Angle between two planes and between a straight line and a plane

Two planes in the space can coincide, they can be parallel or secant. Let's see every case as we define the angle between them:

• If two planes coincide or are parallel, they form an angle of $$0^\circ$$.
• If two planes are secant, they determine four dihedral angles, equals two to two. The smallest is defined to be the angle between the planes.

To calculate the angle between both planes we will determine the angle between the normal vectors of each plane.

It is convenient to remember that given a plane $$\pi$$ with equation $$\pi: Ax + by + cz + D = 0$$, a perpendicular vector (normal vector to the plane) is $$\vec{n}=(A, B, C)$$

Therefore, if we have:

• plane $$\pi_1$$ with governing vector $$\vec{n}_1$$.
• plane $$\pi_2$$ with governing vector $$\vec{n}_2$$.

Then: $$\cos(\widehat{\pi_1 \ \pi_2})=\cos\alpha= |\cos(\widehat{\vec{n}_1 \ \vec{n}_2})|= \Big| \dfrac{\vec{n}_1\cdot\vec{n}_2}{|\vec{n}_1||\vec{n}_2|}\Big|= \dfrac{|\vec{n}_1\cdot\vec{n}_2|}{|\vec{n}_1||\vec{n}_2|}$$$Therefore, $$\alpha =\arccos\Big(\dfrac{|\vec{n}_1\cdot\vec{n}_2|} {|\vec{n}_1||\vec{n}_2|} \Big) \qquad \alpha \in [0,\dfrac{\pi}{2}]$$$

If, as we said earlier, we take:

$$\vec{n}_1=(A_1,B_1,C_1)$$

$$\vec{n}_2=(A_2,B_2,C_2)$$

The previous formula remains:

$$\cos(\alpha)=\dfrac{|A_1 A_2+B_1 B_2+C_1 C_2|} {\sqrt{A_1^2+B_1^2+C_1^2}\sqrt{A_2^2+B_2^2+C_2^2}}$$$Note: It is important to bear in mind that if we have a plane in vector or parametric form we can obtain a normal vector by doing the vector product between the governing vectors of the plane. It is also remarkable that we can completely determine one plane by means of a normal vector and one point. To do it we will use the general equation, where we only have $$D$$ as unknown, and which we will find by replacing the point and solving. Considering the planes: $$\pi_1: 3x-y+2z+1=0 \qquad \pi_2: 2x+y-5z-1=0$$$

Find the angle that they form.

First we find the normal vectors:

$$\vec{n}_1= (3, -1, 2)$$

$$\vec{n}_2= (2, 1, -5)$$

and we can already apply the formula: $$\begin{array}{rl} \alpha =& \arccos \Big( \dfrac{|A_1 A_2+B_1 B_2+C_1 C_2|} {\sqrt{A_1^2+B_1^2+C_1^2}\sqrt{A_2^2+B_2^2+C_2^2}} \Big) \\ =& \arccos\Big( \dfrac{|3\cdot2+(-1)\cdot1+2\cdot(-5)|} {\sqrt{3^2+1^2+2^2}\sqrt{2^2+1^2+(-5)^2}} \Big) \\ =& \arccos \Big( \dfrac{5}{\sqrt{420}}\Big)= \arccos(0.244)= 75.88^\circ \end{array}$$$A straight line can be on the plane, can be parallel to him, or can be secant. Let's see how the angle between them is defined in every case: • If the straight line is included on the plane (it is on the plane) or both are parallel, the straight line and the plane form an angle of $$0^\circ$$. • If the straight line and the plane are secant, we define the angle $$\alpha$$ between the straight line and the plane as the angle that the straight line forms with its orthogonal projection on the plane. Notice that defining the angle between the straight line and the plane as the angle between the straight line and its orthogonal projection on the plane, we can know the angle from the governing vector of the straight line and the normal vector to the plane. This is due to the fact that 3 vectors (straight line, projection and normal to the plane) are coplanar - they are in the same plane - and also the angle between the normal vector and the governing vector of the projection are straight and constant. Therefore, if we have: • $$\pi$$: plane. • $$r$$: straight line. • $$s$$: perpendicular line (perpendicular to the plane). • $$\vec{v}_r$$: governing vector of $$r$$. • $$\vec{v}_p$$: governing vector of the projection of $$r$$. • $$\vec{n}$$: normal vector of the plane (and governing vector of $$s$$). Then: $$\sin(\widehat{r\ \pi})=\sin\alpha=\cos(\widehat{r\ s}) =|\cos(\widehat{\vec{n} \ \vec{v}_r})|= \Big| \dfrac{\vec{v}_r\cdot\vec{n}}{|\vec{v}_r||\vec{n}|}\Big|$$$

$$\alpha =\arcsin\Big(\dfrac{|\vec{v}_r\cdot\vec{n}|}{|\vec{v}_r||\vec{n}|} \Big) \qquad \alpha \in [0,\dfrac{\pi}{2}]$$$Also, if $$\vec{v}_r=(v_1,v_2,v_3)$$ and $$\vec{n}=(A,B,C)$$, we can express the previous formula in components as follows: $$\sin\alpha =\dfrac{|v_1 A+v_2 B+v_3 C|} {\sqrt{v_1^2+v_2^2+v_3^2}\sqrt{A^2+B^2+C^2}}$$$

Compute the angle between the line $$r$$ and the plane $$\pi$$:

$$r:\left\{ \begin{array}{l} x=3+k \\ y=-2+k \\ z=5 \end{array} \right. \qquad \pi: 3x-4y+5z-1=0$$$We look for a governing vector of the straight line, and for the normal vector to the plane: $$\vec{v}=(1,1,0)$$ $$\vec{n}=(3,-4,5)$$ Therefore: $$\begin{array}{rl} \alpha=& \arcsin\Big( \dfrac{|v_1 A+v_2 B+v_3 C|} {\sqrt{v_1^2+v_2^2+v_3^2}\sqrt{A^2+B^2+C^2}} \Big) \\ =& \arcsin\Big( \dfrac{|1\cdot3+1\cdot(-4)+0\cdot5|} {\sqrt{1^2+1^2+0^2}\sqrt{3^2+(-4)^2+5^2}} \Big) \\ =& \arcsin\Big( \dfrac{1}{\sqrt{2}\sqrt{50}} \Big)= \arcsin{0.1}=5.74^\circ \end{array}$$$