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Question
Find the angle between the planes.
2x − y + z = 4 and x + y + 2z = 3
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Solution
` \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 } `
\[\cos \theta = \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{{a_1}^2 + {b_1}^2 + {c_1}^2} \sqrt{{a_2}^2 + {b_2}^2 + {c_2}^2}}\]
\[\text{ So, the angle between } 2x - y + z = 4 \text{ and } x + y + 2z = 3 \text{ is given by }\]
\[\cos \theta = \frac{\left( 2 \right) \left( 1 \right) + \left( - 1 \right) \left( 1 \right) + \left( 1 \right) \left( 2 \right)}{\sqrt{2^2 + \left( - 1 \right)^2 + 1^2} \sqrt{1^2 + 1^2 + 2^2}} = \frac{2 - 1 + 2}{\sqrt{4 + 1 + 1} \sqrt{1 + 1 + 4}} = \frac{3}{\sqrt{6} \sqrt{6}} = \frac{3}{6} = \frac{1}{2}\]
\[ \Rightarrow \theta = \cos^{- 1} \left( \frac{1}{2} \right) = \frac{\pi}{3}\]
