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 $3x - 3y + 6z - 1 = 0$.
(A) The direction ratios of the normals to the planes $L_1: a_1x + b_1y + c_1z + d_1 = 0$ and $L_2: a_2x + b_2y + c_2z + d_2 = 0$ are $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ respectively.
$L_1 \parallel L_2$ if $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
$L_1 \perp L_2$ if $a_1a_2 + b_1b_2 + c_1c_2 = 0$.
The angle $\theta$ between $L_1$ and $L_2$ is given by $\cos \theta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \cdot \sqrt{a_2^2 + b_2^2 + c_2^2}}$.
For the given planes $2x - 2y + 4z + 5 = 0$ and $3x - 3y + 6z - 1 = 0$:
$a_1 = 2, b_1 = -2, c_1 = 4$ and $a_2 = 3, b_2 = -3, c_2 = 6$.
Checking for parallelism:
$\frac{a_1}{a_2} = \frac{2}{3}, \frac{b_1}{b_2} = \frac{-2}{-3} = \frac{2}{3}, \frac{c_1}{c_2} = \frac{4}{6} = \frac{2}{3}$.
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{2}{3}$,the given planes are parallel to each other.
$L_1 \parallel L_2$ if $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
$L_1 \perp L_2$ if $a_1a_2 + b_1b_2 + c_1c_2 = 0$.
The angle $\theta$ between $L_1$ and $L_2$ is given by $\cos \theta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \cdot \sqrt{a_2^2 + b_2^2 + c_2^2}}$.
For the given planes $2x - 2y + 4z + 5 = 0$ and $3x - 3y + 6z - 1 = 0$:
$a_1 = 2, b_1 = -2, c_1 = 4$ and $a_2 = 3, b_2 = -3, c_2 = 6$.
Checking for parallelism:
$\frac{a_1}{a_2} = \frac{2}{3}, \frac{b_1}{b_2} = \frac{-2}{-3} = \frac{2}{3}, \frac{c_1}{c_2} = \frac{4}{6} = \frac{2}{3}$.
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{2}{3}$,the given planes are parallel to each other.
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$P_1: x-y+z=1$
$P_2: x+y-z=-1$
$P_3: x-3y+3z=2$
Let $L_1, L_2, L_3$ be the lines of intersection of the planes $P_2$ and $P_3$,$P_3$ and $P_1$,and $P_1$ and $P_2$,respectively.
$STATEMENT-1$: At least two of the lines $L_1, L_2$ and $L_3$ are non-parallel.
$STATEMENT-2$: The three planes do not have a common point.
$P_1: x-y+z=1$
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$P_3: x-3y+3z=2$
Let $L_1, L_2, L_3$ be the lines of intersection of the planes $P_2$ and $P_3$,$P_3$ and $P_1$,and $P_1$ and $P_2$,respectively.
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