If \(\alpha=\hat{\mathbf{i}}-3 \hat{\mathbf{j}}, \beta=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}\), then express \(\beta\) in the form \(\beta=\beta_1+\beta_2\) where \(\beta_1\) is parallel to \(\alpha\) and \(\beta_2\) is perpendicular to \(\alpha\), then \(\beta_1\) is given by

Given \(\alpha=\hat{\mathbf{i}}-3 \hat{\mathbf{j}}, \boldsymbol{\beta}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}\)
\[
\beta=\beta_1+\beta_2
\]
\(\beta_1\) is parallel to \(\alpha\)
\[
\begin{aligned}
\Rightarrow \quad \beta_1 & =\lambda \alpha \Rightarrow \beta_1=\lambda \hat{\mathbf{i}}-3 \lambda \hat{\mathbf{j}} \\
\Rightarrow \text { Also, } \beta & =\beta_1+\beta_2 \\
\beta_2 & =\beta-\boldsymbol{\beta}_1=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})-(\lambda \hat{\mathbf{i}}-3 \lambda \hat{\mathbf{j}}) \\
\boldsymbol{\beta}_2 & =(1-\lambda) \hat{\mathbf{i}}+(2-3 \lambda) \hat{\mathbf{j}}-\hat{\mathbf{k}}
\end{aligned}
\]
It is given \(\beta_2\) is perpendicular to \(\alpha\).
\[
\begin{array}{cc}
\therefore(1-\lambda)(1)+(2+3 \lambda)(-3)+(-1)(0)=0 \\
\Rightarrow & 1-\lambda-6-9 \lambda+0=0 \\
\Rightarrow & -5-10 \lambda=0 \\
\Rightarrow & \lambda=-\frac{1}{2} \\
\beta_1=\frac{-1}{2} \hat{\mathbf{i}}+\frac{3}{2} \hat{\mathbf{j}}
\end{array}
\]