The sum of the intercepts on the coordinate axes of the plane passing through the point (-2, -2, 2)

Q: Wait, why do we need to find two vectors? Can't we just use the line equation directly?

You absolutely can use the line equation! However, in 3D geometry, treating the line as two distinct points is often the most robust way to avoid algebraic errors. By finding two vectors $\vec{AB}$ and $\vec{AP}$, you are essentially building the plane's 'foundation' from the ground up. It is a foolproof method that works every single time.

Q: Does the order of the cross product matter? What if I get $\vec{n}$ as $(1, -3, -6)$ instead of $(-1, 3, 6)$?

That is a fantastic question! The order of the cross product determines the direction of the normal vector (it will point in the opposite direction). However, the *equation* of the plane remains identical because the normal vector is just a scaling factor. Whether you use $\vec{n}$ or $-\vec{n}$, the final plane equation $x - 3y - 6z + 8 = 0$ will be the same. You are safe!

Q: Why do we convert to the intercept form at the end? Is there a faster way?

The intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ is the most elegant way to extract the intercepts directly. While you could set $y=0$ and $z=0$ in the general equation to find the x-intercept, and repeat for others, the intercept form gives you all three values in one clean step. It is less prone to calculation errors and shows a deeper understanding of the plane's geometry.

Question

The sum of the intercepts on the coordinate axes of the plane passing through the point $(-2, -2, 2)$ and containing the line joining the points $(1, -1, 2)$ and $(1, 1, 1)$, is

Options:

(A) 4

(B) $-4$

(C) 12

(D) $-8$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional void. You have a line stretching out before you, defined by two points, $A(1, -1, 2)$ and $B(1, 1, 1)$. Floating nearby is a third point, $P(-2, -2, 2)$.

Your mission is to find the unique, perfectly flat surface—the plane—that contains both the line and this point. We know that three non-collinear points uniquely define a plane. By using $A$, $B$, and $P$, we have everything we need to anchor our surface in space.

### Constructing the Vectors

To define the orientation of our plane, we need its normal vector—a vector that stands perfectly perpendicular to the surface. We first construct two vectors that lie flat on the plane.

We define vector $\vec{AB}$ by subtracting the position vector of $A$ from $B$, and vector $\vec{AP}$ by subtracting $A$ from $P$:

$$\vec{AB} = (1-1)\hat{i} + (1-(-1))\hat{j} + (1-2)\hat{k} = 0\hat{i} + 2\hat{j} - 1\hat{k}$$

$$\vec{AP} = (-2-1)\hat{i} + (-2-(-1))\hat{j} + (2-2)\hat{k} = -3\hat{i} - 1\hat{j} + 0\hat{k}$$

These two vectors, $\vec{AB} = (0, 2, -1)$ and $\vec{AP} = (-3, -1, 0)$, are now our tools. They are embedded in the plane, and their cross product will yield the direction of the normal vector.

### The Normal Vector

The cross product is where the orientation is determined. We set up the determinant:

$$\vec{n} = \vec{AB} 	imes \vec{AP} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 2 & -1 \ -3 & -1 & 0 \end{vmatrix}$$

Expanding this, we get:

$$\vec{n} = \hat{i}(0 - 1) - \hat{j}(0 - 3) + \hat{k}(0 - (-6)) = -1\hat{i} + 3\hat{j} + 6\hat{k}$$

For the sake of cleaner arithmetic, we can use the vector $\vec{n} = (1, -3, -6)$, which is parallel to our calculated normal. This vector acts as the "spine" of our plane.

### Building the Equation

With the normal vector $\vec{n} = (1, -3, -6)$ and a point $A(1, -1, 2)$ on the plane, we use the point-normal form:

$$a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$$

Substituting our values, we get:

$$1(x - 1) - 3(y - (-1)) - 6(z - 2) = 0$$

Expanding this, we find $x - 1 - 3y - 3 - 6z + 12 = 0$, which simplifies to the general equation of our plane:

$$x - 3y - 6z + 8 = 0$$

### Final Calculation: The Intercepts

The problem asks for the sum of the intercepts. We transform our equation into the intercept form:

$$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$$

Rearranging $x - 3y - 6z = -8$ and dividing by $-8$, we obtain:

$$\frac{x}{-8} + \frac{-3y}{-8} + \frac{-6z}{-8} = 1 \Rightarrow \frac{x}{-8} + \frac{y}{8/3} + \frac{z}{4/3} = 1$$

The intercepts are $a = -8$, $b = \frac{8}{3}$, and $c = \frac{4}{3}$. Summing these values, we get:

$$-8 + \frac{8}{3} + \frac{4}{3} = -8 + \frac{12}{3} = -8 + 4 = -4$$

The final result is **$-4$**.

Animated Solution for Mathematics - Three Dimensional Geometry: The sum of the intercepts on the coordinate axes of the plane passing through the point $(- 2, - 2, 2)$ and containing the line joining the points $(1, - 1, 2)$ and $(1, 1, 1)$ , is

Select Answer:

Visualized Solution

Visualizing the Given Points

Defining the Plane

Finding Vectors in the Plane

Calculating Vector $A B$

Calculating Vector $A P$

The Normal Vector $n$

Setting up the Cross Product

Evaluating the Normal Vector

Equation of the Plane

Substituting Values

Simplifying the Equation

The Intercept Form

Converting to Intercept Form

Calculating the Final Sum

The Sigma Insight: Equation of a Plane

Analyzing the Setup

Constructing the Vectors

The Normal Vector

Building the Equation

Final Calculation

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