Given points $A=[3; -1; -2]$, $B=[5; 2; -3]$, $C=[2; -4; 3]$, $K=[-2; 3; 4]$, and $L=[-3; 1; 6]$. Let the plane $\alpha=\overleftrightarrow{ABC}$ and the straight line $p = \overrightarrow{KL}$. Using the direction vector of the line and the normal vector of the plane, determine whether the line $p$ intersects the plane $\alpha$. Solve the task without calculating their intersection.
Elizabeth solved this task in the following steps:
(1) Firstly, she wrote:
- If the normal vector of the plane and the direction vector of the line are perpendicular, then the line is parallel to the plane, or the line lies in the plane.
- If the normal vector of the plane and the direction vector of the line are not perpendicular, then the line intersects the plane.
(2) She determined the direction vector of the line $p$: $$\overrightarrow{u}=\overrightarrow{KL}= L - K = (-1; -2; 2)$$ (3) Then, she determined vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$: $$\begin{aligned} \overrightarrow{AB}&= B - A = (2; 3; -1)\cr\cr \overrightarrow{AC}&= C - A = (-1; -3; 5) \end{aligned}$$ and the normal vector of the plane $\alpha$: $$\overrightarrow{n_{\alpha}}=\overrightarrow{AB}\times\overrightarrow{AC}$$
$$\begin{aligned} \overrightarrow{n_{\alpha}}&=(2\cdot(-3)-(-1)\cdot3; 3\cdot5-(-3)\cdot(-1);-1\cdot(-1)-5\cdot2)=\cr &=(-6+3;15-3;1-10) = (-3;12;-9)\end{aligned}$$
(4) Next, she calculated the scalar product $\overrightarrow{u}\cdot\overrightarrow{n_{\alpha}}$: $$\overrightarrow{u}\cdot\overrightarrow{n_{\alpha}}=-1\cdot(-3)+(-2)\cdot12 + 2\cdot(-9)=3-24-18 =-45$$
(5) Finally, she concluded that $\overrightarrow{u}$ and $\overrightarrow{n_{\alpha}}$ are not perpendicular $(\overrightarrow{u}\cdot\overrightarrow{n_{\alpha}}\neq0)$, therefore the line $p$ intersects the plane $\alpha$.
Is Elizabeth's solution correct? If not, determine where Elizabeth made a mistake.
Elizabeth's solution is correct.
The mistake is in step (1). The statements written in this step are not true.
The mistake is in step (3). Elizabeth incorrectly determined the vector $\overrightarrow{n_{\alpha}}$.
The mistake is in step (4). Elizabeth incorrectly determined the scalar product $\overrightarrow{u}\cdot\overrightarrow{n_{\alpha}}$.
Elizabeth incorrectly determined the vector $\overrightarrow{n_{\alpha}}$. The correct procedure is as follows:
After determining vectors $\overrightarrow{AB}= (2; 3; -1)$ and $\overrightarrow{AC}= (-1; -3; 5)$, we obtain the normal vector of the plane $\alpha$ as the cross product:
$$\overrightarrow{n_{\alpha}}=\overrightarrow{AB}\times\overrightarrow{AC}$$
$$\begin{aligned}
\overrightarrow{n_{\alpha}}&=(3\cdot5-(-3)\cdot(-1); -1\cdot(-1)-5\cdot2;2\cdot(-3)- (-1)\cdot3)=\cr
&=(15-3;1-10;-6+3) = (12;-9;-3)
\end{aligned}$$
Then, the scalar product $\overrightarrow{u}\cdot\overrightarrow{n_{\alpha}}$ is calculated as:
$$\overrightarrow{u}\cdot\overrightarrow{n_{\alpha}}= -1\cdot12 + (-2)\cdot(-9) + 2\cdot(-3) = 0$$
This implies that the normal vector of the plane and the direction vector of the line are perpendicular $(\overrightarrow{u}\cdot\overrightarrow{n_{\alpha}}=0)$. Therefore, the line $p$ is parallel to the plane $\alpha$ or the line $p$ lies in the plane $\alpha$.