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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 27 Straight Line in Space Exercise 27.2 Question 13 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 27 Straight Line in Space  Exercise 27.2 Question 13 Maths Textbook Solution.

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Answer: the equation of the lines will be \vec{r}=(-\hat{i}+2 \hat{j}+\hat{k})+\lambda(2 \hat{i}+2 / 3 \hat{j}-3 \hat{k})

Given : Find the equation of the line passing through the point (-1,2,1) and parallel to the line

\frac{2x-1}{2}=\frac{3y+5}{2}=\frac{2-z}{3}

Hint:\vec{r}=\vec{a}+\lambda \vec{b}

Solution:\vec{a}=\vec{-i}+2\hat{j}+\hat{k}

Now: \frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c} \\

=\frac{x-1 / 2}{2}=\frac{y-(-5 / 3)}{2 / 3}=\frac{z-2}{-3} \\

\vec{R}=(-\hat{i}+2 \hat{j}+\hat{k})+\lambda(2 \hat{i}+2 / 3 \hat{j}-3 \hat{k})

So the equation will be \vec{R}=(-\hat{i}+2 \hat{j}+\hat{k})+\lambda\left(2 \hat{i}+\frac{2}{3} \hat{j}-3 \hat{k}\right)

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