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  • Provide solution for RD Sharma maths class 12 chapter 27 Straight Line in Space exercise 27.1 question 6 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 27 Straight Line in Space exercise 27.1 question 6 maths textbook solution

Answers (1)

Answer :-

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

Hint :-

\begin{aligned} &\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a}) \\ &\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}} \end{aligned}

Given :-

Two points A (1 , 2 , -1) and B ( 2 , 1 , 1)

Solution :-

Position vector of A (1 , 2 , -1)

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

Position vector of B (2 , 1 , 1)

\begin{aligned} &\vec{b}=2 \hat{\imath}+\hat{\jmath}+\hat{k} \\ &\vec{b}-\vec{a}=\hat{\imath}-\hat{\jmath}+2 \hat{k} \end{aligned}

Vector Equation passes through A (1 , 2 , -1) and B ( 2 , 1 , 1)

\begin{aligned} &\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a}) \\ \end{aligned} ,  where ? is parameter

    =\left ( \hat{\imath}-2\hat{\jmath}+ \hat{k} \right )+\lambda \left ( \hat{\imath}-\hat{\jmath}+2 \hat{k} \right )    

Cartisian Equation passes through A (1 , 2 , -1) and B ( 2 , 1 , 1)

\begin{aligned} &\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}} \\ &\frac{x-1}{1}=\frac{y-2}{-1}=\frac{z+1}{2} \end{aligned}

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