The shortest distance from the plane 12x + 4y + 3z = 327 to the sphere x² + y² + z² + 4x - 2y - 6z =

Q: Wait, why do we subtract the radius from the distance? Could it be $R - d$?

That is a fantastic question! Think about the geometry. The distance $d$ is the perpendicular length from the center of the sphere to the plane. If $d > R$, the sphere is entirely on one side of the plane. The shortest distance is the gap between the surface of the sphere and the plane, which is the total distance $d$ minus the radius $R$. If $d < R$, the plane would intersect the sphere, and the shortest distance would be zero. So, $d - R$ is the correct measure of the gap.

Q: I always mess up the sphere equation coefficients. Is there a trick?

The trick is to be methodical. Always rewrite the equation in the form $x^2 + 2gx + y^2 + 2fy + z^2 + 2hz + c = 0$. The center is $(-g, -f, -h)$. Don't rush the signs! A single sign error here propagates through the entire problem. Take a breath, write it out, and double-check your coefficients before moving to the next step.

Q: Why is the denominator $\sqrt{12^2 + 4^2 + 3^2}$? Where does that come from?

That is the magnitude of the normal vector to the plane. The plane equation $12x + 4y + 3z = 327$ tells us the normal vector is $\vec{n} = 12\hat{i} + 4\hat{j} + 3\hat{k}$. The distance formula relies on projecting the vector from the center to any point on the plane onto this normal vector. The denominator is simply the length of that normal vector, which normalizes our projection. It is the geometric key to measuring distance in 3D space!

Question

The shortest distance from the plane $12x + 4y + 3z = 327$ to the sphere $x^2 + y^2 + z^2 + 4x - 2y - 6z = 155$ is

Options:

(A) 39

(B) 26

(C) $11 \frac{4}{13}$

(D) 13

Sigma Percentile · Accepted Answer

### The Geometry of the Shortest Path

Imagine you are standing in a vast, three-dimensional space. Before you floats a perfect sphere, and cutting through the void is a flat, infinite plane. Your mission is to find the absolute shortest distance between this sphere and the plane.

The shortest path between any point and a plane is always the perpendicular line. When we extend this to a sphere, the shortest path must lie along the line dropped perpendicularly from the sphere's center directly to the plane.

If we find the total distance $d$ from the center to the plane and then subtract the sphere's radius $R$, we are left with the exact, minimal gap between the sphere's surface and the plane.

### Phase 1: Unmasking the Sphere

We are given the sphere's equation: $x^2 + y^2 + z^2 + 4x - 2y - 6z = 155$. To understand this object, we need its center and radius.

We compare this to the general form $x^2 + y^2 + z^2 + 2gx + 2fy + 2hz + c = 0$. By matching coefficients, we find $2g = 4$, $2f = -2$, and $2h = -6$. This gives us $g = 2$, $f = -1$, and $h = -3$.

The center $C$ of the sphere is defined as $(-g, -f, -h)$, which lands us at $C(-2, 1, 3)$.

Now, for the radius $R$, we use the formula:

$$R = \sqrt{g^2 + f^2 + h^2 - c}$$

Plugging in our values, we get:

$$R = \sqrt{2^2 + (-1)^2 + (-3)^2 - (-155)}$$

Be careful here—the constant $c$ is $-155$, so subtracting it becomes adding $155$. We get:

$$R = \sqrt{4 + 1 + 9 + 155} = \sqrt{169} = 13$$

Our sphere is centered at $(-2, 1, 3)$ with a radius of $13$ units.

### Phase 2: The Plane's Barrier

Now, we turn our attention to the plane: $12x + 4y + 3z = 327$, or $12x + 4y + 3z - 327 = 0$. We need the perpendicular distance $d$ from our center $C(-2, 1, 3)$ to this plane.

The formula for the distance from a point $(x_1, y_1, z_1)$ to a plane $Ax + By + Cz + D = 0$ is:

$$d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$$

The denominator, $\sqrt{12^2 + 4^2 + 3^2}$, represents the magnitude of the plane's normal vector. Calculating this, we get:

$$\sqrt{144 + 16 + 9} = \sqrt{169} = 13$$

Now for the numerator:

$$|12(-2) + 4(1) + 3(3) - 327| = |-24 + 4 + 9 - 327| = |-338| = 338$$

Thus, the distance $d$ is:

$$d = \frac{338}{13} = 26$$

### Phase 3: The Final Leap

We have arrived at the climax of our journey. We know the center of the sphere is $26$ units away from the plane.

We also know the sphere extends $13$ units from its center in every direction. The shortest distance is the total distance $d$ minus the radius $R$.

$$26 - 13 = 13$$

The shortest distance between the sphere and the plane is exactly **$13$ units**. It is elegant, it is precise, and it is the result of visualizing the geometry before diving into the algebra.

Animated Solution for Mathematics - Three Dimensional Geometry: The shortest distance from the plane $12 x + 4 y + 3 z = 327$ to the sphere $x^{2} + y^{2} + z^{2} + 4 x - 2 y - 6 z = 155$ is

Select Answer:

Visualized Solution

Visualizing the Geometry

The Shortest Distance Concept

Sphere Equation Analysis

Finding the Center

Finding the Radius

Distance from Center to Plane

Substituting Values

Calculating the Numerator

Calculating the Denominator

Finalizing Distance $d$

Calculating Shortest Distance

The Sigma Insight: Equation of a Plane

The Geometry of the Shortest Path

Phase 1

Phase 2

Phase 3

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Animated Solution for Mathematics - Three Dimensional Geometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The shortest distance from the plane 12x+4y+3z=327 to the sphere x2+y2+z2+4x−2y−6z=155 is