The equation of the plane mid-parallel to the | vedclass

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(A) The given planes are $P_1: 2x - 3y + 6z + 21 = 0$ and $P_2: 2x - 3y + 6z - 14 = 0$.Since the planes are parallel,the mid-parallel plane will have

Vedclass · Accepted Answer

$4x - 6y + 12z + 7 = 0$

Vedclass · Answer

(A) The given planes are $P_1: 2x - 3y + 6z + 21 = 0$ and $P_2: 2x - 3y + 6z - 14 = 0$.Since the planes are parallel,the mid-parallel plane will have the same normal vector $(2, -3, 6)$.Let the equation of the mid-parallel plane be $2x - 3y + 6z + d = 0$.The constant term $d$ of the mid-parallel plane is the average of the constant terms of the two given planes,provided the coefficients of $x, y, z$ are identical.First,ensure the coefficients are the same. Here they are already the same.The constant $d$ is given by $d = \frac{d_1 + d_2}{2} = \frac{21 + (-14)}{2} = \frac{7}{2} = 3.5$.Thus,the equation is $2x - 3y + 6z + 3.5 = 0$.Multiplying by $2$ to clear the fraction,we get $4x - 6y + 12z + 7 = 0$.

The equation of the plane mid-parallel to the planes $2x - 3y + 6z + 21 = 0$ and $2x - 3y + 6z - 14 = 0$ is given by

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