The distance between the planes 2x + 2y − z + 2 = 0 and 4x + 4y − 2z + 5 = 0 is

Question

KollegeApply · Accepted Answer

$\frac{1}{6}$ \text{ Multiplying the first equation of the plane by } 4x + 4y - 2z + 4 = 0 4x + 4y - 2z = - 4 . . . \left( 1 \right) \text{ The second equation of the plane is } 4x + 4y - 2z + 5 = 0 4x + 4y - 2z = - 5 . . . \left( 2 \right) \text{ We know that the distance between two planes } ax + by + cz = d_1 \text{ and }ax + by + cz = d_2 \text{ is }\frac{\left| d_2 - d_1 \right|}{$\sqrt$(a^2 + b^2 + c^2)} \text{ So, the required distance } =\frac{\left| - 5 + 4 \right|}{\sqrt{4^2 + 4^2 + \left( - 2 \right)^2}} =\frac{\left| - 1 \right|}{$\sqrt$(16 + 16 + 4)} = 1/$\sqrt$(36) = $\frac{1}{6}$ \text{ units }

AChoose Your Answer