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  • Provide Solution For R.D. Sharma Maths Class 12 Chapter 23 Scalar or Dot Products Exercise 23.1 Question 17 Maths Textbook Solution.

Provide Solution For  R.D. Sharma Maths Class 12 Chapter 23 Scalar or Dot Products  Exercise 23.1 Question 17 Maths Textbook Solution.

Answers (1)

Answer: \overrightarrow{B}_{1}= \frac{1}{5}\left ( 3\hat{i}+4\hat{j}+5\hat{k} \right )

\overrightarrow{B}_{2}=\frac{1}{5}\left ( 13\hat{i}+9\hat{j}-15\hat{k} \right )

Hint: you must know the rule of solving vectors

Given: if \vec{a}=3\hat{i}+4\hat{j}+5\hat{k} \: \: \: \vec{B}=2\hat{i}+\hat{j}-4\hat{k}

Then express \vec{B} in the form of\vec{B}=\overrightarrow{B}_{1}+\overrightarrow{B}_{2}  where \overrightarrow{B}_{1} is parallel to \vec{a} and \overrightarrow{B_{2}} is perpendicular to \vec{a}

Solution: given that \begin{aligned} &\vec a=3 \hat{\imath}+4 \hat{\jmath}+5 \hat{k}, \ \ B=2 \hat{\imath}+\hat{\jmath}-4 \hat{k} \\ \end{aligned}

Also,

\begin{aligned} &\vec{B}=\overrightarrow{B_{1}}+\overrightarrow{B_{2}} \\ &\overrightarrow{B_{2}}=\vec{B}-\overrightarrow{B_{1}} \quad 1 \end{aligned}

Since \overrightarrow{B_{1}}parallel to\overrightarrow{a}

\begin{aligned} &\overrightarrow{B_{1}}=t \vec{a} \\ &\overrightarrow{B_{1}}=t(3 \hat{\imath}+4 \hat{\jmath}+5 \hat{k}) \\ &=3 t \hat{\imath}+4 t \hat{\jmath}+5 t \hat{k} \end{aligned}

Substituting the value of \overrightarrow{B_{1}} and  \vec{a}

\begin{aligned} &\overrightarrow{B_{2}}=2 \hat{\imath}+\hat{\jmath}-4 \hat{k}-(3 t \hat{\imath}+4 t \hat{\jmath}+5 t \hat{k}) \\\\ &=(2-3 t) \hat{\imath}+(1-4 t) \hat{\jmath}+(-4-5 t) \hat{k} \end{aligned}

Since \overrightarrow{B} is \perp to \vec{a}

\begin{aligned} &\overrightarrow{B_{2}} \cdot \vec{a}=0 \\ &\Rightarrow[(2-3 t) \hat{\imath}+(1-4 t) \hat{\jmath}+(-4-5 t) \hat{k}] \cdot[3 \hat{\imath}+4 \hat{\jmath}+5 \hat{k}]=0 \\ &\Rightarrow 3(2-3 t)+4(1-4 t)+5(-4-5 t)=0 \\ &\Rightarrow 6-9 t+4-16 t-20-25 t=0 \\ &\Rightarrow-50 t=10 \\ &t=\frac{-1}{5} \end{aligned}

\therefore we get

\begin{aligned} &\overrightarrow{B_{1}}=-\frac{1}{5}(3 \hat{\imath}+4 \hat{\jmath}+5 \hat{k}) \\ &\overrightarrow{B_{2}}=\frac{1}{5}(13 \hat{\imath}+9 \hat{j}-15 \hat{k}) \\ &=\frac{1}{5}(13 \hat{\imath}+9 \hat{\jmath}-15 \hat{k}) \end{aligned}

 

 

 

Posted by

infoexpert21

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