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  • Please Solve R.D.Sharma class 12 Chapter 27 Straight Line in Space Exercise 27.2 Question 8 Sub Question 3 Maths Textbook Solution.

Please Solve R.D.Sharma class 12 Chapter 27 Straight Line in Space Exercise 27.2 Question 8 Sub Question 3 Maths Textbook Solution.

Answers (1)

Answer: the angle between the points of lines will be \frac{\pi }{3}

Given: find the angle between the points of lines

                    \begin{aligned} &\vec{r}=\lambda(\hat{i}+\hat{j}+2 \hat{k}) \\ &\vec{r}=2 \hat{j}+\mu(\sqrt{3}-1) \hat{\mathrm{i}}-(\sqrt{3}+1) \hat{j}+4 \hat{k} \end{aligned}

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

Solution:

\begin{aligned} &\vec{r}=2 \hat{j}+\mu[(-1) \hat{i}-(1) \hat{j}+4 \hat{k}] \\\\ &\overrightarrow{b_{1}}=\sqrt{1^{2}+1^{2}+2^{2}}=\sqrt{6} \\\\ &\overrightarrow{b_{2}}=\sqrt{(\sqrt{3}-1)^{2}+(\sqrt{3}+1)^{2}+4^{2}}=\sqrt{24}=2 \sqrt{6} \\\\ &\therefore \cos \theta=\frac{(\hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}})[(\sqrt{3-1})-(\sqrt{3+1})+4 \hat{k}}{\sqrt{6} \times 2 \sqrt{6}} \\ &\qquad \begin{array}{c} \end{array} \end{aligned}

                              \begin{aligned} \cos \theta=\frac{1}{2} \therefore \theta=\frac{\pi}{3} \\\\ \Rightarrow \cos \theta=\cos 60^{\circ} \end{aligned}

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