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  • Please solve RD Sharma class 12 chapter Straight Line in Space exercise 27.5 question 3 sub question (i) maths textbook solution

Please solve RD Sharma class 12 chapter Straight Line in Space exercise 27.5 question 3 sub question (i) maths textbook solution

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Answer: Given Lines are not interesting

Hint: Using the equation    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: \vec{\gamma}=(\hat{\imath}-\hat{\jmath})+\lambda(2 \hat{\imath}+\hat{k}) \text { and } \overrightarrow{\mathrm{\gamma}}=(2 \hat{\imath}-\hat{\jmath})+\mu(\hat{\imath}+\hat{\jmath}+\hat{k})

Solution:

We know that,

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

\begin{aligned} &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=(0-1) \hat{\imath}-(-2-1) \hat{\jmath}+(2-0) \hat{k} \\\\ &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=-\hat{\imath}+3 \hat{\jmath}+2 \hat{k} \\\\ &\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|=\sqrt{(-1)^{2}+(3)^{2}+(2)^{2}} \Rightarrow \sqrt{14} \\\\ &\left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|=|(\hat{\imath}+0 \hat{\jmath}+0 \hat{k})(-\hat{\imath}+3 \hat{\jmath}+2 \hat{k})| \Rightarrow 1 \end{aligned}

Putting three values in the above mentioned equation,

\begin{aligned} &d=\frac{1}{\sqrt{14}} \\\\ &d=\frac{1}{\sqrt{14}} \text { units } \end{aligned}

∴ Shortest distance d between the lines is not 0, so lines are not intersecting.

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