 \hat{k}$
$\vec{\alpha} \cdot \vec{\beta}_2=0\left[\because \vec{\beta}_2 \perp \vec{\alpha}\right]$
$3(2-3 \lambda)-(1+\lambda)=0$
$\lambda=\frac{1}{2}$
$\vec{\beta}_1=\frac{3}{2} \hat{i}-\frac{1}{2} \hat{j}$
$\vec{\beta}_2=\frac{1}{2} \hat{i}+\frac{3}{2} \hat{j}-3 \hat{k}$

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