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  • Provide solution for RD Sharma maths class 12 chapter 27 Straight Line in Space exercise 27.1 question 17 maths textbook solution

Provide solution for RD Sharma maths class 12 chapter 27 Straight Line in Space exercise 27.1 question 17 maths textbook solution

Answers (1)

Answer :-

\frac{x+\frac{1}{3}}{2}=\frac{y-\frac{1}{3}}{1}=\frac{z-1}{-6}…cartesian equation of line

\left(-\frac{1}{3} \hat{\imath}+\frac{1}{3} \hat{\jmath}+\hat{k}\right)+\lambda(2 \hat{\imath}+\hat{\jmath}-\widehat{6 k})…vector equation of line

\left(\frac{-1}{3}, \frac{1}{3}, 1\right)…the fixed point through which the line passes   ;    

(2,1,-6)….direction ratios of the line

Hint :-

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

Given :-

3x+1=6y-2=1-z

Solution :-

The given line can be written as \frac{x+\frac{1}{3}}{\frac{1}{3}}=\frac{y-\frac{2}{6}}{\frac{1}{6}}=\frac{z-1}{-1} 

\Rightarrow \frac{x+\frac{1}{3}}{2}=\frac{y-\frac{1}{3}}{1}=\frac{z-1}{-6}

Thus the line passes through the point \left(\frac{-1}{3}, \frac{1}{3}, 1\right) 

vector equation of the point will be \left(\overrightarrow{a}=-\frac{1}{3} \hat{\imath}+\frac{1}{3} \hat{\jmath}+\hat{k}\right)

The direction ratios are proportional to    (2,1,-6)

and its vector equation will be \overrightarrow{b}=2 \hat{\imath}+\hat{\jmath}-\widehat{6 k}

Vector Equation of line

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

       =\left(-\frac{1}{3} \hat{\imath}+\frac{1}{3} \hat{\jmath}+\hat{k}\right)+\lambda(2 \hat{\imath}+\hat{\jmath}-\widehat{6 k})   

 

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