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  • Explain solution for RD Sharma maths class 12 chapter 26 Direction Cosines and Direction Ratios exercise 26.1 question 15 maths textbook solution

Explain solution for RD Sharma maths class 12 chapter 26 Direction Cosines and Direction Ratios exercise 26.1 question 15 maths textbook solution

Answers (1)

Answer: \left(\pm \frac{1}{\sqrt{6}}, \pm \frac{1}{\sqrt{6}}, \pm \frac{2}{\sqrt{6}}\right) \text { and }\left(\pm \frac{1}{\sqrt{6}}, \pm \frac{2}{\sqrt{6}}, \pm \frac{1}{\sqrt{6}}\right)

Given: I+m+n=0 and 2Im+2In-mn=0 . Find direction cosine of line

Solution: we have

           I+m+n=0                                                    ………………. (1)

          I=-m-n                                                       ………………. (2)

          \begin{aligned} &\Rightarrow 2 \mid m+2 \ln -m n=0 \quad[1=-m-x] \\ &\Rightarrow 2(-m-n) m+2(-m-n) n-m n=0 \\ &\Rightarrow-2 m^{2}-2 m n-2 m n-2 n^{2}-m n=0 \\ &\Rightarrow 2 m^{2}+2 n^{2}+5 m n=0 \\ &\Rightarrow(m+2 n)(2 m+n)=0 \end{aligned}              

Given   m+2 n=0 \Rightarrow m=-2 n

                             Or

              2 \mathrm{~m}+\mathrm{n}=0 \Rightarrow \mathrm{m}=\frac{-\mathrm{n}}{2}

             If m=-2 n, then from (1), \mid=n

             If m=\frac{-n}{2}, then from (1), 1=\frac{-n}{2}

Thus, direction ratios are proportional to (n,-2 n, n) \text { and }\left(\frac{-n}{2}, \frac{-n}{2}, n\right)

              \Rightarrow(1,-2,1) \text { and }\left(-\frac{1}{2},-\frac{1}{2}, 1\right)

Now direction cosines are

              \left(\pm \frac{1}{\sqrt{6}}, \pm \frac{1}{\sqrt{6}}, \pm \frac{2}{\sqrt{6}}\right) \text { and }\left(\pm \frac{1}{\sqrt{6}}, \pm \frac{2}{\sqrt{6}}, \pm \frac{1}{\sqrt{6}}\right)

 

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