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  • Please solve RD Sharma class 12 chapter Straight Line in Space exercise 27.5 question 7 sub question (i) maths textbook solution

Please solve RD Sharma class 12 chapter Straight Line in Space exercise 27.5 question 7 sub question (i) maths textbook solution

Answers (1)

Answer: shortest distance =\frac{3}{\sqrt{2}}

Hints: using expression   \mathrm{D}=\left|\frac{\left.\overrightarrow{\left(a_{2}\right.}-\overrightarrow{\left.a_{1}\right)} \overrightarrow{\left(b_{1}\right.} \times \overrightarrow{b_{2}}\right)}{\left(b_{1} \times \overline{b_{2}}\right)}\right|

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

Solution:

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

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

\begin{aligned} &=\hat{\imath}(-3)-\hat{\jmath}(0)+\hat{k}(1+2) \\\\ &=-3 \hat{\imath}+3 \hat{k} \\\\ &\mathrm{SD}=\frac{(2 \hat{i}-\hat{j}-\hat{k})(-3 \hat{i}+3 \hat{k})}{\sqrt{9+9}} \end{aligned}

\begin{aligned} &=\left|\frac{6-3}{\sqrt{18}}\right| \\\\ &=\frac{9}{3 \sqrt{2}} \\\\ &=\frac{3}{\sqrt{2}} \quad \text { units } \end{aligned}

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