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  • Need solution for RD Sharma maths class 12 chapter Straight Line in Space exercise 27.5 question 1sub question (iii)

Need solution for RD Sharma maths class 12 chapter Straight Line in Space exercise 27.5 question 1sub question (iii)

Answers (1)

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

Hint: Using formula 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}+2 \hat{\jmath}+3 \hat{k})+\lambda(2 \hat{\imath}+3 \hat{\jmath}+4 \hat{k}) \text { and } \overrightarrow{\mathrm{\gamma}}=(2 \hat{\imath}+4 \hat{\jmath}+5 \hat{k})+\mu(3 \hat{\imath}+4 \hat{\jmath}+5 \hat{k})

Solution:

By the given pairs of equation,

\begin{aligned} &\overrightarrow{a_{1}}=\hat{\imath}+2 \hat{\jmath}+3 \hat{k} \\\\ &\overrightarrow{a_{2}}=2 \hat{\imath}+4 \hat{j}+5 \hat{k} \\\\ &\overrightarrow{b_{1}}=2 \hat{\imath}+3 \hat{\jmath}+5 \hat{k} \\\\ &\overrightarrow{b_{2}}=3 \hat{\imath}+4 \hat{\jmath}+5 \hat{k} \end{aligned}

The shortest distance between the lines is,

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

\begin{aligned} &\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| \\\\ &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\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}} \\\\ &\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|=\sqrt{6} \\\\ &\left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|=|(\hat{\imath}+2 \hat{\jmath}+2 \hat{k})(-\hat{\imath}+22 \hat{\jmath}-\hat{k})| \Rightarrow 1 \end{aligned}

Putting three values in the expression,

\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=\frac{1}{\sqrt{6}} \text { units } \end{aligned}

 

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