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  • Provide Solution for RD Sharma Class 12 Chapter 23 Scalar or Dot Products Exercise Very Short Answer Question 38

Provide Solution for RD Sharma Class 12 Chapter 23 Scalar or Dot Products Exercise Very Short Answer Question 38

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Answer:   2

Given:  \overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}-2 \hat{\mathrm{\jmath}}+\hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{\jmath}}-2 \hat{\mathrm{k}} \text { and } \overrightarrow{\mathrm{c}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}}

Hint: Find  \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}  and then apply the formula of projection.

Solution:  \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=\hat{\mathrm{\imath}}+2 \hat{\mathrm{\jmath}}-2 \hat{\mathrm{k}}+2 \hat{\mathbf{i}}-\hat{\mathrm{\jmath}}+4 \hat{\mathrm{k}}

\Rightarrow \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=3 \hat{\mathrm{i}}+\hat{\jmath}+2 \hat{\mathrm{k}}

Projection of   \vec{b}+\vec{c} \text { on } \vec{a}=(\vec{b}+\vec{c}) \times \frac{\vec{a}}{|\vec{a}|}

=\frac{(3 \hat{i}+\hat{j}+2 \hat{k})(2 \hat{i}-2 \hat{j}+\hat{k})}{|2 \hat{\imath}-2 \hat{\jmath}+\widehat{k}|}

\begin{aligned} &=\frac{6-2+2}{\sqrt{2^{2}+(-2)^{2}+1^{2}}} \\ & \end{aligned}

=\frac{6}{\sqrt{4+4+1}}=\frac{6}{\sqrt{9}}=\frac{6}{3}

\Rightarrow Projectio of   (\vec{b}+\vec{c}) \text { on } \vec{a}=2

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