If the plane 2x - y + 2z + 3 = 0 has the distances 1/3 and 2/3 units from the planes 4x - 2y + 4z +

Q: Wait, why did we divide the equation of the second plane by 2? Can't we just use the coefficients as they are?

That is a fantastic question! The formula for the distance between parallel planes, $d = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}$, assumes that the normal vector $(A, B, C)$ is identical for both planes. If you don't normalize them, the denominator $\sqrt{A^2 + B^2 + C^2}$ will be different for each plane, leading to an incorrect result. Always ensure your $A, B,$ and $C$ coefficients are perfectly aligned before calculating the distance!

Q: Why do we get two values for $\lambda$ and two for $\mu$? Does the problem imply there are multiple planes?

Exactly! The distance formula involves an absolute value, $|D_1 - D_2| = d \cdot \sqrt{A^2 + B^2 + C^2}$. Geometrically, this means there is one plane at a distance $d$ on one side of the reference plane, and another plane at the same distance $d$ on the opposite side. Both are valid candidates, which is why we must consider both cases to find the maximum sum.

Q: Is it possible for the planes to be the same plane? What if $\lambda$ or $\mu$ makes the equation identical to the reference plane?

Yes, if the distance were zero, the planes would be coincident. However, here the distances are $1/3$ and $2/3$, so the planes are distinct and parallel. The math naturally handles this; if the distance were zero, you would simply find $D_1 = D_2$.

Q: Why are we looking for the maximum value of $\lambda + \mu$? Is there a constraint I missed?

There is no hidden constraint! The problem asks for the maximum value because $\lambda$ and $\mu$ each have two possible values. Since we want to maximize the sum, we simply pick the largest possible value for $\lambda$ and the largest possible value for $\mu$ from our calculated sets.

Q: I keep getting confused with the signs inside the absolute value. Does it matter if I do $|D_1 - D_2|$ or $|D_2 - D_1|$?

Great catch! Mathematically, $|D_1 - D_2|$ is identical to $|D_2 - D_1|$ because the absolute value swallows the negative sign. You can subtract in either order, as long as you are consistent. Just keep your focus on the constant terms, and the absolute value will take care of the directionality for you.

Question

If the plane $2x - y + 2z + 3 = 0$ has the distances $1/3$ and $2/3$ units from the planes $4x - 2y + 4z + \lambda = 0$ and $2x - y + 2z + \mu = 0$, respectively, then the maximum value of $\lambda + \mu$ is equal to :

Options:

(A) 15

(B) 5

(C) 13

(D) 9

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a three-dimensional space, looking at a flat, infinite sheet—a plane defined by the equation $2x - y + 2z + 3 = 0$.

Now, imagine two other planes floating in this space, both perfectly parallel to the first. They never touch, they never intersect; they simply glide alongside each other like layers of a cosmic sandwich.

Our goal today is to find the positions of these planes, governed by the parameters $\lambda$ and $\mu$, and ultimately find the maximum sum of these values. This isn't just algebra; it is the art of understanding spatial relationships.

### The Normal Vector: The DNA of a Plane

Every plane has a 'normal vector'—a line segment perpendicular to its surface that defines its orientation. For our reference plane $P_1$, the normal vector is $\vec{n} = 2\hat{i} - \hat{j} + 2\hat{k}$.

When we look at the other two planes, $P_2: 4x - 2y + 4z + \lambda = 0$ and $P_3: 2x - y + 2z + \mu = 0$, we notice something beautiful. The coefficients of $x, y,$ and $z$ in $P_3$ are identical to $P_1$.

For $P_2$, the coefficients are exactly double those of $P_1$. This confirms they are all parallel. To work with them, we must normalize $P_2$ by dividing the entire equation by $2$, transforming it into:

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

Now, all our planes share the same 'DNA'—the same normal vector.

### The Distance Formula: Bridging the Gap

The distance $d$ between two parallel planes $Ax + By + Cz + D_1 = 0$ and $Ax + By + Cz + D_2 = 0$ is given by the elegant formula:

$$d = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}$$

Here, our denominator is $\sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = 3$. This constant $3$ acts as our scaling factor.

For the first pair of planes ($P_1$ and $P_2$), we are told the distance is $1/3$. Plugging this into our formula, we get:

$$\frac{|3 - \frac{\lambda}{2}|}{3} = \frac{1}{3}$$

This simplifies beautifully to $|3 - \frac{\lambda}{2}| = 1$. This absolute value is the heartbeat of the problem.

It tells us that $3 - \frac{\lambda}{2}$ can be either $1$ or $-1$. Solving these two cases, we find $\lambda = 4$ or $\lambda = 8$. We have successfully pinned down the possible locations for our second plane!

### The Final Convergence

We repeat this logic for the third plane, $P_3$. The distance between $P_1$ and $P_3$ is $2/3$. Thus:

$$\frac{|3 - \mu|}{3} = \frac{2}{3}$$

This leads us to $|3 - \mu| = 2$. Again, we branch into two possibilities: $3 - \mu = 2$, which gives $\mu = 1$, or $3 - \mu = -2$, which gives $\mu = 5$.

We now have a set of possibilities: $\lambda \in \{4, 8\}$ and $\mu \in \{1, 5\}$. To maximize the sum $\lambda + \mu$, we simply choose the largest values from each set: $8 + 5 = 13$.

The final result is **13**.

Animated Solution for Mathematics - Three Dimensional Geometry: If the plane $2 x - y + 2 z + 3 = 0$ has the distances $1/3$ and $2/3$ units from the planes $4 x - 2 y + 4 z + λ = 0$ and $2 x - y + 2 z + μ = 0$ , respectively, then the maximum value of $λ + μ$ is equal to :

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Visualized Solution

Reference Plane $P_{1}$

Plane $P_{2}$ and Distance

Normalizing Plane $P_{2}$

Distance Between Parallel Planes

Distance Equation for $P_{1}$ & $P_{2}$

Solving for $λ$

Plane $P_{3}$ and Distance

Distance Equation for $P_{1}$ & $P_{3}$

Solving for $μ$

Maximum Value of $λ + μ$

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Normal Vector

The Distance Formula

The Final Convergence

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Visualized Solution

Reference Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P1​

Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P2​ and Distance

Normalizing Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P2​