#### Need solution for RD Sharma Maths Class 12 Chapter 24 Vector Or Cross Product Exercise Very Short Answer Question, question 22.

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HINT:

Simplify  $\overrightarrow{a}\times \widehat{i}=\widehat{j}$ and then find $\overrightarrow{a}. \widehat{i}$

GIVEN:

$\overrightarrow{a}$ is a unit vector

SOLUTION:

\begin{aligned} \text { so }|\vec{a}| &=1 \\ \text { Let, } \vec{a} &=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k} \\ \vec{a} \times \hat{i} &=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ 1 & 0 & 0 \end{array}\right| \\ &=\hat{i}(0-0)-\hat{j}\left(0-a_{3}\right)+\hat{k}\left(0-a_{2}\right) \\ \vec{a} \times \hat{i} &=a_{3} \hat{j}-a_{2} \hat{k} \\ \text { Given, } \vec{a} \times \hat{i} &=\hat{j} \\ & \therefore \hat{j}=a_{3} \hat{j}-a_{2} \hat{k} \end{aligned}

i.e.,

$0 \hat{i}+\hat{j}+0 \hat{k}=a_{3} \hat{j}-a_{2} \hat{k}+0 \hat{i}$

Comparing co-efficient of $\widehat{i},\widehat{j}\text{and } \widehat{k}$

\begin{aligned} &a_{3}=1 \\ &a_{2}=0 \\ &a_{1}=0 \\ &\therefore \vec{a}=a \hat{i}+0+\hat{k} \\ &\vec{a}=0+\hat{k} \\ &\therefore \vec{a} .=\hat{k} \end{aligned}

\begin{aligned} \text { Now, } \vec{a} \hat{i} &=\hat{k} \hat{i} \\ &=0 \quad[\because \text { Dot product is zero, } \hat{i} \hat{j}=\hat{k} \hat{i}] \end{aligned}