A plane P meets the coordinate axes at A, B and C respectively. The centroid of ΔABC is given to be

Q: Wait, why do we use the intercept form of the plane? Can't we just use the general form $Ax + By + Cz = D$?

You absolutely can! But the intercept form is a powerful shortcut when you know the points of intersection with the axes. It saves time and reduces the chance of algebraic errors. It directly maps the geometry of the intercepts to the algebra of the plane.

Q: How do I know the direction ratios of the line are the same as the normal vector of the plane?

By definition, a normal vector $\vec{n}$ is perpendicular to every line lying on the plane. If our line is perpendicular to the plane, it must be parallel to the normal vector. Therefore, they share the same direction ratios. It is a beautiful geometric consequence!

Q: Is the centroid formula for a 3D triangle different from a 2D one?

Not at all! The beauty of 3D geometry is that it extends naturally from 2D. You just add the average of the $z$-coordinates to the average of the $x$ and $y$ coordinates. It is the same principle of finding the center of mass.

Question

A plane $P$ meets the coordinate axes at $A, B$ and $C$ respectively. The centroid of $\Delta ABC$ is given to be $(1, 1, 2)$. Then the equation of the line through this centroid and perpendicular to the plane $P$ is:

Options:

(A) $\frac{x-1}{2} = \frac{y-1}{2} = \frac{z-2}{1}$

(B) $\frac{x-1}{1} = \frac{y-1}{1} = \frac{z-2}{2}$

(C) $\frac{x-1}{1} = \frac{y-1}{2} = \frac{z-2}{2}$

(D) $\frac{x-1}{2} = \frac{y-1}{1} = \frac{z-2}{1}$

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of the Corner

Imagine you are standing in the corner of a room. This corner is our origin $(0, 0, 0)$, and the edges of the walls are our $x$, $y$, and $z$ axes.

A plane $P$ slices through this corner, cutting the axes at three distinct points: $A$, $B$, and $C$. This forms a triangle in 3D space.

Our mission is to find the equation of a line that passes through the centroid $G(1, 1, 2)$ of this triangle and is perpendicular to the plane $P$.

### Phase 1: Determining the Intercepts

Since the points $A$, $B$, and $C$ lie exactly on the coordinate axes, their coordinates are elegantly simple. Let the intercepts be $a$, $b$, and $c$.

Thus, $A$ is $(a, 0, 0)$, $B$ is $(0, b, 0)$, and $C$ is $(0, 0, c)$. The problem provides the centroid $G(1, 1, 2)$.

The centroid formula for a 3D triangle is given by:

$$G = \left( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3} \right)$$

Substituting our points, we get $G = (\frac{a}{3}, \frac{b}{3}, \frac{c}{3})$. Equating this to $(1, 1, 2)$, we find:

$$a=3, \quad b=3, \quad c=6$$

We have successfully unlocked the geometry of the plane.

### Phase 2: The Plane's Identity

With the intercepts $a=3$, $b=3$, and $c=6$, we use the intercept form of the plane:

$$\frac{x}{3} + \frac{y}{3} + \frac{z}{6} = 1$$

To simplify this expression, we multiply the entire equation by $6$, yielding:

$$2x + 2y + z = 6$$

This is the general form of our plane. The coefficients of $x$, $y$, and $z$ represent the components of the normal vector $\vec{n} = (2, 2, 1)$. This vector is the key to defining our perpendicular line.

### Phase 3: The Final Line Equation

We require a line passing through $G(1, 1, 2)$ with the direction of the normal vector $\vec{n} = (2, 2, 1)$. The symmetric form of a line is given by:

$$\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n}$$

Substituting our point $(1, 1, 2)$ and direction ratios $(2, 2, 1)$, we obtain the final equation:

$$\frac{x-1}{2} = \frac{y-1}{2} = \frac{z-2}{1}$$

**The resulting equation of the line is $\frac{x-1}{2} = \frac{y-1}{2} = \frac{z-2}{1}$.**

Animated Solution for Mathematics - Three Dimensional Geometry: A plane $P$ meets the coordinate axes at $A, B$ and $C$ respectively. The centroid of $Δ A B C$ is given to be $(1, 1, 2)$ . Then the equation of the line through this centroid and perpendicular to the plane $P$ is:

Select Answer:

Visualized Solution

The 3D Setup

Defining the Intercepts

The Centroid of $Δ A B C$

The Centroid Formula

Substituting the Coordinates

Calculating Intercepts $a, b, c$

Intercept Form of a Plane

Substituting Intercepts

Simplifying the Plane Equation

Identifying the Normal Vector

Equation of a Line in 3D

The Final Line Equation

The Sigma Insight: Equation of a Plane

Analyzing the Geometry of the Corner

Phase 1

Phase 2

Phase 3

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