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

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

Answers (1)

Answer: d=\frac{8}{\sqrt{29}} \text { units }

Hint: Consider the given equation as (1) and (2)

Given: \frac{x-1}{-1}=\frac{y+2}{1}=\frac{z-3}{-2} \rightarrow(1) \text { and } \frac{x-1}{1}=\frac{y+1}{2}=\frac{z+1}{-2} \rightarrow(2)

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

Here,

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

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

Here,

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

Now,

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

\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|=\sqrt{(2)^{2}+(-4)^{2}+(-3)^{2}} \Rightarrow \sqrt{4+16+9} \Rightarrow \sqrt{29} \text { and }

The shortest distance between the lines

\begin{aligned} &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|} \\\\ &d=\left|\frac{8}{\sqrt{29}}\right| \Rightarrow \frac{8}{\sqrt{29}} \text { units } \end{aligned}

Posted by

infoexpert26

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