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  • Provide solution for RD Sharma maths class 12 chapter 27 Straight Line in Space exercise 27.1 question 13 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 27 Straight Line in Space exercise 27.1 question 13 maths textbook solution

Answers (1)

Answer :-

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

Hint :-

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

Given :-

The line passes through the point \hat{\imath}-2 \hat{\jmath}-3 \hat{k} and parallel to the line joining with position vectors   \hat{\imath}- \hat{\jmath}+4 \hat{k} and 2\hat{\imath}+ \hat{\jmath}+2 \hat{k}

Solution :-

The points of position vector \hat{\imath}- \hat{\jmath}+4 \hat{k} and 2\hat{\imath}+ \hat{\jmath}+2 \hat{k} are P(1 , -1 , 4) and Q(2 , 1 , 2)

D.R of PQ =Q-P= (1 , 2 , -2)       ,       \overrightarrow{b}=\hat{\imath}+2 \hat{\jmath}-2 \hat{k}

Points of P.V  \overrightarrow{b}=\hat{\imath}-2 \hat{\jmath}-3 \hat{k} are (1 , -2 , -3)

The Cartesian equation of the line having the point (1 , -2 , -3)  and

D.R (1 , 2 , -2)   are

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

And vector equation are

\begin{aligned} \vec{r} &=\vec{a}+\lambda \vec{b} \\ &=(\hat{\imath}-2 \hat{\jmath}-3 \hat{k})+\lambda(\hat{\imath}+2 \hat{\jmath}-2 \hat{k}) \end{aligned}

                        Where \lambda is constant parameter.

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