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  • Provide solution for RD Sharma maths class 12 chapter 27 Straight Line in Space exercise 27 point 1 question 9 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 27 Straight Line in Space exercise 27.1 question 9 maths textbook solution

Answers (1)

Answer :-

(5 \hat{\imath}-4 \hat{\jmath}+6 \hat{k})+\lambda(3 \hat{\imath}+7 \hat{\jmath}+2 \hat{k})

Hint :-

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

Given :-

\begin{aligned} &\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2} \end{aligned}

Solution :-

$$ \begin{aligned} &\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}=\lambda(\text { say }) \\ &\therefore x=3 \lambda+5, \quad y=7 \lambda-4, \quad z=2 \lambda+6 \end{aligned}

We know that

\begin{aligned} \vec{r} &=x \hat{\imath}-y \hat{\jmath}+z \hat{k} \\ &=(3 \lambda+5) \hat{\imath}+(7 \lambda-4) \hat{\jmath}+(2 \lambda+6) \hat{k} \\ &=(5 \hat{\imath}-4 \hat{\jmath}+6 \hat{k})+\lambda(3 \hat{\imath}+7 \hat{\jmath}+2 \hat{k}) \end{aligned}….where  ?  is constant parameter.

 

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