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  • Explain Solution R.D.Sharma Class 12 Chapter 27 Straight Line in Space Exercise 27.2 Question 24 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 27 Straight Line in Space Exercise 27.2 Question 24 Maths Textbook Solution.

Answers (1)

Answer : the direction cosines will be \frac{2}{7},\frac{3}{7},\frac{-6}{7} and the equation will be \vec{r}=\left ( \hat{i}-2\hat{j}+3\hat{k} \right )+\lambda \left ( 2\hat{i}+3\hat{j}-6\hat{k} \right )

Given: find the direction cosines of the lines \frac{x+2}{2}=\frac{2y-7}{6}=\frac{5-z}{6} and also find the vector equation of the line through the point (-1,2,3) and parallel to the given line

Hint:

Solution: \frac{x+2}{2}=\frac{y-7/2}{3}=\frac{z-5}{-6}

Direction ratio =2,3,-6

Direction cosines =  \frac{2}{\sqrt{49}}=\frac{3}{\sqrt{49}}=\frac{6}{\sqrt{49}}

\mathrm{DC}=\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}, \frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}},=\frac{2}{7}, \frac{3}{7}, \frac{-6}{7}, \frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}

Now, vector along the line; 2\hat{i}+3\hat{j}-6\hat{k}

\vec{r}=\hat{i}-2\hat{j}+3\hat{k} +\lambda \left ( 2\hat{i}+3\hat{j}-6\hat{k} \right )

So the answer will be

\vec{r}=\hat{i}-2\hat{j}+3\hat{k} +\lambda \left ( 2\hat{i}+3\hat{j}-6\hat{k} \right ) and cosines will be \frac{2}{7},\frac{3}{7},\frac{-6}{7}

 

 

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