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  • Please Solve RD Sharma Class 12 Chapter 27 Straight Line in Space Exercise Very Short Answer Question 15 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 27 Straight Line in Space Exercise Very Short Answer Question 15 Maths Textbook Solution.

Answers (1)

Answer:

Required answer is   \frac{\sqrt{3}}{\sqrt{55}}, \frac{4}{\sqrt{55}}, \frac{6}{\sqrt{55}}

Hint:

Use equation of line in space

Given:

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

Solution:

We have,

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

The equation of the given line can be rewritten as,

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

\frac{x-\frac{1}{2}}{\sqrt{3}}=\frac{y+2}{4}=\frac{z-3}{6}

The direction ratios of the line parallel to AB are proportional to  \sqrt{3},4,6

The direction cosines of the line parallel to AB are proportional to,

\begin{aligned} &\frac{\sqrt{3}}{\sqrt{(\sqrt{3})^{2}+4^{2}+6^{2}}}, \frac{4}{\sqrt{(\sqrt{3})^{2}+4^{2}+6^{2}}}, \frac{6}{\sqrt{(\sqrt{3})^{2}+4^{2}+6^{2}}} \\ & \end{aligned}

=\frac{\sqrt{3}}{\sqrt{55}}, \frac{4}{\sqrt{55}}, \frac{6}{\sqrt{55}}

Posted by

infoexpert27

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