Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • 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

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE extends single girl child scholarship registration deadline till November 20
anam-khan

CBSE reminds schools to submit registration data of Class 9, 11 students by October 16
anam-khan

CBSE reopens online portal for submission of LOC for Class 10, 12 board exams 2026; submit forms by Oct 8

Latest Question