Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is (a) 3/2 (b) 5/2 (c)

Q: Wait, why do we have to divide the second equation by 2? Can't we just use the coefficients as they are?

That is a fantastic question! The distance formula $d = \frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}$ relies on the normal vector $(a, b, c)$ being the same for both planes. If you don't normalize the coefficients, the denominator $\sqrt{a^2 + b^2 + c^2}$ would be different for each plane, which breaks the geometric logic of the formula. By dividing, we ensure we are measuring the gap along the same normal direction.

Q: Does the order of $d_1$ and $d_2$ in the numerator matter? What if I get a negative value?

Don't worry about the order! The formula uses the absolute value $|d_1 - d_2|$, which ensures the distance is always positive. Distance is a scalar quantity representing a physical gap, so it cannot be negative. Whether you calculate $d_1 - d_2$ or $d_2 - d_1$, the modulus will give you the same correct result.

Q: What if the planes weren't parallel? Would this formula still work?

If the planes were not parallel, they would eventually intersect. In that case, the perpendicular distance between them is not constant—it varies from zero at the line of intersection to infinity. This specific formula is exclusively for parallel planes where the distance is constant everywhere. If the normal vectors weren't proportional, you would know immediately that they aren't parallel!

Question

Distance between two parallel planes $2x + y + 2z = 8$ and $4x + 2y + 4z + 5 = 0$ is

Options:

(A) $\frac{3}{2}$

(B) $\frac{5}{2}$

(C) $\frac{7}{2}$

(D) $\frac{9}{2}$

Sigma Percentile · Accepted Answer

### The Geometry of Parallel Worlds

Imagine you are standing in a vast, empty 3D space. Before you, two perfectly flat, infinite sheets of paper are suspended in the air. They are parallel, never touching, stretching out into the void.

Your mission is to find the exact, shortest distance between these two sheets. This is the essence of our problem today: finding the distance between the parallel planes $2x + y + 2z = 8$ and $4x + 2y + 4z + 5 = 0$.

### Phase 1: The Parallelism Check

Before we dive into the algebra, we must confirm our intuition. How do we know these planes are parallel?

Look at the coefficients of $x, y,$ and $z$. For the first plane, they are $(2, 1, 2)$. For the second, they are $(4, 2, 4)$.

Notice that the second set is exactly double the first. This confirms that the normal vectors—the arrows pointing straight out from the surfaces—are pointing in the same direction. They are indeed parallel.

### Phase 2: The Normalization Process

Here is where many students stumble. To use the elegant distance formula, we need the equations to speak the same language. The standard form of a plane is $ax + by + cz + d = 0$.

Currently, our planes are:
1. $2x + y + 2z - 8 = 0$
2. $4x + 2y + 4z + 5 = 0$

To fix the discrepancy in coefficients, we divide the entire second equation by $2$. This transforms it into:

$$2x + y + 2z + \frac{5}{2} = 0$$

Now, both planes share the same normal vector $(2, 1, 2)$. We have successfully standardized our coordinate system.

### Phase 3: The Distance Formula

Now, we reach for our most powerful tool: the distance formula for parallel planes:

$$d = \frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}$$

Think of this formula as a bridge. The numerator, $|d_1 - d_2|$, measures the difference in the 'offsets' of the planes from the origin. The denominator, $\sqrt{a^2 + b^2 + c^2}$, acts as a scaling factor that accounts for the orientation of the planes in space.

### Phase 4: The Final Execution

Let's plug in our values. We have $d_1 = -8$ and $d_2 = \frac{5}{2}$. Our normal vector components are $a=2, b=1, c=2$.

Substituting these into our formula:

$$d = \frac{|-8 - \frac{5}{2}|}{\sqrt{2^2 + 1^2 + 2^2}}$$

First, let's tackle the numerator: $|-8 - 2.5| = |-10.5| = \frac{21}{2}$.

Next, the denominator: $\sqrt{4 + 1 + 4} = \sqrt{9} = 3$.

Finally, we combine them:

$$d = \frac{\frac{21}{2}}{3} = \frac{21}{6} = \frac{7}{2}$$

And there it is! The distance between these two infinite sheets is exactly **$\frac{7}{2}$ units**. It is a beautiful, clean result.

Remember, geometry is not just about memorizing formulas; it is about visualizing the space and ensuring your tools are calibrated to the problem at hand. You have mastered the distance between planes—keep that momentum going!

Animated Solution for Mathematics - Three Dimensional Geometry: Distance between two parallel planes $2 x + y + 2 z = 8$ and $4 x + 2 y + 4 z + 5 = 0$ is

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Visualized Solution

$A na l y z e t h e P l an e E q u a t i o n s$

$C h ec k f or P a r a l l e l i s m$

$S t an d a r d i z e P l an e 1$

$S t an d a r d i z e P l an e 2$

$E x t r a c tP a r am e t er s$

$T h eD i s t an ce F or m u l a$

$S u b s t i t u t e t h e V a l u es$

$S im pl i f y t h e N u m er a t or$

$S im pl i f y t h eD e n o mina t or$

$F ina l C a l c u l a t i o n$

The Sigma Insight: Equation of a Plane

The Geometry of Parallel Worlds

Phase 1

Phase 2

Phase 3

Phase 4

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