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Provide solution for RD Sharma maths class 12 chapter 23 Scaler and Dot Product exercise Multiple choice question 2

Answers (1)

Answer:

Option (b) \hat{i}

Hint:

You must know about the concept of dot product

Given:

\vec{a} \cdot \hat{i}=\vec{a} \cdot(\hat{i}+\hat{j})=\vec{a} \cdot(\hat{i}+\hat{j}+\hat{k})=1

Solution:

\vec{a} \cdot \hat{i}=\vec{a} \cdot(\hat{i}+\hat{j})=\vec{a} \cdot(\hat{i}+\hat{j}+\hat{k})=1            ............(i)

\begin{aligned} & \vec{a} \cdot \hat{i}=a_{1} \\\\ \Rightarrow & \vec{a} \cdot \hat{i}+\vec{a} \cdot \hat{j}=a_{1}+a_{2} \\\\ & \vec{a} \cdot \hat{i}+\vec{a} \cdot \hat{j}+\vec{a} \cdot \hat{k}=a_{1}+a_{2}+a_{3} \end{aligned}

Putting in (i)

a_{1}=a_{1}+a_{2}=a_{1}+a_{2}+a_{3}=1                    ...........(ii)

Now,a_{1}=a_{1}+a_{2} \Rightarrow a_{2}=0\; \; and\\\\ a_{1}+a_{2}=a_{1}+a_{2}+a_{3} \Rightarrow a_{3}=0

Put in (ii)

\begin{aligned} &a_{1}=a_{1}=a_{1}=1 \quad\left[\because a_{2}=0, a_{3}=0\right] \\\\ &\Rightarrow a_{1}=\hat{i} \end{aligned}

Hence, option (b) is correct

 

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