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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 23 Scalar or Dot Products Exercise 23.1 Question 36 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 23 Scalar or Dot  Products  Exercise 23.1 Question 36 Maths Textbook Solution.

Answers (1)

Answer:  \left ( -\frac{1}{2}\hat{i}-\hat{j}+\frac{1}{2} \hat{k}\right )+\frac{5}{2}\left ( \hat{i}+\hat{k} \right )

Hints: You must know the rules of solving vectors.

Given: Express 2\hat{i}-\hat{j}+3\hat{k} as the sum of a vector parallel and a vector perpendicular to 2\hat{i}+4\hat{k}-2\hat{k}.

Solution:

Let, \vec{a}=2\hat{i}-\hat{j}+3\hat{k}

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

And \vec{x} and \vec{y} be such that,

                                                \Rightarrow \vec{a}=\vec{x}+\vec{y}

\Rightarrow \vec{y}=\vec{a}-\vec{x}

Since \vec{x} is parallel to \vec{b}

\begin{aligned} &\Rightarrow \vec{x}=t \vec{b} \\ &\Rightarrow \vec{x}=t(2 \hat{\imath}+4 \hat{\jmath}-2 \hat{k}) \\ &\Rightarrow \vec{x}=2 t \hat{\imath}+4 t \hat{\jmath}-2 t \hat{k} \end{aligned}

Substituting the values of \vec{x}and \vec{a}

\begin{aligned} &\Rightarrow(2 \hat{\imath}-\hat{\jmath}+3 \hat{k})-(2 t \hat{\imath}+4 t \hat{\jmath}-2 t \hat{k}) \\\\ &\Rightarrow(2-2 t) \hat{\imath}+(-1-4 t) \hat{\jmath}+(3+2 t) \hat{k} \\ \end{aligned}

Since y is perpendicular to \vec{b}

\begin{aligned} &\Rightarrow \vec{y} \cdot \vec{b}=0 \\ &\Rightarrow[(2-2 t) \hat{\imath}+(-1-4 t) \hat{\jmath}+(3+2 t) \hat{k}] \cdot(2 \hat{\imath}+4 \hat{\jmath}-2 \hat{k})=0 \\ &\Rightarrow 2(2-2 t)+4(-1-4 t)-2(3+2 t)=0 \\ &\Rightarrow 4-4 t-4-10 t-6-4 t=0 \\ &\Rightarrow-24 t=6 \end{aligned}

\begin{aligned} &\Rightarrow t=\left(\frac{-1}{4}\right) \\ &\therefore \vec{x}=2\left(\frac{-1}{4}\right) \hat{\imath}+4\left(\frac{-1}{4}\right) \hat{\jmath}-2\left(\frac{-1}{4}\right) \hat{k} \\ &\vec{x}=\frac{-1}{2} \hat{\imath}-\hat{\jmath}+\frac{1}{2} \hat{k} \\ &\vec{y}=\left[2-2\left(\frac{-1}{4}\right)\right] \hat{\imath}+\left[-1-4\left(\frac{-1}{4}\right)\right] \hat{\jmath}+\left[3+2\left(\frac{-1}{4}\right)\right] \hat{k} \\ &\vec{y}=\frac{5}{2} \hat{\imath}+\frac{5}{2} \hat{k}=\frac{5}{2}(\hat{\imath}+\hat{k}) \end{aligned}

So,

\vec{a}=\vec{x}+\vec{y}

\vec{a}=\left ( -\frac{1}{2}\hat{i}-\hat{j}+\frac{1}{2} \hat{k}\right )+\frac{5}{2}\left ( \hat{i}+\hat{k} \right )

 

 

Posted by

infoexpert21

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