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  • Please solve RD Sharma class 12 chapter Straight Line in Space exercise 27.5 question 5 sub question (i) maths textbook solution

Please solve RD Sharma class 12 chapter Straight Line in Space exercise 27.5 question 5 sub question (i) maths textbook solution

Answers (1)

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Answer: d=0

 

Hint:  \frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}

 

Given: (0,0,0) \text { and }(1,0,2)

Solution:

Applying the given point in the formula of the straight line

\begin{aligned} &\frac{x-0}{1-0}=\frac{y-0}{0-0}=\frac{z-0}{2-1} \\\\ &\Rightarrow \frac{x}{1}=\frac{y}{0}=\frac{z}{2} \end{aligned}

Solution (ii)

Again applying the points(1,3,0)  and (0,3,0)

The equation of the line passing through the points (1,3,0) is,

\begin{aligned} &\frac{x-1}{0-1}=\frac{y-3}{3-3}=\frac{z-0}{2-1} \\\\ &\Rightarrow \frac{x-1}{-1}=\frac{y-3}{0}=\frac{z}{0} \end{aligned}

Since the first line passes through the points (1,3,0) and has the directions proportional to (1,0,2) its vector is,

\overrightarrow{\mathrm{\gamma}}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}} \rightarrow(1)

Hence,

\begin{aligned} &\vec{a}=0 \hat{\imath}+0 \hat{\jmath}+0 \hat{k} \\\\ &\vec{b}=\hat{i}+0 \hat{\jmath}+2 \hat{k} \end{aligned}

Also, the second line passes through the point (1,3,0)  and has the directions ratios proportional to (-1,0,0)  its vector is,

\overrightarrow{\mathrm{\gamma}}=\overrightarrow{a_{2}}+\mu \overrightarrow{b_{2}} \rightarrow \text { (2) }

Here,

\begin{aligned} &\overrightarrow{a_{2}}=\hat{\imath}+3 \hat{j}+0 \hat{k} \\\\ &\overrightarrow{b_{2}}=\hat{\imath}+0 \hat{\jmath}+0 \hat{k} \end{aligned}

Now,

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

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

The shortest distance between the lines,

\begin{aligned} \overrightarrow{\mathrm{\gamma }} &=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}} \\\\ \overrightarrow{\mathrm{\gamma }} &=\overrightarrow{a_{2}}+\mu \overrightarrow{b_{2}} \end{aligned}

\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{|-6|}{|2|} \\\\ &d=3 \end{aligned}

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