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  • Provide solution for RD Sharma maths class 12 chapter Straight Line in Space exercise 27.5 question 3 sub question (ii)

Provide solution for RD Sharma maths class 12 chapter Straight Line in Space exercise 27.5 question 3 sub question (ii)

Answers (1)

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: \overrightarrow{\mathrm{\gamma }}=(\hat{\imath}+\hat{\jmath}-\hat{k})+\lambda(3 \hat{\imath}-\hat{\jmath}) \text { and } \overrightarrow{\mathrm{\gamma}}=(4 \hat{\imath}-\hat{k})+\mu(2 \hat{\imath}+3 \hat{k})

Solution:

We know that,

\begin{array}{ll} \overrightarrow{a_{1}}=\hat{\imath}+\hat{\jmath}-\hat{k} & \overrightarrow{b_{1}}=3 \hat{\imath}-1 \hat{\jmath}+0 \hat{k} \\\\ \overrightarrow{a_{2}}=4 \hat{\imath}+0 \hat{\jmath}-\hat{k} & \overrightarrow{b_{2}}=2 \hat{\imath}+0 \hat{\jmath}+3 \hat{k} \end{array}

The shortest distance between the lines is,

\begin{aligned} &\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)=(4 \hat{\imath}+0 \hat{\jmath}-\hat{k})-(\hat{\imath}+\hat{\jmath}-\hat{k}) \\\\ &\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)=3 \hat{\imath}-\hat{\jmath}+0 \hat{k} \\\\ &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 3 & -1 & 0 \\ 2 & 0 & 3 \end{array}\right| \end{aligned}

\begin{aligned} &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=(3-0) \hat{\imath}-(9-0) \hat{\jmath}+(0+2) \hat{k} \\\\ &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=-3 \hat{\imath}-9 \hat{j}+2 \hat{k} \\\\ &\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|=\sqrt{(-3)^{2}+(-9)^{2}+(2)^{2}} \Rightarrow \sqrt{94} \\\\ &\left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|=|(3 \hat{\imath}-1 \hat{\jmath}+0 \hat{k})(-3 \hat{\imath}-9 \hat{\jmath}+2 \hat{k})| \Rightarrow 0 \end{aligned}

Putting three values in the above mentioned equation,

\begin{aligned} &d=\frac{0}{\sqrt{94}} \\\\ &d=0 \end{aligned}

∴ Shortest distance d between the lines is 0, so lines are intersecting.

Posted by

infoexpert26

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