Angle of Line and Plane

Given the points $P=[3; 1; -4]$ and $Q=[5; -2; -1]$, and the general equation of the plane $\rho:\ 3x - 2y + z - 4 = 0$. Calculate the angle of the line $PQ$ and the plane $\rho$.

Alice solved this problem in the following steps:

(1) Firstly, she wrote the formula for determining the angle $$\cos\varphi=\left|\frac{\overrightarrow{u}\cdot\overrightarrow{n_{\rho}}}{\left|\ \overrightarrow{u}\ \right|\cdot\left|\overrightarrow{n_{\rho}}\right|}\right|,$$

where:

• $\varphi$ is an angle of the line $PQ$ and the plane $\rho$,
• $\overrightarrow{u}$ is a direction vector of the line $PQ$,
• $\overrightarrow{n_{\rho}}$ is a normal vector of the plane $\rho$.

(2) Then, she calculated the coordinates of the direction vector of the line $PQ$: $$\overrightarrow{u}=\overrightarrow{PQ}= Q - P = (2; -3; 3)$$ and determined the coordinates of the normal vector of the plane $\rho$: $$\overrightarrow{n_{\rho}}= (3; -2; 1).$$

(3) Next, she substituted for $\overrightarrow{u}$ and $\overrightarrow{n_{\rho}}$ into the formula and calculated the value of $\cos\varphi$: \begin{aligned} \cos\varphi&=\left|\frac{\overrightarrow{u}\cdot\overrightarrow{n_{\rho}}}{\left|\ \overrightarrow{u}\ \right|\cdot\left|\overrightarrow{n_{\rho}}\right|}\right|=\left|\frac{(2; -3; 3)\cdot (3; -2; 1)}{\left|(2; -3; 3)\right|\cdot\left|(3; -2; 1)\right|}\right|=\cr &=\left|\frac{2\cdot3 +(-3)\cdot(-2)+3\cdot1}{\sqrt{2^2+(-3)^2+ 3^2}\cdot\sqrt{3^2+(-2)^2+ 1^2}}\right|=\cr &=\left|\frac{6 + 6 + 3}{\sqrt{4 + 9 + 9}\cdot\sqrt{9 + 4 + 1}}\right|=\cr &=\left|\frac{15}{\sqrt{22}\cdot\sqrt{14}}\right|=\frac{15}{\sqrt{308}}. \end{aligned}

(4) Finally, using a calculator, she calculated $\varphi$: $$\varphi= \arccos\left(\frac{15}{\sqrt{308}}\right)= 31^{\circ}16'22''.$$ Is Alice's solution correct? If not, determine where Alice made a mistake in the procedure.