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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 23 Scalar or Dot Products Exercise 23.1 Question 26 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 23 Scalar or Dot  Products  Exercise 23.1 Question 26 Maths Textbook Solution.

Answers (1)

Answer:  2

Hint:  You must know the rules of solving vectors.

Given:  Find the projection of  \vec{b}+\vec{c} on \vec{a}, where \vec{a}=2\hat{i}-2\hat{j}+\hat{k},    \vec{b}=\hat{i}+2\hat{j}-2\hat{k}

                                                                                \vec{c}=2\hat{i}-\hat{j}+4\hat{k}

Solution: Given that,

\vec{a}=2\hat{i}-2\hat{j}+\hat{k}

                                                                                    \vec{b}=\hat{i}+2\hat{j}-2\hat{k} and

                                                                                    \vec{c}=2\hat{i}-\hat{j}+4\hat{k}

\begin{aligned} &\vec{b}+\vec{c}=\hat{\imath}+2 \hat{\jmath}-2 \hat{k}+2 \hat{i}-\hat{\jmath}+4 \hat{k} \\ &\vec{b}+\vec{c}=3 \hat{\imath}+\hat{\jmath}+2 \hat{k} \end{aligned}

Projection of\vec{b}+\vec{c} on \vec{a} is

\begin{aligned} &\frac{(\vec{b}+\vec{c}) \cdot \vec{a}}{|\vec{a}|}=\frac{(3 \hat{\imath}+\hat{\jmath}+2 \hat{k}) \cdot(2 \hat{\imath}-2 \hat{\jmath}+\hat{k})}{\left|\sqrt{(2)^{2}+(-2)^{2}+(1)^{2}}\right|} \\ &\frac{(\vec{b}+\vec{c}) \cdot \vec{a}}{|\vec{a}|}=\frac{6-2+2}{|\sqrt{4+4+1}|} \\ &\frac{(\vec{b}+\vec{c}) \cdot \vec{a}}{|\vec{a}|}=\frac{6}{|\sqrt{9}|} \\ &\frac{(\vec{b}+\vec{c}) \cdot \vec{a}}{|\vec{a}|}=\frac{6}{3} \\ &\frac{(\vec{b}+\vec{c}) \cdot \vec{a}}{|\vec{a}|}=2 \end{aligned}

 

 

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