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  • Provide solution for RD Sharma maths class 12 chapter 27 Straight Line in Space exercise 27.1 question 10 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 27 Straight Line in Space exercise 27.1 question 10 maths textbook solution

Answers (1)

Answer :-

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

Hint :-

\vec{r}=a+\lambda b

Given :-

The line passing through (1 , -1 , 2) and parallel to the line whose equation  \frac{x-3}{1}=\frac{y-1}{2}=\frac{z+1}{-2}

Solution :-

Cartesian Equation of a line passing through a point (1,-1,2) and parallel to line 

\frac{x-3}{1}=\frac{y-1}{2}=\frac{z+1}{-2} is

\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}

\left ( x_{1} ,y_{1},z_{1}\right ) is the point and a,b,c are the direction ratios of the parallel line.)

  \frac{x-1}{1}=\frac{y+1}{2}=\frac{z-2}{-2}is the required cartesian equation of the line.

Let,  \frac{x-1}{1}=\frac{y+1}{2}=\frac{z-2}{-2}=\lambda \text { (say) }

We know that

\begin{aligned} \vec{r} &=x \hat{\imath}-y \hat{\jmath}+z \hat{k} \\ &=(\lambda+1) \hat{\imath}+(2 \lambda-1) \hat{\jmath}+(-2 \lambda+2) \hat{k} \\ &=(\hat{\imath}-\hat{\jmath}+2 \hat{k})+\lambda(\hat{\imath}+2 \hat{\jmath}-2 \hat{k}) \end{aligned}is the required vector equation.

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