Let the equation of the plane P containing the line (x+10)/1 = (8-y)/2 = z be ax + by + 3z = 2(a+b)

Q: Wait, why is the line equation $\frac{8-y}{2}$ problematic? Can't I just use the denominator as the direction vector?

That is a classic trap! The standard form of a line is $\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n}$. Notice the coefficients of $x, y,$ and $z$ must be positive one. In your term $\frac{8-y}{2}$, the coefficient of $y$ is $-1$. You must rewrite it as $\frac{y-8}{-2}$ to correctly identify the direction vector component $m = -2$.

Q: How do we know the line lies on the plane? Is there a simpler way than using two conditions?

Think of it this way: a plane is defined by a point and a normal vector. If a line is on the plane, its entire direction must be 'flat' against the plane, meaning it is perpendicular to the normal vector ($\vec{n} \cdot \vec{v} = 0$). Furthermore, the plane must 'pass through' the line, so any point on the line must also be on the plane. These two conditions are necessary and sufficient to lock the plane's orientation and position in space.

Q: Why is the distance formula $c = \frac{|Ax_1 + By_1 + Cz_1 - D|}{\sqrt{A^2 + B^2 + C^2}}$? Where does it come from?

Great question! This formula is essentially the projection of a vector connecting any point on the plane to your external point $Q$ onto the plane's normal vector $\vec{n}$. The numerator represents the scalar projection, and the denominator normalizes the vector. It is the most efficient way to find the shortest (perpendicular) distance from a point to a plane.

Question

Let the equation of the plane $P$ containing the line $\frac{x+10}{1} = \frac{8-y}{2} = z$ be $ax + by + 3z = 2(a+b)$ and the distance of the plane $P$ from the point $(1, 27, 7)$ be $c$. Then $a^2 + b^2 + c^2$ is equal to ______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional void. You have a flat, infinite sheet of paper representing a plane $P$, and a thin, perfectly straight wire representing a line $L$ that lies flat against that paper.

We are given the equation of the plane $P: ax + by + 3z = 2(a+b)$ and the line $L: \frac{x+10}{1} = \frac{8-y}{2} = z$. Our mission is to find the constants $a$ and $b$ that define this plane, and then calculate the distance $c$ from a specific point $Q(1, 27, 7)$ to this plane.

### Standardizing the Line

Before we can dance with the geometry, we must ensure our tools are sharp. The equation of the line $L$ is given as $\frac{x+10}{1} = \frac{8-y}{2} = z$.

Look closely at the middle term. In the standard form $\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n}$, the variables $x, y,$ and $z$ must have a coefficient of $+1$. The term $\frac{8-y}{2}$ is a clever trap.

By factoring out a $-1$ from the numerator, we transform it into $\frac{y-8}{-2}$. Now, our line is in its true form:
$$\frac{x-(-10)}{1} = \frac{y-8}{-2} = \frac{z-0}{1}$$

From this, we can immediately extract two treasures: a point on the line, $A(-10, 8, 0)$, and the direction vector of the line, $\vec{v} = (1, -2, 1)$.

### The Two Pillars of Containment

Now, how do we force this line to lie on the plane? We need two pillars of logic.

First, if the line is on the plane, then the point $A(-10, 8, 0)$ must also be on the plane. Substituting these coordinates into the plane's equation $ax + by + 3z - 2(a+b) = 0$, we get:
$$a(-10) + b(8) + 3(0) - 2(a+b) = 0$$

Simplifying this, we find $-10a + 8b - 2a - 2b = 0$, which leads to $-12a + 6b = 0$, or simply $b = 2a$. This is our first breakthrough.

Second, the direction vector of the line $\vec{v} = (1, -2, 1)$ must be perpendicular to the normal vector of the plane $\vec{n} = (a, b, 3)$. When two vectors are perpendicular, their dot product is zero.

Thus, $\vec{n} \cdot \vec{v} = (a)(1) + (b)(-2) + (3)(1) = 0$, which simplifies to $a - 2b + 3 = 0$.

### Solving the Mystery

We now have a system of two equations: $b = 2a$ and $a - 2b + 3 = 0$. Substituting the first into the second, we get:
$$a - 2(2a) + 3 = 0$$

This means $a - 4a + 3 = 0$, or $-3a = -3$. Thus, $a = 1$. Consequently, $b = 2(1) = 2$.

Our plane is now fully revealed: $1x + 2y + 3z = 2(1+2)$, or $x + 2y + 3z = 6$.

### The Final Distance

We have the plane $x + 2y + 3z - 6 = 0$ and the point $Q(1, 27, 7)$. The perpendicular distance $c$ from a point $(x_1, y_1, z_1)$ to a plane $Ax + By + Cz + D = 0$ is given by:
$$c = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$$

Plugging in our values, we get:
$$c = \frac{|(1)(1) + (2)(27) + (3)(7) - 6|}{\sqrt{1^2 + 2^2 + 3^2}}$$

The numerator is $|1 + 54 + 21 - 6| = |70| = 70$. The denominator is $\sqrt{1 + 4 + 9} = \sqrt{14}$. So, $c = \frac{70}{\sqrt{14}} = 5\sqrt{14}$.

Finally, the problem asks for $a^2 + b^2 + c^2$. Substituting our values, we get:
$$1^2 + 2^2 + (5\sqrt{14})^2 = 1 + 4 + 25(14) = 5 + 350 = 355$$

We have arrived at the summit. The final answer is **355**.

Enter Numerical Value:

Visualized Solution

Visualizing the Plane and Line

Standardizing the Line Equation

Extracting Point and Direction

The Plane's Normal Vector

Condition 1: Point on Plane

Simplifying Condition 1

Condition 2: Orthogonality

Evaluating the Dot Product

Solving for a and b

The Final Plane Equation

Distance from a Point to a Plane

Substituting into Distance Formula

Calculating Distance c

Final Evaluation

The Sigma Insight: Equation of a Plane

Analyzing the Setup

Standardizing the Line

The Two Pillars of Containment

Solving the Mystery

The Final Distance

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