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  • Explain solution RD sharma class 12 chapter 23 scalar or dot product exercise Fill in the blanks question 18

Explain solution RD sharma class 12 chapter 23 scalar or dot product exercise Fill in the blanks question 18

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\frac{3}{5\sqrt{2}}Hint:

               \vec{a} and \vec{b} projection

Given:

               \vec{a}=2\hat{i}-\hat{j}+2\hat{k}  and \vec{b}=5\hat{i}-3\hat{j}-4\hat{k}

Solution:

\vec{a}=2\hat{i}-\hat{j}+2\hat{k} \\ \vec{b}=5\hat{i}-3\hat{j}-4\hat{k}

We know

Projection of vector  \vec{a} and \vec{b} =\frac{1}{|\vec{b}|} \cdot(\vec{a} \cdot \vec{b})

\begin{aligned} (\vec{a} \cdot \vec{b}) &=(2 \times 5)+(-1 \times-3)+(2 \times-4) \\ &=10+3-8 \\ &=5 \\ \end{aligned}

\begin{aligned} |\vec{b}| &=\sqrt{25+9+16} \\ &=\sqrt{25+25} \\ &=\sqrt{50} \end{aligned}

Projection of vector \vec{a} on \vec{b} =\frac{1}{\sqrt{2}.5} \times 5\\

                                                  =\frac{1}{\sqrt{2}}

 Projection of \vec{b} and \vec{a}   =\frac{1}{|\vec{a}|} \cdot(\vec{b} \cdot \vec{a})

\begin{aligned} \vec{b} \cdot \vec{a}=&(5 \times 2)+(-3 \times-1)+(-4 \times 2) \\ &=10+3-8 \\ &=5 \\ \end{aligned}

\begin{aligned} |\vec{a}|=& \sqrt{2^{2}+1^{2}+2^{2}} \\ =& \sqrt{4+1+4} \\ &=\sqrt{9}\end{aligned}           

        \begin{aligned} &=3 \\ &\frac{\text { Projection of } \vec{a} \text { and } \vec{b}}{\text { Projection of } \vec{b} \text { and } \vec{a}}=\frac{\frac{1}{\sqrt{2}}}{\frac{5}{3}} \\ &\qquad=\frac{3}{5 \sqrt{2}} \end{aligned}

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