Question

Find the angle between the planes.

 x − y + z = 5 and x + 2y + z = 9

Answer 1

` \text{ We know that the angle between the planes }  a_1 x + b_1 y + c_1 z + d_1 = 0 \text{ and } a_2 x + b_2 y + c_2 z + d_2 = 0 \text{ is given by } `
\[\cos \theta = \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{{a_1}^2 + {b_1}^2 + {c_1}^2} \sqrt{{a_2}^2 + {b_2}^2 + {c_2}^2}}\]
\[ \text{ So, the angle between } x - y + z = 5 \text{ and } x + 2y + z = 9 \text{ is given by } \]
\[\cos \theta = \frac{\left( 1 \right) \left( 1 \right) + \left( - 1 \right) \left( 2 \right) + \left( 1 \right) \left( 1 \right)}{\sqrt{1^2 + \left( - 1 \right)^2 + 1^2} \sqrt{1^2 + 2^2 + 1^2}} = \frac{1 - 2 + 1}{\sqrt{1 + 1 + 1} \sqrt{1 + 4 + 1}} = \frac{0}{\sqrt{3} \sqrt{6}} = 0\]
\[ \Rightarrow \theta = \cos^{- 1} \left( 0 \right) = \frac{\pi}{2}\]