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  • Provide solution for RD Sharma maths class 12 chapter Straight Line in Space exercise 27.5 question 1 sub question (ii)

Provide solution for RD Sharma maths class 12 chapter Straight Line in Space exercise 27.5 question 1 sub question (ii)

Answers (1)

Answer: D=\frac{64}{\sqrt{62}} \text { units }

Hint: Using formula    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 \overline{b_{2}}\right)\right|}

 

Given: \overrightarrow{\mathrm{\gamma}}=(3 \hat{\imath}+5 \hat{\jmath}+7 \hat{k})+\lambda(\hat{\imath}-2 \hat{\jmath}+7 \hat{k}) \text { and } \overrightarrow{\mathrm{\gamma}}=(-\hat{\imath}-\hat{\jmath}-\hat{k})+\mu(7 \hat{\imath}-6 \hat{\jmath}+\hat{k})

 

Solution:

\begin{aligned} &\overrightarrow{a_{1}}=3 \hat{\imath}+5 \hat{\jmath}+7 \hat{k} \\\\ &\overrightarrow{a_{2}}=-\hat{\imath}-\hat{\jmath}-\hat{k} \\\\ &\overrightarrow{b_{1}}=\hat{\imath}-2 \hat{\jmath}+7 \hat{k} \\\\ &\overrightarrow{b_{2}}=7 \hat{\imath}-6 \hat{\jmath}+\hat{k} \end{aligned}

The shortest distance between the lines is

\begin{gathered} d=\frac{\left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overline{b_{2}}\right)\right|}{\left|\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|} \\\\ \left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)=(-\hat{\imath}-\hat{\jmath}-\hat{k})-(3 \hat{\imath}+5 \hat{\jmath}+7 \hat{k}) \\\\ =-4 \hat{\imath}-6 \hat{\jmath}-8 \hat{k} \end{gathered}

\begin{aligned} \left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right) &=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & -2 & 7 \\ 7 & -6 & 1 \end{array}\right| \\\\ \left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right) &=(-2+42)+[-(1-49) \hat{j}]+(-6+14 \hat{k}) \\\\ &=40 \hat{\imath}+48 \hat{\jmath}+8 \hat{k} \end{aligned}

\begin{aligned} \left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right| &=\sqrt{(40)^{2}+(48)^{2}+(8)^{2}} \\\\ &=8 \sqrt{62} \end{aligned}

\begin{aligned} \left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right| &=|(-4 \hat{\imath}-6 \hat{\jmath}-8 \hat{k})(40 \hat{\imath}+48 \hat{\jmath}+8 \hat{k})| \Rightarrow \\\\ \left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right| &=|-512| \\\\ &=512 \end{aligned}

Putting the values in the above mentioned formula,

\begin{aligned} &d=\frac{512}{8 \sqrt{62}} \\\\ &=\frac{64}{\sqrt{62}} \\\\ &=\frac{64}{\sqrt{62}} \text { units } \end{aligned}

 

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