How do I find the angle between the planes x + 2y − z + 1 = 0 and x − y + 3z + 4 = 0?

Jan 4, 2016

$\alpha = {119.50}^{\circ}$

Explanation:

To find the angle between two planes, one has to find the angle between normal vectors of these planes.

Explanation of why the angle between two vectors, each one normal to a plane, gives the angle between the two involved planes

Normal Vectors
For the plane 1 , $x + 2 y - z + 1 = 0$
$N \vec{1} = \hat{i} + 2 \cdot \hat{j} - 1 \cdot \hat{k}$

For the plane 2 , $x - y + 3 z + 4 = 0$
$N \vec{2} = \hat{i} - \hat{j} + 3 \cdot \hat{k}$

Angle between the 2 vectors
$N \vec{1} \cdot N \vec{2} = | N \vec{1} | \cdot | N \vec{2} | \cdot \cos \alpha$ => $\cos \alpha = \frac{N \vec{1} \cdot N \vec{2}}{| N \vec{1} | \cdot | N \vec{2} |}$
$N \vec{1} \cdot N \vec{2} = 1 \cdot 1 + 2 \cdot \left(- 1\right) + \left(- 1\right) \left(3\right) = 1 - 2 - 3 = - 4$
$| N \vec{1} | = \sqrt{{1}^{2} + {2}^{2} + {\left(- 1\right)}^{2}} = \sqrt{1 + 4 + 1} = \sqrt{6}$
$| N \vec{2} | = \sqrt{{1}^{2} + {\left(- 1\right)}^{2} + {3}^{2}} = \sqrt{1 + 1 + 9} = \sqrt{11}$
cos alpha=-4/(sqrt(6)*sqrt(11) => $\cos \alpha = - \frac{4}{\sqrt{66}}$ => $\alpha = {119.50}^{\circ}$