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

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

Answers (1)

Answer: d=\frac{10}{\sqrt{59}} \text { units }

Hint: Using the expression 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{\jmath}+\hat{k}) \text { and } \vec{\gamma}=(2 \hat{\imath}+\hat{\jmath}-\hat{k})+\mu(3 \hat{\imath}-5 \hat{\jmath}+2 \hat{k})

Solution: We know that,

\begin{array}{ll} \overrightarrow{a_{1}}=\hat{\imath}+\hat{\jmath} & \overrightarrow{b_{1}}=2 \hat{\imath}-\hat{\jmath}+\hat{k} \\\\ \overrightarrow{a_{2}}=2 \hat{\imath}+\hat{\jmath}-\hat{k} & \overrightarrow{b_{2}}=3 \hat{\imath}-5 \hat{\jmath}+2 \hat{k} \end{array}

It can be written as,

\begin{aligned} &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & -1 & 1 \\ 3 & -5 & 2 \end{array}\right| \\\\ &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=(-2+5) \hat{\imath}-(4-3) \hat{\jmath}+(-10+3) \hat{k} \\\\ &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=3 \hat{\imath}-\hat{\jmath}-7 \hat{k} \end{aligned}

\begin{aligned} &\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|=\sqrt{(3)^{2}+(-1)^{2}+(-7)^{2}} \Rightarrow \sqrt{9+1+49} \Rightarrow \sqrt{59} \\\\ &\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)=(2-1) \hat{\imath}-(1-1) \hat{\jmath}+(-1-0) \hat{k}=\hat{\imath}+0 \hat{\jmath}-\hat{k} \end{aligned}

We can write as,

\begin{aligned} \left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right) \cdot\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) &=(3 \hat{\imath}-\hat{\jmath}-7 \hat{k}) \cdot(\hat{\imath}+0 \hat{\jmath}-\hat{k}) \\\\ &=(3 \times 1)+[(-1) \times 0]+([(-7) \times(-1)\\\\ &=3+0+7 \Rightarrow 10 \end{aligned}

The shortest distance between the given lines is,

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|} \Rightarrow \frac{10}{\sqrt{59}} \text { units }

Posted by

infoexpert26

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