# 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:

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

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

\[ \overrightarrow{n_{1}} \parallel \overrightarrow{n_{2}} \implies a_{1} : b_{1} : c_{1} = a_{2} : b_{2} : c_{2}. \]

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

\[ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}},\quad \frac{c_{1}}{c_{2}} \neq \frac{d_{1}}{d_{2}}. \]

## 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, respectively, then we have

\[ \overrightarrow{n_{1}}=(a,2,b),\quad \overrightarrow{n_{2}}=(2,3,-4).\]

Then since the two vectors are parallel, we have

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

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, respectively, then we have

\[ \overrightarrow{n_{1}}=(4,6,a),\quad \overrightarrow{n_{2}}=(8,12,-4).\]

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

\[ 2(x-3) + 1(y+2) -3(z-4)=0 \implies 2x+y-3z+8 = 0 . \ _\square \]

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 one another, then what is \( a+b+c+d ?\)

Let \( \overrightarrow{n_{1}}, \overrightarrow{n_{2}}, \overrightarrow{n_{3}}\) be the normal vectors of the planes \(\alpha,\beta,\gamma,\) 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

\[ \frac{a}{3} = \frac{b}{6} = \frac{4}{2} \implies a = 6,\ b = 12. \]

From \( (2)\) and \( (3) ,\) we have

\[ \frac{3}{-1} = \frac{6}{c} = \frac{2}{d} \implies c=-2,\ d=-\frac{2}{3} . \]

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 the normal vector of the plane \( \alpha ?\)

Since the two planes \( \alpha\) and \( \beta \) are parallel, it follows that

\[ \frac{3}{a} = \frac{b}{2} = \frac{1}{2} \implies a=6,\ b=1 .\]

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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