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  • Need solution for RD Sharma maths class 12 chapter Straight Line in Space exercise 27.5 question 7 sub question (iii)

Need solution for RD Sharma maths class 12 chapter Straight Line in Space exercise 27.5 question 7 sub question (iii)

Answers (1)

best_answer

Answer: d=\left|\frac{3}{\sqrt{19}}\right|

Hint: using  \frac{\left.\overrightarrow{\left(b_{1}\right.} \times \overrightarrow{\left.b_{2}\right)} \overrightarrow{\left(a_{2}\right.}-\overrightarrow{a_{1}}\right)}{\left(b_{1} \times \overline{\left.b_{2}\right)}\right.}

Given: \overrightarrow{8}=\hat{\imath}+2 \hat{\jmath}+3 \hat{k}+\lambda(\hat{\imath}-3 \hat{\jmath}+2 \hat{k}) and

            \overrightarrow{8}=4 \hat{\imath}+5 \hat{\jmath}+6 \hat{k}+\mu(2 \hat{\imath}+3 \hat{\jmath}+\hat{k})

Solution:

Shortest distance between the lines with vector equations

\overrightarrow{8}=\overrightarrow{a_{1}}+\overrightarrow{b_{1}} \text { and } \overrightarrow{8}=\overrightarrow{a_{2}}+\overrightarrow{b_{2}} \text { is }\left|\frac{\overrightarrow{\left(b_{1}\right.} \times \overrightarrow{\left.b_{2}\right)}}{\left.\overline{\left(b_{1}\right.} \times \overline{a_{2}}-\overrightarrow{a_{1}}\right)}\right|

Now

\overrightarrow{8}=\hat{\imath}+2 \hat{\jmath}+3 \hat{k}+\lambda(\hat{\imath}-3 \hat{\jmath}+2 \hat{k})                           \overrightarrow{8}=(4 \hat{\imath}+5 \hat{\jmath}+6 \hat{k}+\mu(2 \hat{\imath}+3 \hat{\jmath}+\hat{k})

Comparing with \overrightarrow{8}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}}                                         Comparing with \overrightarrow{8}=\overrightarrow{a_{2}}+\mu \overrightarrow{b_{2}}

\overrightarrow{a_{1}}=\hat{i}+2 \hat{j}+3 \widehat{k}                                                           \overrightarrow{a_{2}}=4 \hat{i}+5 \hat{j}+6 \widehat{k}

and                                                                                      and

\overrightarrow{b_{1}}=\hat{i}-3 \hat{j}+2 \widehat{k}                                                            \overrightarrow{b_{2}}=2 \hat{i}-3 \hat{j}+1 \widehat{k}

 

\begin{aligned} &\left.\overrightarrow{\left(a_{2}\right.}-\overrightarrow{a_{1}}\right)=(4 \hat{\imath}+5 \hat{\jmath}+6 \hat{k}-(1 \hat{\imath}+2 \hat{\jmath}+3 \hat{k}) \\ &=(4-1) \hat{\imath}+(5-2) \hat{\jmath}+(6-3) \hat{k} \\ &=3 \hat{\imath}+3 \hat{\jmath}+3 \hat{k} \end{aligned}

\overrightarrow{\left(b_{1}\right.} x \overrightarrow{\left.b_{2}\right)}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & -3 & 2 \\ 2 & 3 & 1 \end{array}\right|

\begin{aligned} &=\hat{\imath}[(-3 \times 1)-(3 \times 2)]-\hat{\jmath}[(1 \times 1)-(2 \times 2)]+\hat{k}[(1 \times 3)-(2 \times-3)] \\ &=\hat{\imath}(-9)-\hat{\jmath}(-3)+\hat{k}(9) \\ &=9 \hat{\imath}+3 \hat{\jmath}+9 \hat{k} \end{aligned}

Magnitude of \overrightarrow{\left(b_{1}\right.} \times \overrightarrow{\left.b_{2}\right)}=\sqrt{(-9)^{2}+3^{2}+9^{2}}

                                      \begin{aligned} &=\sqrt{81+9+81} \\ &=\sqrt{171} \end{aligned}

                                      \begin{aligned} &=\sqrt{9 \times 29} \\ &=3 \sqrt{19} \end{aligned}

Also

\begin{aligned} &\left.\overrightarrow{\left(b_{1}\right.} \times \overrightarrow{\left.b_{2}\right)} \overrightarrow{\left(a_{2}\right.}-\overrightarrow{a_{1}}\right)=(-9(+3 \hat{\jmath}+9 \hat{k})(3 \hat{\imath}+3 \hat{\jmath}+3 \hat{k}) \\ &=(-9 \times 3)+(3 \times 3)+(9+3) \\ &=-27+9+27 \\ &=9 \end{aligned}

Shortest Distance \begin{aligned} &=\left|\frac{9}{3 \sqrt{19}}\right| \\\\ \end{aligned}

                            =\left|\frac{3}{\sqrt{19}}\right|

                                    

 

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