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  • Provide solution for RD Sharma maths class 12 chapter 27 Straight Line in Space exercise 27.1 question 4 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 27 Straight Line in Space exercise 27.1 question 4  maths textbook solution

Answers (1)

Answer :-

\begin{aligned} &\vec{r}=(2 \hat{\imath}-3 \hat{\jmath}+4 \hat{k})+\lambda(3 \hat{\imath}+4 \hat{\jmath}-5 \hat{k}) \\ &\frac{x-2}{3}=\frac{y+3}{4}=\frac{z-4}{-5} \end{aligned}

Hint :-

\vec{r}=\overrightarrow{a}+\lambda \overrightarrow{b}

Given :-

Position vector of the point =2 \hat{\imath}-3 \hat{\jmath}+4 \hat{k}  and  in the direction of  3 \hat{\imath}+4 \hat{\jmath}-5 \hat{k}

Solution :-

Position vector of point

a=2 \hat{\imath}-3 \hat{\jmath}+4 \hat{k}  ,  point = ( 2 , -3 , 4)

And position vector of parallel line(in the direction)

b=3 \hat{\imath}+4 \hat{\jmath}-5 \hat{k}   ,    Direction ratio = (3 , 4 , -5 )

Vector Equation :-

\vec{r}=\overrightarrow{a}+\lambda \overrightarrow{b}  ,  where ? is constant parameter

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

Cartesian Equation :-

\begin{aligned} &\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c} \\ &\frac{x-2}{3}=\frac{y+3}{4}=\frac{z-4}{-5} \end{aligned}

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