Question

Find the angle between the planes.

 2x + y − 2z = 5 and 3x − 6y − 2z = 7

 

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 }`
\[\cos \theta = \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{{a_1}^2 + {b_1}^2 + {c_1}^2} \sqrt{{a_2}^2 + {b_2}^2 + {c_2}^2}}\]
\[\text{ So, the angle between}  2x + y - 2z = 5 \text{ and } 3x - 6y - 2z = 7 \text{ is given by } \]
\[\cos \theta = \frac{\left( 2 \right) \left( 3 \right) + \left( 1 \right) \left( - 6 \right) + \left( - 2 \right) \left( - 2 \right)}{\sqrt{2^2 + 1^2 + \left( - 2 \right)^2} \sqrt{3^2 + \left( - 6 \right)^2 + \left( - 2 \right)^2}} = \frac{6 - 6 + 4}{\sqrt{4 + 1 + 4} \sqrt{9 + 36 + 4}} = \frac{4}{\left( 3 \right) \left( 7 \right)} = \frac{4}{21}\]
\[ \Rightarrow \theta = \cos^{- 1} \left( \frac{4}{21} \right)\]