Find the angle between lines $p$ and $q$, where $$p\colon y=2x-1\ \mbox{ and }\ q\colon 3x+7y-4=0.$$
Michael's solution:
(1) The general form of the equation of line $p$ is $$2x-y-1=0.$$
(2) Normal vectors of both lines are: $$\overrightarrow{n}_p=(2;-1),\ \overrightarrow{n}_q=(3;7).$$
(3) The angle between the lines $p$ and $q$ is calculated by the formula $$\cos\varphi=\frac{\overrightarrow{n}_p\cdot\overrightarrow{n}_q}{|\overrightarrow{n}_p|\cdot|\overrightarrow{n}_q|}.$$
(4) After substituting to the formula we obtain: $$\cos\varphi=\frac{2\cdot3+(-1)\cdot7}{\sqrt{2^2+(-1)^2}\cdot\sqrt{3^2+7^2}}=\frac{-1}{\sqrt{5}\cdot\sqrt{58}}=\frac{-1}{\sqrt{290}}.$$
(5) $\varphi=\arccos\left(\frac{-1}{\sqrt{290}}\right)\cong 93^\circ\, 22´$ (using calculator).
Michael's solution is wrong. Where did Michael make a mistake in his solution?
The mistake is in the step (2). Normal vectors of both lines are: $$\overrightarrow{n}_p=(1;2), \ \overrightarrow{n}_q=(7;-3).$$
The mistake is in the step (3). The angle between the lines $p$ and $q$ is calculated by the formula $$\cos\varphi=\frac{|\overrightarrow{n}_p\cdot\overrightarrow{n}_q|}{|\overrightarrow{n}_p|\cdot|\overrightarrow{n}_q|}.$$
The mistake is in the step (4). By substituting we obtain: $$\cos\varphi=\frac{2\cdot3+(-1)\cdot7}{|2^2+(-1)^2|\cdot|3^2+7^2|}=\frac{-1}{5\cdot58}=\frac{-1}{290}.$$
The mistake is in the step (5). The angle value is incorrectly calculated on the calculator. The correct solution is: $\varphi\cong 3^\circ\, 22´$.
$$\begin{aligned} &\overrightarrow{n}_p=(2;-1)\mbox{,}\qquad\overrightarrow{n}_q=(3;7)\cr &\cos\varphi=\frac{|\overrightarrow{n}_p\cdot\overrightarrow{n}_q|}{|\overrightarrow{n}_p|\cdot|\overrightarrow{n}_q|}=\frac{|2\cdot3+(-1)\cdot7|}{\sqrt{2^2+(-1)^2}\cdot\sqrt{3^2+7^2}}=\frac{|-1|}{\sqrt{5}\cdot\sqrt{58}}=\frac{1}{\sqrt{290}}\implies\varphi\cong86^\circ\, 38´\textrm{.} \end{aligned}$$