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### Answers (1)

Answer:$\frac{3}{2}\left ( \hat{i}+\hat{j} \right )$

Given: The vector$\overrightarrow{b}=3\hat{i}+4\hat{k}$is to be written as the sum of a vector$\overrightarrow{a}$parallel to$\overrightarrow{a}=\hat{i}+\hat{j}$and$\overrightarrow{\beta }$

perpendicular to $\overrightarrow{a}$. Then $\overrightarrow{a}=$

Hint: If$\overrightarrow{a }\parallel \overrightarrow{a }\Rightarrow \overrightarrow{a }$and if $\overrightarrow{a }\perp \overrightarrow{a }\Rightarrow \overrightarrow{a }.\overrightarrow{a }=0$

Explanation:

Here $\overrightarrow{a}=\hat{i}+\hat{j}$,  $\overrightarrow{b}=3\hat{i}+4\hat{k}$

Also $\overrightarrow{b}=\overrightarrow{a}+\overrightarrow\beta$                                                         ……………. (1)    where $\overrightarrow{a}\parallel \overrightarrow{a}$and $\overrightarrow{\beta }\perp \overrightarrow{a}$

As $\vec{\alpha} \| \vec{a} \Rightarrow \vec{\alpha}=\lambda \vec{a} \Rightarrow \vec{\alpha}=\lambda(\hat{i}+\hat{j}) \Rightarrow \vec{\alpha}=\lambda \hat{i}+\lambda \hat{j} \quad[\because \vec{a} \| \vec{b} \Rightarrow \vec{a}=k \vec{b}]$

By (1)

\begin{aligned} &\vec{\beta}=\vec{b}-\vec{\alpha} \\ &\Rightarrow \vec{\beta}=3 \hat{i}+4 \hat{k}-\lambda \hat{i}-\lambda \hat{j} \\ &\Rightarrow \vec{\beta}=(3-\lambda) \hat{i}-\lambda \hat{j}+4 \hat{k} \end{aligned}

Now $\vec{\beta}.\vec{a}=0$                                                             ………….. (2)                     $\because \vec{\beta}\perp \vec{a}$

\begin{aligned} &\Rightarrow \vec{\beta} \cdot \vec{a}=(3-\lambda)(1)-\lambda(1)+4(0) \\ &=3-\lambda-\lambda \\ &=3-2 \lambda \end{aligned}

\begin{aligned} &\because \text { By (2) } \quad 3-2 \lambda=0 \Rightarrow 2 \lambda=3 \Rightarrow \lambda=\frac{3}{2} \\ &\Rightarrow \vec{\alpha}=\frac{3}{2} \vec{a}=\frac{3}{2}(\hat{i}+\hat{j}) \\ &\Rightarrow \vec{\alpha}=\frac{3}{2}(\hat{i}+\hat{j}) \end{aligned}

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