Question

diatance between planes x+2y-2=-1 and 2x+4y-4=-5 is​

Answer 1

Question:

Find the distance between the planes x + 2y - 2z = -1 and 2x + 4y - 4z = -5

Answer:

$\bigstar{\bold{Distance=\dfrac{\sqrt{3} }{4}\:units}}$

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• Equation of plane 1 = x + 2y - 2z = -1
• Equation of plane 2 = 2x + 4y - 4z = -5

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• The distance between the planes

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ The equations of the planes as given are :

  x + 2y - 2z + 1 = 0

 2x + 4y -4z + 5 = 0

→ Taking the direction ratios,

  $\dfrac{1}{2} ,\dfrac{2}{4},\dfrac{-2}{-4}=\dfrac{1}{2}$

→ Since the direction ratios of the equations are same, hence the planes are parallel to each other.

→ Consider a point P ( x₁ , y₁ , z₁ ) on the plane x + 2y - 2z + 1 = 0

  Let P be any point on the plane, the

  x₁ + 2y₁ - 2z₁ + 1 = 0

  x₁ + 2y₁ - 2z₁ = -1 -------equation 1

→ Now, find the perpendicular distance from point P to the plane 2 =  2x + 4y - 4z = -5

→ The equation from a point to a plane is given by

   $distance=|\dfrac{ax_1+by_1+cz_1-d}{\sqrt{a^{2}+b^{2}+c^{2} } }|$

  where a = 2, b = 4, c = -4 , d = -5

  $distance =| \dfrac{2x_1+4y_1-4z_1+5}{\sqrt{4^{2}+4^{2}+(-4^{2}) } } |$

  $distance=|\dfrac{2 (x_1+2y_1-2z_1)+5}{\sqrt{48} }|$

→ Substitute equation 1 in above equation

  $distance=|\dfrac{2\times -1+5}{\sqrt{48} }|$

  $distance=|\dfrac{3}{\sqrt{48} }|=|\dfrac{4\sqrt{3} }{16} |=\dfrac{\sqrt{3} }{4}$

$\boxed{\bold{Distance=\dfrac{\sqrt{3} }{4}\:units}}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ Distance between a point and a plane is given by

  $distance=|\dfrac{ax_1+by_1+cz_1-d}{\sqrt{a^{2}+b^{2}+c^{2} } }|$