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Q.
The angle between the straight lines $x - 1 =\frac{2y+3}{3}=\frac{z+5}{2}$ and $x = 3r + 2; y = -2r - 1; z= 2$, where $ r$ is a parameter, is
Introduction to Three Dimensional Geometry
Solution:
Equation of first line is $\frac{x-1}{1}=\frac{y+\frac{3}{2}}{\left(\frac{3}{2}\right)}= \frac{z+5}{2}$
equation of second line is $\frac{x-2}{3 }=\frac{y+1}{-2}=\frac{z-2}{0}$
cos $\theta = \frac{1.3 + \left(\frac{3}{2}\right)\left(-2\right)+2.0}{\sqrt{\left(1\right)^{2 }+\left(\frac{3}{2}\right)^{2 }+\left(2\right)^{2} }\sqrt{\left(3\right)^{2} + \left(-2\right)^{2} + 10^{2}}}$
cos $\theta$ = 0 $\Rightarrow $ = 90$^\circ$