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  • Explain solution RD Sharma class 12 Chapter 27 Striaght Line in Space Exercise Very Short Answer question 6

Explain solution RD Sharma class 12 Chapter 27 Striaght Line in Space Exercise Very Short Answer question 6

Answers (1)

Answer:

Required answer is  \left(\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}\right)

Hint:

Use properties of vector

Given:

6 x-2=3 y+1=2 z-4

Solution:

We have

6 x-2=3 y+1=2 z-4

The equation of the given line can be rewritten as,

\begin{aligned} &\frac{x-\frac{1}{3}}{\frac{1}{6}}=\frac{y+\frac{1}{3}}{\frac{1}{3}}=\frac{z-2}{\frac{1}{2}} \\ & \end{aligned}

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

The direction ratios of the line parallel to AB are proportional to  1,2,3

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

\begin{aligned} &\frac{1}{\sqrt{1^{2}+2^{2}+3^{2}}}, \frac{2}{\sqrt{1^{2}+2^{2}+3^{2}}}, \frac{3}{\sqrt{1^{2}+2^{2}+3^{2}}} \\ & \end{aligned}

=\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}

Posted by

infoexpert27

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