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  • Provide solution for RD Sharma maths class 12 chapter 27 Straight Line in Space exercise 27.1 question 16 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 27 Straight Line in Space exercise 27.1 question 16 maths textbook solution

Answers (1)

Answer :-

\begin{aligned} &\frac{x-1}{-2}=\frac{y-2}{14}=\frac{z-3}{3} \\ &(\hat{\imath}+2 \hat{\jmath}+3 \hat{k})+\lambda(\widehat{-2} \imath+14 \hat{\jmath}+3 \hat{k}) \end{aligned}

Hint :-

\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}

Given :-

The line passes through the point ( 1 , 2 , 3) and parallel to the line

\begin{aligned} &\frac{-x-2}{1}=\frac{y+3}{7}=\frac{2z-6}{3} \\ \end{aligned} 

Solution :-

The P.V of the point (1 , 2 , 3) will be \overrightarrow{r}=\hat{i}+2\hat{j}+3\hat{k}

The line \begin{aligned} &\frac{-x-2}{1}=\frac{y+3}{7}=\frac{2z-6}{3} \\ \end{aligned} can be written as \begin{aligned} &\frac{x+2}{-1}=\frac{y+3}{7}=\frac{z-3}{\frac{3}{2}} \\ \end{aligned}    ---------- (1)

So the direction ratios can be  -1,7,\frac{3}{2} or -2,14,3 PV=\widehat{-2}i+14\hat{j}+3\hat{k}

Cartesian equation of the line passing through ( 1 , 2 , 3) and parallel to (1)

\begin{aligned} &\frac{x-1}{-2}=\frac{y-2}{14}=\frac{z-3}{3} \\ \end{aligned}

Vector Equation

\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}

      =\begin{aligned} &(\hat{\imath}+2 \hat{\jmath}+3 \hat{k})+\lambda(\widehat{-2} \imath+14 \hat{\jmath}+3 \hat{k}) \end{aligned}

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