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  • Provide solution for RD Sharma maths class 12 chapter 27 Straight Line in Space exercise 27.1 question 1 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 27 Straight Line in Space exercise 27.1 question 1  maths textbook solution

Answers (1)

Answer :-

\begin{aligned} &\vec{r}=5 \hat{\imath}+2 \hat{\jmath}-4 \hat{k}+\lambda(3 \hat{\imath}+2 \hat{\jmath}-8 \hat{k}): \\ &\frac{x-5}{3}=\frac{y-2}{2}=\frac{z+4}{-8} \end{aligned}

Hint :-

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

Given :- 

Line is through point (5 , 2 , -4) and parallel to vector 3\hat{i}+2\hat{j}-8\hat{k}

Solution :- 

Position vector of (5 , 2 , -4) is

\begin{aligned} &\vec{a}=5 \hat{\imath}+2 \hat{\jmath}-4 \hat{k} \\ &\text { and } \vec{b}=3 \hat{\imath}+2 \hat{\jmath}-8 \hat{k} \end{aligned}              

Direction Ratio = (3 , 2 , -8 )

Vector equation of line which passes through (5 , 2 , -4 ) and parallel to vector 3\hat{i}+2\hat{j}-8\hat{k}  is \begin{aligned} &\vec{a}=5 \hat{\imath}+2 \hat{\jmath}-4 \hat{k} \\ &\text { and } \vec{b}=3 \hat{\imath}+2 \hat{\jmath}-8 \hat{k} \end{aligned}   , where ? is constant parameter

                       =(5 \hat{\imath}+2 \hat{\jmath}-4 \hat{k})+\lambda(3 \hat{\imath}+2 \hat{\jmath}-8 \hat{k})

Cartesian Equation :-

                    \begin{gathered} \frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c} \\ \Rightarrow \frac{x-5}{3}=\frac{y-2}{2}=\frac{z+4}{-8} \end{gathered}

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