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

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

Answers (1)

Answer: d=\frac{3}{\sqrt{2}} \text { units }

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: \overrightarrow{\mathrm{\gamma}}=(2 \hat{\imath}-\hat{\jmath}-\hat{k})+\lambda(2 \hat{\imath}-5 \hat{\jmath}+2 \hat{k}) \text { and } \overrightarrow{\mathrm{\gamma}}=(\hat{\imath}+2 \hat{\boldsymbol{j}}+\hat{k})+\mu(\hat{\imath}-\hat{\jmath}+\hat{k})

Solution:

\begin{array}{ll} \overrightarrow{a_{1}}=2 \hat{\imath}-\hat{\jmath}-\hat{k} & \overrightarrow{b_{1}}=2 \hat{\imath}-5 \hat{\jmath}+2 \hat{k} \\\\ \overrightarrow{a_{2}}=\hat{\imath}+2 \hat{\jmath}+\hat{k} & \overrightarrow{b_{2}}=\hat{\imath}-\hat{\jmath}+\hat{k} \end{array}

The shortest distance between the lines is,

\begin{aligned} &\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)=(\hat{\imath}+2 \hat{\jmath}+\hat{k})-(2 \hat{\imath}-\hat{\jmath}-\hat{k}) \\\\ &\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)=-\hat{\imath}+3 \hat{\jmath}+2 \hat{k} \end{aligned}

\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & -5 & 2 \\ 1 & -1 & 1 \end{array}\right|

\begin{aligned} &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=(-5+2) \hat{\imath}-(2-2) \hat{\jmath}+(-2+5) \hat{k} \\\\ &\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|=\sqrt{(-3)^{2}+(0)^{2}+(3)^{2}} \\\\ &\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|=3 \sqrt{2} \end{aligned}

\left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|=|(-\hat{\imath}+3 \hat{\jmath}+2 \hat{k})(-3 \hat{\imath}+0 \hat{\jmath}+3 \hat{k})| \Rightarrow 9

Putting three values in the above mentioned equation,

\begin{aligned} &d=\frac{9}{3 \sqrt{2}} \\\\ &d=\frac{3}{\sqrt{2}} \text { units } \end{aligned}

 

Posted by

infoexpert26

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