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  • Please solve RD Sharma class 12 chapter Straight Line in Space exercise 27.5 question 2 sub question (i) maths textbook solution

Please solve RD Sharma class 12 chapter Straight Line in Space exercise 27.5 question 2 sub question (i) maths textbook solution

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Answer: d=\frac{1}{\sqrt{6}} \text { units }

Hint: Using the expression to find 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-2}{3}=\frac{z-3}{4} \rightarrow(1) \text { and } \frac{x-2}{3}=\frac{y-3}{4}=\frac{z-5}{4} \rightarrow(2)

Solution: Since the given line (1) passes through the point (1,2,3) and has direction ratios proportional to (2,3,4)  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}}=2 \hat{\imath}+3 \hat{\jmath}+4 \hat{k} \end{aligned}

Also line (2) passes through the point (2,3,5) and has direction ratios proportional to (3,4,5)  its vector equation is \overrightarrow{\mathrm{\gamma}}=\overrightarrow{a_{2}}+\mu \overrightarrow{b_{2}}

Here,

\begin{aligned} &\overrightarrow{a_{2}}=2 \hat{\imath}+3 \hat{\jmath}+5 \hat{k} \\\\ &\overrightarrow{b_{2}}=3 \hat{\imath}+4 \hat{\jmath}+5 \hat{k} \end{aligned}

Now,

\begin{aligned} \left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) &=\hat{\imath}+\hat{j}+2 \hat{k} \\\\ \left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right) &=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{array}\right| \\\\ &=-\hat{\imath}+2 \hat{\jmath}-\hat{k} \end{aligned}

\begin{aligned} &\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|=\sqrt{(-1)^{2}+(2)^{2}+(-1)^{2}} \Rightarrow \sqrt{6} \text { and } \\\\ &\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=(\hat{\imath}+\hat{\jmath}+2 \hat{k})(-\hat{\imath}+2 \hat{\jmath}-\hat{k}) \Rightarrow-1+2-2 \Rightarrow-1 \end{aligned}

The shortest distance between the given lines is,

d=\left|\frac{-1}{\sqrt{6}}\right| \Rightarrow \frac{1}{\sqrt{6}} \text { units }

 

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infoexpert26

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