If the distance between the plane Ax - 2y + z = d and the plane containing the lines (x-1)/2 = (y-2)

Q: Wait, why do we assume the planes are parallel? The problem doesn't explicitly say so.

That is a brilliant observation! The problem states that the distance between the two planes is $\sqrt{6}$. If two planes were not parallel, they would intersect, and the distance between them would be zero at the line of intersection. Since a specific, non-zero distance is given, they must be parallel.

Q: How do I know which point to use for the plane equation? Both lines have points.

You can use any point that lies on the plane! Since both lines lie on the plane, any point from $L_1$ (like $(1, 2, 3)$) or $L_2$ (like $(2, 3, 4)$) will satisfy the equation of the plane. The result will be identical regardless of your choice.

Q: Why did we get $A=1$ so quickly?

Because the planes are parallel, their normal vectors must be proportional. The normal vector of the first plane is $(1, -2, 1)$, and the normal vector of the second plane is $(A, -2, 1)$. For these to be parallel, the ratios of their components must be equal: $\frac{A}{1} = \frac{-2}{-2} = \frac{1}{1}$. This forces $A$ to be $1$.

Question

If the distance between the plane $Ax - 2y + z = d$ and the plane containing the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $\frac{x-2}{3} = \frac{y-3}{4} = \frac{z-4}{5}$ is $\sqrt{6}$, then find $|d|$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a 3D coordinate space. You have two lines, $L_1$ and $L_2$, floating in front of you. These lines are coplanar, meaning they define a single, flat surface—a plane we shall call $P_1$.

Our mission is to find the distance between this plane $P_1$ and another plane $P_2$, defined by $Ax - 2y + z = d$. We are told this distance is $\sqrt{6}$.

### Defining the Plane $P_1$

To define a plane, we need a point on the plane and a normal vector perpendicular to it. Looking at $L_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$, we see it passes through $P(1, 2, 3)$ with a direction vector $\vec{v}_1 = 2\hat{i} + 3\hat{j} + 4\hat{k}$.

Similarly, $L_2: \frac{x-2}{3} = \frac{y-3}{4} = \frac{z-4}{5}$ passes through $Q(2, 3, 4)$ with direction vector $\vec{v}_2 = 3\hat{i} + 4\hat{j} + 5\hat{k}$. The normal vector $\vec{n}$ of our plane $P_1$ must be perpendicular to both $\vec{v}_1$ and $\vec{v}_2$.

We find this using the cross product: $\vec{n} = \vec{v}_1 	imes \vec{v}_2$. Setting up the determinant, we calculate the components:

$$
\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 4 \ 3 & 4 & 5 \end{vmatrix} = \hat{i}(3 	imes 5 - 4 	imes 4) - \hat{j}(2 	imes 5 - 3 	imes 4) + \hat{k}(2 	imes 4 - 3 	imes 3)
$$

This yields $\vec{n} = -1\hat{i} + 2\hat{j} - 1\hat{k}$. For simplicity, we can use the normal vector $\vec{n} = (1, -2, 1)$.

### The Equation of $P_1$

Now, we use the point-normal form: $1(x-1) - 2(y-2) + 1(z-3) = 0$. Expanding this, we get:

$$
x - 1 - 2y + 4 + z - 3 = 0
$$

This simplifies beautifully to $x - 2y + z = 0$. This is the equation of our first plane.

### The Parallelism Insight

We are given $P_2: Ax - 2y + z = d$. Since the distance between $P_1$ and $P_2$ is a constant $\sqrt{6}$, they must be parallel. This implies their normal vectors are proportional.

Comparing $x - 2y + z = 0$ and $Ax - 2y + z = d$, we see that the coefficients of $y$ and $z$ match, implying $A$ must be $1$. Now we have two parallel planes: $x - 2y + z = 0$ and $x - 2y + z = d$.

### Final Calculation

The distance $D$ between two parallel planes $ax + by + cz = d_1$ and $ax + by + cz = d_2$ is given by the formula:

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

Substituting our values, we have:

$$
\sqrt{6} = \frac{|d - 0|}{\sqrt{1^2 + (-2)^2 + 1^2}}
$$

This simplifies to $\sqrt{6} = \frac{|d|}{\sqrt{6}}$. Multiplying both sides by $\sqrt{6}$, we find $|d| = 6$.

We have navigated the geometry, conquered the cross product, and arrived at the elegant solution. The value of **$|d|$ is 6**.

Animated Solution for Mathematics - Three Dimensional Geometry: If the distance between the plane $A x - 2 y + z = d$ and the plane containing the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ is $6$ , then find $∣ d ∣$ .

Enter Numerical Value:

Visualized Solution

Visualize the Geometry

Analyze Line $L_{1}$

Analyze Line $L_{2}$

Finding the Normal Vector

Computing the Normal Vector

The Final Normal Vector

Equation of the Plane $P_{1}$

Simplifying the Equation

Comparing Parallel Planes

Distance Formula for Parallel Planes

Substituting the Values

Final Calculation for $∣ d ∣$

The Sigma Insight: Equation of a Plane

Analyzing the Setup

Defining the Plane $P_{1}$

The Equation of $P_{1}$

The Parallelism Insight

Final Calculation

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Enter Numerical Value:

Visualized Solution

Visualize the Geometry

Analyze Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​

Analyze Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Finding the Normal Vector

Computing the Normal Vector

The Final Normal Vector

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

Simplifying the Equation

Comparing Parallel Planes

Distance Formula for Parallel Planes

Substituting the Values

Final Calculation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣d∣

The Sigma Insight: Equation of a Plane

Analyzing the Setup

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

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

The Parallelism Insight

Final Calculation

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Analyze Line $L_{1}$

Analyze Line $L_{2}$

Equation of the Plane $P_{1}$

Final Calculation for $∣ d ∣$

Defining the Plane $P_{1}$

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