9.4 Angles and Distance Formulas in Space

Subsection 8.4.1 Angle Between a Line and a Plane

Theorem 8.4.1. Formula for Line-Plane Angle.

Let \(\theta\) be the acute angle between a line with direction vector \(\mathbf{v}\) and a plane with normal vector \(\mathbf{n}\text{.}\) Then

\begin{equation*} \sin\theta=\frac{|\mathbf{v}\cdot\mathbf{n}|}{\|\mathbf{v}\|\,\|\mathbf{n}\|}. \end{equation*}

🔗

Proof.

Let \(\varphi\) be the angle between \(\mathbf{v}\) and \(\mathbf{n}\text{.}\) Since the line-plane angle is complementary to this angle, \(\theta=\left|\frac{\pi}{2}-\varphi\right|\text{.}\) Therefore

\begin{equation*} \sin\theta=|\cos\varphi|=\frac{|\mathbf{v}\cdot\mathbf{n}|}{\|\mathbf{v}\|\,\|\mathbf{n}\|}. \end{equation*}

🔗

described in detail following the image — Figure 8.4.2. Angle Between a Line and a Plane
🔗

🔗

Subsection 8.4.2 Angle Between Two Planes

Proposition 8.4.3. Plane-Plane Angle via Normals.

The acute angle \(\phi\) between two planes equals the acute angle between their normal vectors \(\mathbf{n}_1\) and \(\mathbf{n}_2\text{:}\)

\begin{equation*} \cos\phi=\frac{|\mathbf{n}_1\cdot\mathbf{n}_2|}{\|\mathbf{n}_1\|\,\|\mathbf{n}_2\|}. \end{equation*}

🔗

Example 8.4.4. Computing a Plane-Plane Angle.

Find the angle between planes

\begin{equation*} \Pi_1: x+2y-z+1=0,\qquad \Pi_2: 2x-y+2z-3=0. \end{equation*}

🔗

Solution.

Normal vectors are \(\mathbf{n}_1=(1,2,-1)\) and \(\mathbf{n}_2=(2,-1,2)\text{.}\) Then

\begin{equation*} \mathbf{n}_1\cdot\mathbf{n}_2=1\cdot2+2\cdot(-1)+(-1)\cdot2=-2. \end{equation*}

So

\begin{equation*} \cos\phi=\frac{|-2|}{\sqrt{1^2+2^2+(-1)^2}\,\sqrt{2^2+(-1)^2+2^2}}=\frac{2}{\sqrt{6}\cdot3}=\frac{2}{3\sqrt{6}}. \end{equation*}

🔗

Subsection 8.4.3 Distance Formulas in Space

Theorem 8.4.5. Distance from a Point to a Plane.

For point \(P(x_0,y_0,z_0)\) and plane \(Ax+By+Cz+D=0\text{,}\)

\begin{equation*} d(P,\Pi)=\frac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^2+B^2+C^2}}. \end{equation*}

🔗

Theorem 8.4.6. Distance Between Two Skew Lines.

For skew lines

\begin{equation*} \ell_1:\mathbf{r}=\mathbf{r}_1+t\mathbf{v}_1,\qquad \ell_2:\mathbf{r}=\mathbf{r}_2+s\mathbf{v}_2, \end{equation*}

the distance is

\begin{equation*} d(\ell_1,\ell_2)=\frac{|(\mathbf{r}_2-\mathbf{r}_1)\cdot(\mathbf{v}_1\times\mathbf{v}_2)|}{\|\mathbf{v}_1\times\mathbf{v}_2\|}. \end{equation*}

🔗