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  • Provide solution for RD Sharma maths class 12 chapter Straight Line in Space exercise 27.5 question 4 sub question (ii)

Provide solution for RD Sharma maths class 12 chapter Straight Line in Space exercise 27.5 question 4 sub question (ii)

Answers (1)

Answer: d=0

Hint:Using the equation  d=\frac{\left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|}{\left|\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|}    Let the given lines L1 and L2

Given: \overrightarrow{\mathrm{\gamma}}=(\hat{\imath}+\hat{\jmath})+\lambda(2 \hat{\imath}-\hat{\jmath}+\hat{k}) \text { and } \overrightarrow{\mathrm{\gamma }}=(2 \hat{\imath}+\hat{\jmath}-\hat{k})+\mu(4 \hat{\imath}-2 \hat{\jmath}+2 \hat{k})

Solution:

\begin{aligned} &L_{1}=a_{1}+\lambda b_{1} \text { and } L_{2}=a_{2}+\mu b_{2} \\\\ &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & -1 & 1 \\ 4 & -2 & 2 \end{array}\right| \Rightarrow \hat{\imath}(0)-\hat{\jmath}(0)+\hat{k} \\\\ &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=0 \end{aligned}

So the shortest distance is =0

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