Question

Find the angle between the planes.
 2x − 3y + 4z = 1 and − x + y = 4

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 } `

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

\[ \text{ So, the angle between } 2x - 3y + 4z = 1 \text{ and }  -x + y + 0z = 4 \text{ is given by } \]

\[\cos \theta = \frac{\left( 2 \right) \left( - 1 \right) + \left( - 3 \right) \left( 1 \right) + \left( 4 \right) \left( 0 \right)}{\sqrt{2^2 + \left( - 3 \right)^2 + 4^2} \sqrt{\left( - 1 \right)^2 + 1^2 + 0^2}} = \frac{- 2 - 3 + 0}{\sqrt{4 + 9 + 16} \sqrt{1 + 1 + 0}} = \frac{- 5}{\sqrt{29} \sqrt{2}} = \frac{- 5}{\sqrt{58}}\]

\[ \Rightarrow \theta = \cos^{- 1} \left( \frac{- 5}{\sqrt{58}} \right)\]