# 3D Coordinate Geometry - Parallel Planes

#### Contents

## Summary

Parallel planes are planes in the same three-dimensional space that never meet. If we let two planes \( \alpha\) and \( \beta \) be as follows:

\[ \begin{align} \alpha : a_{1}x+b_{1}y+c_{z}+d_{1} &= 0 \\ \beta : a_{2}x+b_{2}y+c_{z}+d_{2} &= 0, \end{align} \]

then the normal vectors of the planes, \(\overrightarrow{n_{1}}\) and \(\overrightarrow{n_{2}},\) respectively, are

\[ \begin{align} \overrightarrow{n_{1}} &= (a_{1}, b_{1}, c_{1}) \\ \overrightarrow{n_{2}} &= (a_{2}, b_{2}, c_{2}) . \end{align} \]

Since the two planes \( \alpha\) and \(\beta\) are parallel, their normal vectors are also parallel. This implies

\[ \begin{align} \overrightarrow{n_{1}} &\parallel \overrightarrow{n_{2}} \\ \Rightarrow a_{1} : b_{1} : c_{1} &= a_{2} : b_{2} : c_{2}. \end{align} \]

However, when \( \frac{c_{1}}{c_{2}}= \frac{d_{1}}{d_{2}} ,\) the two planes coincide.

Hence, the condition for the two planes to be parallel to each other is \[ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}, \frac{c_{1}}{c_{2}} \neq \frac{d_{1}}{d_{2}}. \]

## Example Problems

## If the following two planes \( \alpha\) and \(\beta \) are parallel, what is \( a+b:\)

\[ \begin{align} \alpha : ax+2y+bz+5=0 \\ \beta : 2x+3y-4z+1=0? \end{align} \]

If two planes are parallel, their normal vectors are also parallel. So, if we let \( \overrightarrow{n_{1}}\) and \(\overrightarrow{n_{2}} \) be the normal vectors of the planes, then we have \[ \overrightarrow{n_{1}}=(a,2,b), \overrightarrow{n_{2}}=(2,3,-4).\]

Then since the two vectors are parallel, we have

\[ \begin{align} \frac{a}{2} &= \frac{2}{3} = \frac{b}{-4} \\ \Rightarrow a&=\frac{4}{3}, b= -\frac{8}{3}. \end{align} \]

Hence, \[a+b=\frac{4}{3} + \left (-\frac{8}{3} \right) = -\frac{4}{3}. \ _\square \]

## If the following two planes \( \alpha\) and \(\beta \) are parallel, what is \( a:\)

\[ \begin{align} \alpha : 4x+6y+az+5&=0\\ \beta : 8x+12y-4z+1&=0? \end{align} \]

Since the two planes are parallel, their normal vectors are also parallel. If we let \( \overrightarrow{n_{1}}\) and \(\overrightarrow{n_{2}} \) be the normal vectors of the planes, then we have \[ \overrightarrow{n_{1}}=(4,6,a), \overrightarrow{n_{2}}=(8,12,-4).\]

Then since the two vectors are parallel, it follows that

\[ \begin{align} \frac{4}{8} &= \frac{6}{12} = \frac{a}{-4} \\ \Rightarrow a&=-2. \ _\square \end{align} \]

## If a plane \(\alpha\) which passes through the point \( A = (3,-2,4) \) is parallel to the plane \( 2x+y-3z=4 ,\) what is the equation of the plane \( \alpha ?\)

Since \( \alpha \) is parallel to the plane \( 2x+y-3z=4,\) the normal vector of \( \alpha \) is parallel with the normal vector of the plane \( 2x+y-3z=4\) which is \((2,1,-3).\) Also, since \( \alpha\) passes through the point \( A=(3,-2,4) ,\) the equation of \( \alpha \) is \[ 2(x-3) + 1(y+2) -3(z-4)=0 \Rightarrow 2x+y-3z+8 = 0 . \ _\square \]

## If three planes \( \alpha, \beta\) and \(\gamma \)

\[ \begin{align} \alpha : ax+by+4z+3=0 \\ \beta : 3x + 6y + 2z + 1 = 0 \\ \gamma : -x + cy + dz + 2 = 0 \end{align} \]

## are parallel to each other, then what is \( a+b+c+d ?\)

Let \( \overrightarrow{n_{1}}, \overrightarrow{n_{2}}, \overrightarrow{n_{3}}\) be the normal vectors of the planes \(\alpha,\beta,\) and \(\gamma,\) respectively. Then we have \[ \begin{align} \overrightarrow{n_{1}}&=(a,b,4) &\qquad (1) \\ \overrightarrow{n_{2}}&=(3,6,2) &\qquad (2) \\ \overrightarrow{n_{3}}&=(-1,c,d). &\qquad (3) \\ \end{align} \]

Since the three planes are parallel, from \( (1) \) and \( (2) ,\) we have \[ \frac{a}{3} = \frac{b}{6} = \frac{4}{2} \Rightarrow a = 6, b = 12. \]

From \( (2)\) and \( (3) ,\) we have \[ \frac{3}{-1} = \frac{6}{c} = \frac{2}{d} \Rightarrow c=-2, d=-\frac{2}{3} . \]

Hence, \[ a+b+c+d = 6 + 12 + (-2) + \left(-\frac{2}{3}\right) = \frac{46}{3}. \ _\square \]

## If two planes \( \alpha\) and \(\beta \)

\[ \begin{align} \alpha : 3x+by+z+3&=0 \\ \beta : ax + 2y + 2z + 1 &= 0 \\ \end{align} \]

## are parallel, then what is normal vector of the plane \( \alpha ?\)

Since the two planes \( \alpha\) and \( \beta \) are parallel, it follows that \[ \frac{3}{a} = \frac{b}{2} = \frac{1}{2} \Rightarrow a=6, b=1 .\]

Thus, the equation of the plane \( \alpha \) is \( 3x + y + z + 3 = 0,\) implying that the normal vector of the plane is \( (3,1,1). \ _\square \)

**Cite as:**3D Coordinate Geometry - Parallel Planes.

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