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  • Explain solution RD Sharma class 12 chapter Straight Line in Space exercise 27.5 question 1 sub question (viii) maths

Explain solution RD Sharma class 12 chapter Straight Line in Space exercise 27.5 question 1 sub question (viii) maths

Answers (1)

Answer: d=14 \text { units }

Hint: Using the expression to find 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 }}=(8+3 \lambda) \hat{\imath}-(9+16 \lambda) \hat{\jmath}+(10+7 \lambda) \hat{k} \text { and } \overrightarrow{\mathrm{\gamma }}=(15 \hat{\imath}+29 \hat{\jmath}+5 \hat{k})+\mu(3 \hat{\imath}+8 \hat{\jmath}-2 \hat{k})

 

Solution: We know that,

\begin{array}{ll} \overrightarrow{a_{1}}=8 \hat{\imath}-9 \hat{\jmath}+10 \hat{k} & \overrightarrow{b_{1}}=3 \hat{\imath}-16 \hat{\jmath}+7 \hat{k} \\\\ \overrightarrow{a_{2}}=15 \hat{\imath}+29 \hat{\jmath}+5 \hat{k} & \overrightarrow{b_{2}}=3 \hat{\imath}+8 \hat{\jmath}-5 \hat{k} \end{array}

The shortest distance between the given lines is,

\begin{aligned} &\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)=(15 \hat{\imath}+29 \hat{\jmath}+5 \hat{k})-(8 \hat{\imath}-9 \hat{\jmath}-10 \hat{k}) \\\\ &\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)=7 \hat{\imath}+38 \hat{\jmath}-5 \hat{k} \\\\ &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 3 & -16 & 7 \\ 3 & -8 & -5 \end{array}\right| \end{aligned}

\begin{aligned} &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=(80-56) \hat{\imath}-(-15-21) \hat{\jmath}+(24+48) \hat{k} \\\\ &\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=24 \hat{\imath}+36 \hat{\jmath}+72 \hat{k} \Rightarrow 4(6 \hat{\imath}+8 \hat{\jmath}+18 \hat{k}) \end{aligned}

\begin{aligned} &\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|=4 \sqrt{(6)^{2}+(9)^{2}+(18)^{2}} \Rightarrow 4 \sqrt{441} \Rightarrow 4 \times 21 \Rightarrow 84 \\\\ &\left|\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right)\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)\right|=|4(7 \hat{\imath}+38 \hat{\jmath}-5 \hat{k})(6 \hat{\imath}+9 \hat{\jmath}+18 \hat{k})| \Rightarrow 4(42+342-90) \Rightarrow 1176 \end{aligned}

Putting three values in the above mentioned equation,

\begin{aligned} &d=\frac{1176}{84} \\\\ &d=14 \text { units } \end{aligned}

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