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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 2 maths textbook solution

Explain solution for RD Sharma maths class 12 chapter 26 Direction Cosines and Direction Ratios exercise 26.1 question 16 sub question 2 maths textbook solution

Answers (1)

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

Given: 

              \begin{aligned} &2 l-m+2 n=0 \\ &m n+n l+\mid m=0 \end{aligned}, Find angle between the lines

Hint: Use \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}}}

Solution: we have

              2 l-m+2 n=0                                            …………….. (1)

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

             

              \Rightarrow m n+n l+\operatorname{lm}=0                                  \quad[m=2 \mid+2 n]

              

              \Rightarrow(2 \mid+2 n) n+n l+l(2 l+2 n)=0

              \Rightarrow\left(2\left|n+2 n^{2}+n\right|+\left.2\right|^{2}+2 n \mid\right)=0

              \left.\Rightarrow 2\right|^{2}+51 n+2 n^{2}=0

             \Rightarrow(1+2 n)(21+n)=0

             \Rightarrow \mid=-2 \mathrm{n}$ or $\mathrm{I}=\frac{-\mathrm{n}}{2}

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

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

Thus, direction ratio of two lines are proportional to

\begin{aligned} &(-2 n,-2 n, n) \text { and }\left(\frac{-n}{2}, n, n\right) \\ &\Rightarrow(-2,-2,1) \text { and }\left(-\frac{1}{2}, 1,1\right) \\ &\Rightarrow\left(a_{1}, b_{1}, c_{1}\right)=(-2,-2,1) \\ &\Rightarrow\left(a_{2}, b_{2}, c_{2}\right)=\left(-\frac{1}{2}, 1,1\right) \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)\left(-\frac{1}{2}\right)+(-2)(1)+(1)(1)}{\sqrt{(-2)^{2}+(-2)^{2}+(1)^{2}} \sqrt{\left(-\frac{1}{2}\right)^{2}+1^{2}+1^{2}}} \\ &\cos \theta=\frac{1-2+1}{\sqrt{4+4+1} \sqrt{\frac{1}{4}+1+1}} \\ &\cos \theta=0 \\ &\theta=\cos ^{-1}(0) \\ \end{aligned}

               \theta=\frac{\pi}{2}

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

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