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Please Solve R.D.Sharma class 12 Chapter 27 Straight Line in Space  Exercise 27.2 Question 8 Sub Question 2 Maths Textbook Solution.

Answers (1)

Answer: the angle between the points of lines will be \cos ^{-1}\frac{19}{21}

Given: find the angle between the points of lines

                 \vec{r}=(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-4 \hat{\mathrm{k}})+8(\hat{\mathrm{i}}+2 \hat{j}+2 \hat{k}) and

                \vec{r}=(5 \hat{j}-2 \hat{k})+\mu(3 \hat{i}+2 \hat{j}+6 \hat{k})

Hint:  \cos \theta=\frac{\overrightarrow{b_{1}} \cdot \overrightarrow{b_{2}}}{\left|\vec{b}_{1}\right|\left|\overrightarrow{b_{2}}\right|}

Solution:

                          Let ‘\theta’ be the angle between points of lines

                            \begin{aligned} &\vec{b}_{1}=\hat{i}+2 \hat{j}+2 \hat{k} \\ &\overrightarrow{b_{2}}=3 \hat{i}+2 \hat{j}+6 \hat{k} \\ \end{aligned}

                            \begin{aligned} &\cos \theta=\frac{3+4+12}{\left|\sqrt{1^{2}+2^{2}+2^{2}} \| \sqrt{3^{2}+2^{2}+6^{2}}\right|} \\ \end{aligned}

                              \begin{aligned} &=\frac{19}{|\sqrt{9} \| \sqrt{49}|} \\ \end{aligned}

                                   \begin{aligned} &=\frac{19}{3 \times 7} \\ \end{aligned}\begin{aligned} &=\frac{19}{3 \times 7} \\ \end{aligned}

                        \begin{aligned} &\cos \theta=\frac{19}{21} \\ \end{aligned}

                             \begin{aligned} &\theta=\cos ^{-1} \frac{19}{21} \end{aligned}

                            So the answer will be \cos ^{-1}\frac{19}{21}

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