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Three Dimensional Geometry — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsThree Dimensional Geometry4 Marks
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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