Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Provide solution for RD Sharma maths class 12 chapter 27 Straight Line in Space exercise 27.1 question 11 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 27 Straight Line in Space exercise 27.1 question 11 maths textbook solution

Answers (1)

Answer :-

\frac{-2}{7}, \frac{6}{7},-\frac{3}{7} ; \vec{r}=(4 \hat{\imath}+\hat{k})+\lambda(-2 \hat{\imath}+6 \hat{\jmath}-3 \hat{k})

Hint :-

\vec{r}=x \hat{\imath}-y \hat{\jmath}+z \hat{k}

Given :-

\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}

Solution :-

\begin{aligned} & \frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3} \\ \Rightarrow & \frac{x-4}{-2}=\frac{y}{6}=\frac{z-1}{-3} \end{aligned}      

Direction ratios are (-2 , 6 , -3)

So, its directional cosines will be

\frac{-2}{\sqrt{(-2)^{2}+6^{2}+(-3)^{2}}}, \frac{6}{\sqrt{(-2)^{2}+6^{2}+(-3)^{2}}}, \frac{-3}{\sqrt{(-2)^{2}+6^{2}+(-3)^{2}}} \\
=\frac{-2}{7}, \frac{6}{7},-\frac{3}{7}

Let \frac{x-4}{-2}=\frac{y}{6}=\frac{z-1}{-3}=\lambda (say)

\Rightarrow x=-2 \lambda+4 \quad, \quad y=6 \lambda, \quad z=-3 \lambda+1

We know that

\vec{r}=x \hat{\imath}+y \hat{\jmath}+z \hat{k} \\

    =(-2 \lambda+4) \hat{\imath}+(6 \lambda) \hat{\jmath}+(-3 \lambda+1) \hat{k} \\

    =(4 \hat{\imath}+\hat{k})+\lambda(-2 \hat{\imath}+6 \hat{\jmath}-3 \hat{k})

                        Where ? is constant parameter.

Posted by

Infoexpert

View full answer

Similar Questions

Latest Question