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

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

Answers (1)

Answer :-

\begin{aligned} &\frac{x-b}{a}=\frac{y-0}{1}=\frac{z-d}{c} \\ &\vec{r}=(b \hat{\imath}+d \hat{k})+\lambda(a \hat{\imath}+\hat{\jmath}+c \hat{k}) \end{aligned}

(a , 1 , c)

Hint :-

Take x=ay+b  and z=cy+d and  find y

Given :-

           x=ay+b  and z=cy+d 

Solution :-

      x=ay+b  and z=cy+d

? x-b=ay          cy=z-d

? y=\frac{x-b}{a}  ----- (1)         ? y=\frac{z-d}{c} ----- (2)

By (1) and (2)

\begin{aligned} &\frac{x-b}{a}=y=\frac{z-d}{c} \\ &\frac{x-b}{a}=\frac{y-0}{1}=\frac{z-d}{c} \end{aligned}

D.R = ( a , 1 , c)

Let  \frac{x-b}{a}=y=\frac{z-d}{c}=\lambda \left ( say \right )

\Rightarrow x=a \lambda+b \quad, \quad y=\lambda \quad, \quad z=c \lambda+d

We know that

\vec{r} =x \hat{\imath}+y \hat{\jmath}+z \hat{k} \\

     =(a \lambda+b) \hat{\imath}+\lambda \hat{\jmath}+(c \lambda+d) \hat{k}

      =(b \hat{\imath}+d \hat{k})+\lambda(a \hat{\imath}+\hat{\jmath}+c \hat{k})

 

 

 

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