The perpendicular distance from the origin to the plane containing the two lines, (x+2)/3 = (y-2)/5

Q: Wait, why do we reject the negative root in the denominator of the distance formula?

The distance formula $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$ uses the absolute value of $D$ in the numerator because distance must always be a non-negative quantity. The denominator $\sqrt{A^2 + B^2 + C^2}$ is the magnitude of the normal vector, which is also inherently positive. We are looking for a physical length, so we don't worry about signs in the square root!

Q: Can I pick any point on the lines to form the plane equation?

Absolutely! Since both lines lie entirely within the plane, any point on $L_1$ or $L_2$ is a valid point on the plane. Just ensure you use the coordinates of a point that satisfies the equation of the line you chose. The final plane equation will be the same regardless of which point you pick.

Q: Why did we divide the normal vector by 7? Is that allowed?

Yes, it is perfectly allowed! The normal vector $\vec{n}$ only defines the *orientation* of the plane. Any scalar multiple of the normal vector (like $\vec{n}/7$) is still a normal vector to the same plane. Scaling it down just makes the arithmetic in the plane equation much cleaner and reduces the chance of calculation errors.

Question

The perpendicular distance from the origin to the plane containing the two lines, $\frac{x+2}{3} = \frac{y-2}{5} = \frac{z+5}{7}$ and $\frac{x-1}{1} = \frac{y-4}{4} = \frac{z+4}{7}$, is:

Options:

(A) $11/\sqrt{6}$

(B) $6\sqrt{11}$

(C) 11

(D) $11\sqrt{6}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, empty room. You have two straight, rigid rods floating in the air, representing our lines $L_1$ and $L_2$.

These lines are not parallel and they define a single, flat surface—a plane—that contains both of them. Our goal is to find the shortest distance from the origin $(0,0,0)$ to this invisible sheet of paper.

### The Directional Foundation

Every line in 3D space is defined by a point it passes through and a direction vector. Looking at our lines:

$$L_1: \frac{x+2}{3} = \frac{y-2}{5} = \frac{z+5}{7}$$

$$L_2: \frac{x-1}{1} = \frac{y-4}{4} = \frac{z+4}{7}$$

We can immediately extract their direction vectors, which act as the 'compass headings' of our lines. For $L_1$, the direction vector is $\vec{b_1} = 3\hat{i} + 5\hat{j} + 7\hat{k}$. For $L_2$, it is $\vec{b_2} = 1\hat{i} + 4\hat{j} + 7\hat{k}$.

### The Magic of the Cross Product

To define a plane, we need a normal vector—a vector that sticks straight out of the surface, perpendicular to every line on it. We find this by calculating the cross product $\vec{n} = \vec{b_1} 	imes \vec{b_2}$.

Setting up the determinant:

$$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 5 & 7 \ 1 & 4 & 7 \end{vmatrix}$$

Expanding this, we get $\vec{n} = \hat{i}(35 - 28) - \hat{j}(21 - 7) + \hat{k}(12 - 5)$, which simplifies to $\vec{n} = 7\hat{i} - 14\hat{j} + 7\hat{k}$.

### Simplifying the Normal

Mathematics is often about finding the most elegant path. Notice that all components of our normal vector are multiples of $7$.

We can divide the entire vector by $7$ to get a simpler normal vector: $\vec{n} = 1\hat{i} - 2\hat{j} + 1\hat{k}$. This vector $(1, -2, 1)$ carries the exact same directional information as the original, but it makes our subsequent algebra much lighter.

### Constructing the Plane

Now, we need a point on the plane. Since both lines lie on the plane, we can pick any point from either line; let's take a point from $L_2$: $P(1, 4, -4)$.

Using the point-normal form of a plane, $A(x-x_1) + B(y-y_1) + C(z-z_1) = 0$, we substitute our normal vector $(1, -2, 1)$ and our point $(1, 4, -4)$:

$$1(x - 1) - 2(y - 4) + 1(z + 4) = 0$$

Expanding this, we get $x - 1 - 2y + 8 + z + 4 = 0$, which simplifies beautifully to the plane equation:

$$x - 2y + z + 11 = 0$$

### The Final Leap

We need the perpendicular distance from the origin $(0,0,0)$ to the plane $x - 2y + z + 11 = 0$. The formula for the distance $d$ from the origin to a plane $Ax + By + Cz + D = 0$ is:

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

Substituting our values ($A=1, B=-2, C=1, D=11$):

$$d = \frac{|11|}{\sqrt{1^2 + (-2)^2 + 1^2}} = \frac{11}{\sqrt{1 + 4 + 1}} = \frac{11}{\sqrt{6}}$$

The geometry of the lines, the power of the cross product, and the elegance of the distance formula have guided us to the final answer: **$\frac{11}{\sqrt{6}}$**.

Animated Solution for Mathematics - Three Dimensional Geometry: The perpendicular distance from the origin to the plane containing the two lines, $\frac{x + 2}{3} = \frac{y - 2}{5} = \frac{z + 5}{7}$ and $\frac{x - 1}{1} = \frac{y - 4}{4} = \frac{z + 4}{7}$ , is:

Select Answer:

Visualized Solution

Visualize the Geometry

Identify Direction Vectors

Finding the Normal Vector $n$

Calculate the Cross Product

Simplify the Normal Vector

Pick a Point on the Plane

Form the Plane Equation

Simplify the Equation

Distance from Origin Formula

Final Calculation

Summary and Takeaway

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Directional Foundation

The Magic of the Cross Product

Simplifying the Normal

Constructing the Plane

The Final Leap

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