Question

Find the angle between the planes.

x + y − 2z = 3 and 2x − 2y + z = 5

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 } `

\[\cos \theta = \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{{a_1}^2 + {b_1}^2 + {c_1}^2} \sqrt{{a_2}^2 + {b_2}^2 + {c_2}^2}}\]

\[\text{ So, the angle between } x + y - 2z = 3 \text{ and } 2x - 2y + z = 5 \text{ is given by } \]

\[\cos \theta = \frac{\left( 1 \right) \left( 2 \right) + \left( 1 \right) \left( - 2 \right) + \left( - 2 \right) \left( 1 \right)}{\sqrt{1^2 + 1^2 + \left( - 2 \right)^2} \sqrt{2^2 + \left( - 2 \right)^2 + 1^2}} = \frac{2 - 2 - 2}{\sqrt{1 + 1 + 4} \sqrt{4 + 4 + 1}} = \frac{- 2}{\sqrt{6} \sqrt{9}} = \frac{- 2}{3\sqrt{6}}\]

\[ \Rightarrow \theta = \cos^{- 1} \left( \frac{- 2}{3\sqrt{6}} \right)\]