# Are the planes x+y+z=1 , x-y+z=1 parallel, perpendicular, or neither? If neither, what is the angle between them?

Apr 20, 2018

$\textcolor{b l u e}{\text{Neither}}$

#### Explanation:

We can find the angle between two planes, by finding the angle between the normals to the plane:

When given the equation of a plane in Cartesian form:

$a x + b y + c z = d$

The normal vector to the plane is:

$\vec{\boldsymbol{n}} = \left(\begin{matrix}a \\ b \\ c\end{matrix}\right)$

Let:

${\Pi}_{1} = x + y + z - 1 = 0$

${\Pi}_{2} = x - y + z - 1 = 0$

Then:

$\vec{{\boldsymbol{n}}_{1}} = \left(\begin{matrix}1 \\ 1 \\ 1\end{matrix}\right)$

$\vec{{\boldsymbol{n}}_{2}} = \left(\begin{matrix}1 \\ - 1 \\ 1\end{matrix}\right)$

We can find the angle between these normals using the Dot Product:

$\boldsymbol{{n}_{1} \cdot {n}_{2}} = | | \boldsymbol{{n}_{1}} | | \cdot | | \boldsymbol{{n}_{2}} | | \cdot \cos \left(\theta\right)$

$\boldsymbol{{n}_{1} \cdot {n}_{2}} = 1$

$| | \boldsymbol{{n}_{1}} | | = \sqrt{{\left(1\right)}^{2} + {\left(1\right)}^{2} + {\left(1\right)}^{2}} = \sqrt{3}$

$| | \boldsymbol{{n}_{2}} | | = \sqrt{{\left(1\right)}^{2} + {\left(- 1\right)}^{2} + {\left(1\right)}^{2}} = \sqrt{3}$

$1 = \sqrt{3} \cdot \sqrt{3} \cdot \cos \left(\theta\right)$

$\cos \left(\theta\right) = \frac{1}{\sqrt{3} \cdot \sqrt{3}} = \frac{1}{3}$

$\theta = \arccos \left(\frac{1}{3}\right) = {70.53}^{\circ}$ 2 d.p.

If the planes were perpendicular, the the angle between them would be ${90}^{\circ}$.

If the planes were parallel, the the angle between them would be ${0}^{\circ}$. Therefore they are neither.

Plot: