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  • Provide solution for RD Sharma maths class 12 chapter 27 straight line in space exercise 27.1 question 3 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 27 straight line in space exercise 27.1 question 3 maths textbook solution

Answers (1)

Answer :-

\begin{aligned} &\vec{r}=5 \hat{\imath}-\hat{2} \hat{\jmath}+4 \hat{k}+\lambda(2 \hat{\imath}-\hat{\jmath}+3 \hat{k}) \\ &\frac{x-5}{2}=\frac{y+2}{-1}=\frac{z-4}{3} \end{aligned}

Hint :-

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

Given :-

Line passes through (5 , -2 , 4) and parallel to 2 \hat{\imath}-\hat{\jmath}+3 \hat{k}

Solution :-

Position Vector of the point (5 , -2 , 4) is 

\vec{a}=5 \hat{\imath}-\hat{2} \hat{\jmath}+4 \hat{k}      and

parallel to \overrightarrow{b}=2 \hat{\imath}-\hat{\jmath}+3 \hat{k}

Direction ratio = (2 , -1 , 3)

Vector equation of the line passes through the point (5 , -2 , 4) and parallel to 5 \hat{\imath}-\hat{2} \hat{\jmath}+4 \hat{k}

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

   =5 \hat{\imath}-\hat{2} \hat{\jmath}+4 \hat{k}+\lambda(2 \hat{\imath}-\hat{\jmath}+3 \hat{k}) \\, where ? is constant

And Cartisian form

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

 

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