Let (x-2)/3 = (y+1)/-2 = (z+3)/-1 lie on the plane px - qy + z = 5, for some p, q ∈ R. The shortest

Q: Wait, why do we need two equations? Can't we just use the point on the line?

That is a fantastic question! A single point on the line only gives us one constraint on the plane's parameters $p$ and $q$. Since we have two unknowns, we need two independent pieces of information. The point gives us the first, and the direction of the line (which must be parallel to the plane) gives us the second. Without both, the plane could 'rotate' around that single point, and we wouldn't have a unique solution.

Q: Why is the dot product of the direction vector and the normal vector zero?

Think about the geometry: the normal vector $\vec{n}$ is, by definition, perpendicular to every vector lying on the plane. Since the line lies on the plane, its direction vector $\vec{v}$ is also on the plane. Therefore, $\vec{v}$ must be perpendicular to $\vec{n}$. In the language of vectors, two perpendicular vectors have a dot product of zero. It is the most elegant way to enforce the 'parallel' condition!

Q: Is there a shortcut for the distance from the origin?

For the distance from the origin $(0,0,0)$ to a plane $Ax + By + Cz + D = 0$, the formula simplifies beautifully to $\frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$. You don't need to plug in zeros for $x, y, z$ every time—just remember that the origin effectively 'kills' the $Ax, By, Cz$ terms, leaving only the constant term in the numerator. Just be careful with the sign of $D$!

Question

Let $\frac{x-2}{3} = \frac{y+1}{-2} = \frac{z+3}{-1}$ lie on the plane $px - qy + z = 5$, for some $p, q \in \mathbb{R}$. The shortest distance of the plane from the origin is:

Options:

(A) $\sqrt{\frac{3}{109}}$

(B) $\sqrt{\frac{5}{142}}$

(C) $\sqrt{\frac{5}{71}}$

(D) $\sqrt{\frac{1}{142}}$

Sigma Percentile · Accepted Answer

### The Geometry of Embedding: A Journey into 3D Space

Imagine you are standing in a vast, three-dimensional coordinate system. You have a flat, infinite plane stretching out before you, and resting perfectly on that surface is a straight line.

This is not just any line; it is a line that is completely 'embedded' in the plane. We are given the line:

$$\frac{x-2}{3} = \frac{y+1}{-2} = \frac{z+3}{-1}$$

And a plane defined by the equation $px - qy + z = 5$. Our mission is to find the parameters $p$ and $q$ that make this embedding possible, and then calculate the shortest distance from the origin to this plane.

### Phase 1: The Point of Contact

When a line lies on a plane, every single point on that line must also satisfy the equation of the plane. This is our first powerful insight.

Looking at the symmetric form of the line, we can immediately extract a point that lies on it. By setting the ratios equal to a parameter, we see that at the point where the numerators are zero, we have $P(2, -1, -3)$.

Since this point is on the line, it must be on the plane. Substituting these coordinates into the plane equation $px - qy + z = 5$, we get:

$$p(2) - q(-1) + (-3) = 5$$

Simplifying this, we arrive at our first beautiful constraint:

$$2p + q = 8$$

### Phase 2: The Vector Dance

Now, we need a second constraint. If the line is on the plane, then the direction of the line must be parallel to the plane.

The plane is defined by its normal vector, $\vec{n} = (p, -q, 1)$, which points straight out of the surface. The line has a direction vector, $\vec{v} = (3, -2, -1)$, which runs along the surface.

If the line is on the plane, its direction vector must be perpendicular to the plane's normal vector. The dot product of two perpendicular vectors is zero:

$$\vec{v} \cdot \vec{n} = 0$$

Calculating this, we get $3(p) + (-2)(-q) + (-1)(1) = 0$, which simplifies to:

$$3p + 2q = 1$$

We now have a system of two linear equations: $2p + q = 8$ and $3p + 2q = 1$.

### Phase 3: Solving the System

Algebra is the language of geometry. To solve for $p$ and $q$, we can multiply the first equation by 2 to get $4p + 2q = 16$.

Now, subtract the second equation ($3p + 2q = 1$) from this. The $q$ terms vanish, leaving us with:

$$p = 15$$

Substituting $p = 15$ back into our first equation, $2(15) + q = 8$, we find $30 + q = 8$, which gives:

$$q = -22$$

We have successfully unlocked the parameters of our plane. The equation of our plane is $15x + 22y + z = 5$, or $15x + 22y + z - 5 = 0$.

### Phase 4: The Final Leap

We have defined the plane. Now, for the final act: finding the shortest distance from the origin $(0,0,0)$ to this plane.

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

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

Plugging in our values, we get:

$$d = \frac{|15(0) + 22(0) + 1(0) - 5|}{\sqrt{15^2 + 22^2 + 1^2}}$$

This simplifies to:

$$d = \frac{|-5|}{\sqrt{225 + 484 + 1}} = \frac{5}{\sqrt{710}}$$

To match standard forms, we can write this as $\sqrt{\frac{25}{710}}$, which reduces to:

**$$d = \sqrt{\frac{5}{142}}$$**

Animated Solution for Mathematics - Three Dimensional Geometry: Let $\frac{x - 2}{3} = \frac{y + 1}{- 2} = \frac{z + 3}{- 1}$ lie on the plane $p x - q y + z = 5$ , for some $p, q \in R$ . The shortest distance of the plane from the origin is:

Select Answer:

Visualized Solution

Visualizing the Setup

Extracting a Point from the Line

Point-in-Plane Condition

Simplifying the First Equation

Directional Constraint

Applying the Dot Product

Simplifying the Second Equation

Solving for $p$

Solving for $q$

The Final Plane Equation

Distance from Origin

Distance Formula Setup

Calculating the Distance

Final Simplification

The Sigma Insight: Intersection of a Line and a Plane

The Geometry of Embedding

Phase 1

Phase 2

Phase 3

Phase 4

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