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Three Dimensional Geometry — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSThree Dimensional Geometry5 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, $L_1: a_1x + b_1y + c_1z = 0,$
are $a_1, b_1, c_1$ and $L_2: a_2x + b_2y + c_2z = 0$ are $a_2, b_2, c_2$​​​​​​​ 
$\text{L}_1||\text{L}_2,\ \text{if }\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\text{L}_1\perp\text{L}_2,\ \text{if}\ \text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2=0$
The angle between $L_1$ and $L_2$ 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, $a_1 = 2, b_1 = -2, c_1 = 4$ and $a_2 = 3, b_2 = -3, c_2 = 6$
$a_1a_2 +b_1b_2 + c_1c_2 = 2 \times 3 + (-2) \times (-3) + 4 \times 6 = 6 + 6 + 24$
$=36\neq0$
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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