In the following cases, determine whether the given planes are pa… - Vidyadip

Question

In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.
2x - 2y + 4z + 5 = 0 and 3x - 3y + 6z - 1 = 0

✓

Answer

The direction ratios of normal to the plane, L₁: a₁x + b₁y + c₁z = 0,
are a₁, b₁, c₁ and L₂: a₂x + b₂y + c₂z = 0 are a₂, b₂, c₂
$\text{L}_1||\text{L}_2,\ \text{if }\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\text{L}_1\perp\text{L}_2,\ \text{if}\ \text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2=0$
The angle between L₁ and L₂ is given by,
$\text{Q}=\cos^{-1}\Bigg|\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{\text{a}_1^2+\text{b}_1^2+\text{c}_1^2}.\sqrt{\text{a}_2^2+\text{b}_2^2+\text{c}_2^2}}$
The equations of the given planes are 2x - 2y + 4z + 5 = 0 and 3x - 3y + 6z - 1 = 0
Here, a₁ = 2, b₁ = -2, c₁ = 4 and a₂ = 3, b₂ = -3, c₂ = 6
a₁a₂ +b₁b₂ + c₁c₂ = 2 × 3 + (-2) × (-3) + 4 × 6 = 6 + 6 + 24
$=36\neq0$
Thus, the given planes are not perpendicular to each other.
$\frac{\text{a}_1}{\text{a}_2}=\frac{2}{3},\ \frac{\text{b}_1}{\text{b}_2}=\frac{-2}{-3}=\frac{2}{3}\text{ and } \frac{\text{c}_1}{\text{c}_2}=\frac{4}{6}=\frac{2}{3}$
$\therefore\ \ \frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
Thus, the given planes are parallel to each other.

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