Question

Calculus 3 Planes and Lines

Answer 1

Step-by-step explanation:

Given planes be

$P_{1} : - 4x + 5y + 2z = 10 \\ P_{2} : - 4x + 5y + 2z = 71$

Since, the given planes are parallel, so the distance between then is given by

$d = \frac{ |71 - 10| }{ \sqrt{( - 4)^{2} + { (5)}^{2} + {(2)}^{2} } }$

$\implies \: d = \frac{ 61 }{ \sqrt{16+ 25 + 4 } }$

$\implies \: d = \frac{ 61 }{ \sqrt{45 } }$

$\implies \: d = \frac{ 61 }{ 3\sqrt{5 } }$