The acute angle between the planes 2x − y + z = 6 and x + y + 2z = 3 is

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KollegeApply · Accepted Answer

$60^\circ$ \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 = 6 \text{ and } x + y + 2z = 3 \text{ is given by} \text{ So } ,\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)} = 3/$\sqrt$(6) $\sqrt$(6) = $\frac{3}{6}$ = $\frac{1}{2}$ \Rightarrow $\theta$ = \cos^{- 1} \left( $\frac{1}{2}$ \right) = {60}^o

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