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  • Explain solution for RD Sharma maths class 12 chapter 26 Direction Cosines and Direction Ratios exercise 26.1 question 16 sub question 4 maths textbook solution

Explain solution for RD Sharma maths class 12 chapter 26 Direction Cosines and Direction Ratios exercise 26.1 question 14 sub question 4 maths textbook solution

Answers (1)

Answer: Angle between two lines is \frac{\pi }{2}

Given: 

              \begin{aligned} &21+2 m-n=0 \\ &m n+\ln +\mid m=0 \end{aligned}.

We have to find angle between lines.

Solution:

              \begin{aligned} &21+2 m-n=0 \\ \end{aligned}                                                           …………….. (1)

              \begin{aligned} &n=21+2 m\\ \end{aligned}                                                                  ………………. (2)

              

               \begin{aligned} &m n+\ln +\mid m=0 \end{aligned}                                                  \begin{aligned} \quad[n=2 \mid+2 m] \\ \end{aligned}

               \begin{aligned} &\Rightarrow m(2 \mid+2 m)+I(2 \mid+2 m)+\mid m=0 \\ &\Rightarrow 2\left|m+2 m^{2}+2\right|^{2}+2 m l+m l=0 \\ &\Rightarrow 2 m^{2}+5 m l+\left.2\right|^{2}=0 \\ &\Rightarrow(2 m+1)(m+2 \mid)=0 \end{aligned}

Either    2 m+1=0 \Rightarrow \mid=-2 m

             \mathrm{m}+2 \mathrm{l}=0 \Rightarrow \mathrm{I}=\frac{-\mathrm{m}}{2}

              If $I=-2 m$, then equ (2), n=-2 m

              If \mathrm{I}=\frac{-\mathrm{m}}{2}, then equ (2), \mathrm{n}=\mathrm{m}

Thus the direction ratios of two lines are proportional to

              \begin{aligned} &(-2 \mathrm{~m}, \mathrm{~m},-2 \mathrm{~m}) \text { and }\left(\frac{-\mathrm{m}}{2}, \mathrm{~m}, \mathrm{~m}\right) \\ &\Rightarrow(-2,1,-2) \text { and }(-1,2,2) \\ &\Rightarrow\left(\mathrm{a}_{1}, \mathrm{~b}_{1}, \mathrm{c}_{1}\right)=(-2,1,-2) \\ &\Rightarrow\left(\mathrm{a}_{2}, \mathrm{~b}_{2}, \mathrm{c}_{2}\right)=(-1,2,2) \end{aligned}

              Angle between two lines is

             \begin{aligned} &\cos \theta=\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}} \\ &\cos \theta=\frac{(-2)(-1)+(1)(2)+(-2)(2)}{\sqrt{(-2)^{2}+(1)^{2}+(2)^{2}} \sqrt{(-1)^{2}+(2)^{2}+2^{2}}} \\ &\cos \theta=\frac{2+2-4}{\sqrt{9} \sqrt{9}} \\ &\cos \theta=0 \\ &\theta=\cos ^{-1}(0) \\ &\theta=\frac{\pi}{2} \end{aligned} 

Therefore, the angle between lines is \frac{\pi}{2}

 

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