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  • Explain solution RD Sharma class 12 chapter Straight Line in Space exercise 27.5 question 7sub question (iv) maths

Explain solution RD Sharma class 12 chapter Straight Line in Space exercise 27.5 question 7sub question (iv) maths

Answers (1)

Answer:  d=9

Hint: using the expression  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|}

Given:

 

\begin{aligned} &\overrightarrow{8}=6 \hat{\imath}+2 \hat{\jmath}+2 \hat{k}+\lambda(\hat{\imath}-2 \hat{\jmath}+2 \hat{k}) \text { and } \\\\ &\overrightarrow{8}=4 \hat{\imath}-\hat{k}+\boldsymbol{\mu}(2 \hat{\imath}+3 \hat{\jmath}+\hat{k}) \end{aligned}

 

Solution:

\begin{aligned} &\overrightarrow{a_{1}}=6 \hat{\imath}+2 \hat{\jmath}+2 \hat{k} \\\\ &\overrightarrow{a_{2}}=4 \hat{\imath}-\hat{k} \\\\ &\overrightarrow{b_{1}}=\hat{\imath}-2 \hat{\jmath}+2 \hat{k} \end{aligned}

\begin{aligned} &\overrightarrow{b_{2}}=\hat{3}_{l}-2 \hat{\jmath}-2 \hat{k} \\\\ &\overrightarrow{a_{2}}-\overrightarrow{a_{1}}=-10 \hat{\imath}-2 \hat{\jmath}+3 \hat{k} \end{aligned}

\text { and } \overrightarrow{b_{1}} \times \overrightarrow{b_{2}}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & -2 & 2 \\ 3 & -2 & -2 \end{array}\right| \Rightarrow 8 \hat{\imath}+8 \hat{\jmath}+4 \hat{k}

\begin{aligned} &\Rightarrow\left|\overrightarrow{b_{1}} \times \overline{b_{2}}\right| \Rightarrow \sqrt{8^{2}+8^{2}+4^{2}} \Rightarrow \sqrt{64+64+16} \\\\ &=\sqrt{144} \Rightarrow 12 \end{aligned}

and

\begin{aligned} &\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=-10 \hat{\imath}-2 \hat{\jmath}+3 \hat{k}(8 \hat{\imath}+8 \hat{\jmath}+4 \hat{k}) \\\\ &=-80-16-12 \\\\ &=-108 \end{aligned}

The shortest distance between the lines

\begin{aligned} &\overrightarrow{8}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}} \text { and } \overrightarrow{8}=\overrightarrow{a_{2}}+\mu b_{2} \\\\ &\mathrm{~d}=\left|\frac{-108}{12}\right| \\\\ &=9 \end{aligned}

 

 

 

 

 

 

Posted by

infoexpert26

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