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  • Provide solution for RD Sharma maths class 12 chapter Straight Line in Space exercise 27.5 question 2 sub question (ii)

Provide solution for RD Sharma maths class 12 chapter Straight Line in Space exercise 27.5 question 2 sub question (ii)

Answers (1)

Answer: d=\frac{3}{\sqrt{59}} \text { units }

Hint: Consider the given equation as (1) and (2) and 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: : \frac{x-1}{2}=\frac{y+1}{3}=z \rightarrow(1) \text { and } \frac{x+1}{3}=\frac{y-2}{1} ; z=2 \rightarrow(2)

Solution:

Since line (1) passes through the point (1,-1,0)  and has direction ratios proportional to (2,3,1) its vector equation is \overrightarrow{\mathrm{\gamma}}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}}

Here,

\begin{aligned} &\overrightarrow{a_{1}}=\hat{\imath}-\hat{\jmath}+0 \hat{k} \\\\ &\overrightarrow{b_{1}}=2 \hat{\imath}+3 \hat{\jmath}+\hat{k} \end{aligned}

Also line (2) passes through the point (-1,2,2) and has direction ratios proportional to (3,1,0) its vector equation is \vec{\gamma}=\overrightarrow{a_{2}}+\mu \overrightarrow{b_{2}}

Here,

\begin{aligned} &\overrightarrow{a_{2}}=-\hat{\imath}+2 \hat{\jmath}+2 \hat{k} \\\\ &\overrightarrow{b_{2}}=3 \hat{\imath}+\hat{\jmath}+0 \hat{k} \end{aligned}

Now,

\begin{aligned} \left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) &=-2 \hat{\imath}+2 \hat{\jmath}+2 \hat{k} \\\\ \left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right) &=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 3 & 1 \\ 3 & 1 & 0 \end{array}\right| \\\\ &=-\hat{\imath}+3 \hat{\jmath}-7 \hat{k} \end{aligned}

\begin{aligned} &\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|=\sqrt{(-1)^{2}+(3)^{2}+(-7)^{2}} \Rightarrow \sqrt{1+9+49} \Rightarrow \sqrt{59} \text { and } \\\\ &\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=(-2 \hat{\imath}+3 \hat{\jmath}+2 \hat{k})(-\hat{\imath}+3 \hat{\jmath}-7 \hat{k}) \Rightarrow 2+9-14 \Rightarrow-3 \end{aligned}

The shortest distance between the lines \overrightarrow{\mathrm{\gamma}}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}} \text { and } \overrightarrow{\mathrm{V}}=\overrightarrow{a_{2}}+\mu \overrightarrow{b_{2}}  is given by

d=\left|\frac{-3}{\sqrt{59}}\right| \Rightarrow \frac{3}{\sqrt{59}} \text { units }

 

Posted by

infoexpert26

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