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  • Explain solution RD Sharma class 12 Chapter 23 Scalar or Dot Products Exercise Very Short Answer question 34

Explain solution RD Sharma class 12 Chapter 23 Scalar or Dot Products Exercise Very Short Answer question 34

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Answer:  \lambda= 5

Given:  \vec{a}=\lambda \hat{i}+\hat{\jmath}+4 \hat{k} \& \vec{b}=2 \hat{i}+6 \hat{\jmath}+3 \hat{k}   and projection of  a on  \vec{a} \text { on } \vec{b}=4 \text { units. }

Hint: you must know about the concept of projection of a on  \vec{a} \text { on } \vec{b}

Solution: projection of  a on  \vec{a} \text { on } \vec{b}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}

\begin{aligned} &\Rightarrow 4=\frac{(\lambda \hat{i}+\hat{j}+4 \hat{k}) \cdot(2 \hat{i}+6 \hat{j}+3 \hat{k})}{\sqrt{2^{2}+6^{2}+3^{2}}} \\ & \end{aligned}

\Rightarrow 4=\frac{2 \lambda+6+12}{\sqrt{4+36+9}}

\begin{aligned} &\Rightarrow \quad 4=\frac{2 \lambda+18}{\sqrt{49}} \\ & \end{aligned}

\Rightarrow 4=\frac{2 \lambda+18}{\sqrt{7}}

\begin{aligned} &\Rightarrow 28=2 \lambda+18 \\ & \end{aligned}

\Rightarrow 28-18=2 \lambda \\

\Rightarrow 2 \lambda=10 \\

\Rightarrow \lambda=5

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