3D Coordinate Geometry - Parallel Planes

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

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

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

$$\begin{aligned} \overrightarrow{n_{1}} &= (a_{1}, b_{1}, c_{1}) \\ \overrightarrow{n_{2}} &= (a_{2}, b_{2}, c_{2}) . \end{aligned}$$

ParallelAngleBisector

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

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

If the following two planes $\alpha$ and $\beta$ are parallel, what is $a+b?$

$$\begin{aligned} \alpha : ax+2y+bz+5&=0 \\ \beta : 2x+3y-4z+1&=0 \end{aligned}$$

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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

$$\begin{aligned} \frac{a}{2} = \frac{2}{3} &= \frac{b}{-4} \\\\ \Rightarrow a&=\frac{4}{3},\quad b= -\frac{8}{3}. \end{aligned}$$

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 $a?$

$$\begin{aligned} \alpha : 4x+6y+az+5&=0\\ \beta : 8x+12y-4z+1&=0 \end{aligned}$$

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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

$$\begin{aligned} \frac{4}{8} = \frac{6}{12} &= \frac{a}{-4} \\\\ \Rightarrow a&=-2. \ _\square \end{aligned}$$

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 $\alpha ?$

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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 $\gamma$

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

are parallel to one another, then what is $a+b+c+d ?$

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Let $\overrightarrow{n_{1}}, \overrightarrow{n_{2}}, \overrightarrow{n_{3}}$ be the normal vectors of the planes $\alpha,\beta,\gamma,$ respectively. Then we have

$$\begin{aligned} \overrightarrow{n_{1}}&=(a,b,4) &\qquad (1) \\ \overrightarrow{n_{2}}&=(3,6,2) &\qquad (2) \\ \overrightarrow{n_{3}}&=(-1,c,d). &\qquad (3) \end{aligned}$$

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 $\beta$

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

are parallel, then what is the normal vector of the plane $\alpha ?$

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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 $(3,1,1). \ _\square$