Question

Find the angle between the lines x=1,z-y=0 and 2x-y=-1, z=1.

Answer 1

Answer on Question #45083 – Math – Analytic Geometry

Question

Find the angle between the lines $x=1, z-y=0$ and $2x-y=-1$, $z=1$.

Solution

The equation of the first line could be rewritten as $\frac{x-1}{0} = \frac{y}{1} = \frac{z}{1}$, and the equation of the second line could be rewritten as $\frac{x + \frac{1}{2}}{\frac{1}{2}} = \frac{y}{1} = \frac{z - 1}{0}$. Hence the direction vector of the first line is $(0,1,1)$ and the direction vector of the second line is $\left(\frac{1}{2}, 1, 0\right)$. The angle between lines equals the angle between direction vectors, hence $\cos \alpha = \frac{0\frac{1}{2} + 1\cdot 1 + 1\cdot 0}{\sqrt{0^2 + 1^2 + 1^2}\sqrt{\left(\frac{1}{2}\right)^2 + 1^2 + 0^2}} = \frac{1}{\sqrt{2}\cdot\sqrt{\frac{1}{4} + 1}} = \sqrt{\frac{2}{5}}$, $\alpha = \arccos \sqrt{\frac{2}{5}}$ (approximately $51{}^\circ$).

Answer: $\alpha = \arccos \sqrt{\frac{2}{5}}$.

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