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Question Number 37661 by Rio Mike last updated on 16/Jun/18

$$\mathrm{Given}\mathrm{the}\mathrm{lines}$$$$l_1:\mathrm{r}=-5i+2j+s(3i-j)$$$$l_2:r=-2i+j+t(2i+j)$$$$F\mathrm{ind}\mathrm{the}\mathrm{cosine}\mathrm{of}\mathrm{the}\mathrm{angle}$$$$\mathrm{between}{l}_{1}and{l}_2.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 16/Jun/18

$$ix+yj+5i-2j=s(3i-j)$$$$i(x+5)+j(y-2)=3si-sj$$$$\frac{x+5}{3}=\frac{y-2}{-1}$$$$ix+jy=-2i+j+t(2i+j)$$$$i(x+2)+j(y-1)=2ti+tj$$$$\frac{x+2}{2}=\frac{y-1}{1}$$$$cos\theta =\frac{3\times 2-1\times 1}{\sqrt{3^2+{(-1)}^2\times \sqrt{2^2+1^2}}}$$$$cos\theta =\frac{5}{\sqrt{10}\times \sqrt{5}}=\frac{\sqrt{5}}{\sqrt{5}\times \sqrt{2}}=\frac{1}{\sqrt{2}}=cos\frac{\mathrm{\Pi }}{4}$$$$\theta =\frac{\mathrm{\Pi }}{4}$$

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