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Question Number 53034 by gopikrishnan005@gmail.com last updated on 16/Jan/19

$$\mathrm{Angle}\mathrm{between}\mathrm{the}\mathrm{lines}\frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}-1}{1}=\frac{\mathrm{z}-1}{2}\mathrm{and}\frac{\mathrm{x}-1}{-\sqrt{3-1}}=\frac{\mathrm{y}-1}{\sqrt{3-1}}=\frac{\mathrm{z}-1}{4}\mathrm{is}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 16/Jan/19

$$cos\theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\times \sqrt{a_2^2+b_2^2+c_2^2}}$$$$usethisformula$$$$\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$$$$\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$$

Answered by ajfour last updated on 16/Jan/19

$${\overline{b}}_1=\hat{i}+\hat{j}+2\hat{k}$$$${\overline{b}}_2=-(\sqrt{3}+1)\hat{i}+(\sqrt{3}-1)\hat{j}+4\hat{k}$$$$\cos \theta =\frac{-\sqrt{3}-1+\sqrt{3}-1+8}{\sqrt{6}\sqrt{{(\sqrt{3}+1)}^2+{(\sqrt{3}-1)}^2+16}}$$$$=\frac{\sqrt{6}}{\sqrt{24}}=\frac{1}{2}$$$$\Rightarrow \theta =\frac{\pi }{3}.$$

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