With reference to the right handed system of mutually perpendicul… - Vidyadip

Question

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

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Answer

Let, $\overrightarrow{\beta_1}=\lambda \vec{\alpha}$ is a scalar,
i.e., $\overrightarrow{\beta_1}=3 \lambda \vec{i}-\lambda \vec{j}$
Now, $\overrightarrow{\beta_2}=\vec{\beta}-\overrightarrow{\beta_1}=(2-3 \lambda) \hat{i}+(1+\lambda) \hat{j}-3 \hat{k}$
Now, since $\overrightarrow{\beta_2}$ is to be perpendicular $\vec{\alpha}$
we should have $\vec{\alpha} \cdot \overrightarrow{\beta_2}=0$.
i.e., $3(2-3 \lambda)-(1+\lambda)=0$
OR $\lambda=\frac{1}{2}$.
Therefore, $\overrightarrow{\beta_1}=\frac{3}{2} \widehat{i}-\frac{1}{2} \widehat{j}$ and $\overrightarrow{\beta_2}=\frac{1}{2} \widehat{i}+\frac{3}{2} \widehat{j}-3 \widehat{k}$

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