직선과 평면의 방정식

(a) One side of the triangle is $\overrightarrow{AB}=\left< 1,0,1\right> - \left< 0,1,1\right> = \left< 1,-1,0\right>$. A second side of the triangle is $\overrightarrow{AC}=\left< 1,3,0\right> - \left< 0,1,1\right> = \left< 1,2,-1\right>$. If the angle between $\overrightarrow{AB}$ and $\overrightarrow{AC}$ is $\theta$ and if we take $\overrightarrow{AB}$ as the base of the triangle, then the triangle has base length $b=|\overrightarrow{AB}|$ and height $h=|\overrightarrow{AC}|\sin\theta$ and hence

$$\text{area}=\tfrac{1}{2} bh =\tfrac{1}{2} |\overrightarrow{AB}|\,|\overrightarrow{AC}|\sin\theta =\tfrac{1}{2}|\overrightarrow{AB}\times\overrightarrow{AC}| =\tfrac{1}{2}|\left< 1,-1,0\right> \times \left< 1,2,-1\right>|$$

As

$$\left< 1,-1,0\right>\times\left< 1,2,-1\right> =\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ 1 & -1 & 0 \\ 1 & 2 & -1 \end{matrix}\right] =\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}+3\hat{\mathbf{k}}$$

we have

$$\text{area}=\tfrac{1}{2}|\left< 1,1,3\right>|=\tfrac{1}{2}\sqrt{11}\approx 1.658$$

(b) A unit vector perpendicular to the plane containing the triangle is

$$\hat{\mathbf{n}}=\frac{\overrightarrow{AB}\times\overrightarrow{AC}} {|\overrightarrow{AB}\times\overrightarrow{AC}|} =\frac{\left< 1,1,3\right>}{\sqrt{11}}$$

The distance from $P$ to the plane containing the triangle is the length of the projection of $\overrightarrow{AP}=\left< 5,-10,2\right>-\left< 0,1,1\right> =\left< 5,-11,1\right>$ on $\hat{\mathbf{n}}$. If $\theta$ the angle between $\overrightarrow{AP}$ and $\hat{\mathbf{n}}$, then this is

$$\text{distance}= |\overrightarrow{AP}|\,|\cos\theta| =\big|\overrightarrow{AP}\cdot\hat{\mathbf{n}}\big| =\left|\left< 5,-11,1\right>\cdot\frac{\left< 1,1,3\right>}{\sqrt{11}}\right| =\frac{3}{\sqrt{11}}\approx 0.9045$$