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  • Please solve RD Sharma class 12 chapter Straight Line in Space exercise 27.5 question 4 sub question (i) maths textbook solution

Please solve RD Sharma class 12 chapter Straight Line in Space exercise 27.5 question 4 sub question (i) maths textbook solution

Answers (1)

Answer: d=\sqrt{26} \text { units }

Hint: Let in L2  these is λ draw \mid \mid  Line

                    

Given: \overrightarrow{\mathrm{\gamma }}=(1 \hat{\imath}+2 \hat{\jmath}+3 \hat{k})+\lambda(1 \hat{\imath}-1 \hat{\jmath}+\hat{k}) \text { and } \overrightarrow{\mathrm{\gamma }}=(2 \hat{\imath}-\hat{\jmath}-\hat{k})+\mu(-\hat{\imath}+\hat{\jmath}-\hat{k})

Solution:

\begin{aligned} &L_{1}(1,2,3), \quad \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-3}{1} \\\\ &L_{2}(2,-1,-1), \frac{x-2}{1}=\frac{y+1}{-1}=\frac{z+1}{1}=\lambda \end{aligned}

Line (2) becomes (\lambda+2,-\lambda-1, \lambda-1)

\begin{aligned} &=(\lambda+2-1) \hat{\imath}+(-\lambda-1-2) \hat{\jmath}+(\lambda-1-3) \hat{k} \\\\ &=(\lambda+1) \hat{\imath}+(-\lambda-3) \hat{\jmath}+(\lambda-4) \hat{k} \end{aligned}

\begin{aligned} &=(\lambda+1)(1)+(-\lambda-3)(-1)+(\lambda-4)(1) \\\\ &=\lambda+1+\lambda+3+\lambda-4 \\\\ &=3 \lambda=0 \end{aligned}

\therefore \lambda=0

The shortest distance between the \mid \mid  Line is 2,-1,-1 \; and \; (1,2,3)

 

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