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  • Provide solution for RD Sharma maths class 12 chapter 27 Straight Line in Space exercise 27.1 question 7 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 27 Straight Line in Space exercise 27.1 question 7 maths textbook solution

Answers (1)

Answer :-

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

Hint :-

\begin{aligned} &\vec{r}=\overrightarrow{a}+\lambda \overrightarrow{b} \end{aligned}

Given :-

The line passes through (1 , 2 , 3) and parallel to the vector \hat{\imath}+2 \hat{\jmath}+3 \hat{k} 

Solution :-

Position vector of (1 , 2 , 3)\Rightarrow \overrightarrow{a}=\left ( \hat{\imath}+2 \hat{\jmath}+3 \hat{k} \right )

\overrightarrow{b}=\hat{\imath}-2 \hat{\jmath}+3 \hat{k}

Direction Ratio (1 , -2 , 3)

Vector equation of the line passes through (1 , 2 , 3) and

parallel to \hat{\imath}-2 \hat{\jmath}+3 \hat{k}

\begin{aligned} &\vec{r}=\overrightarrow{a}+\lambda \overrightarrow{b} \end{aligned} , where ? is parameter

    =\left ( \hat{\imath}+2 \hat{\jmath}+3 \hat{k} \right )+\lambda \left ( \hat{\imath}-2 \hat{\jmath}+3 \hat{k} \right )

Cartesian Equation

\begin{aligned} &\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c} \\ &\frac{x-1}{1}=\frac{y-2}{-2}=\frac{x-3}{3} \end{aligned}

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