Let P be the plane, passing through the point (1, -1, -5) and perpendicular to the line joining the

Q: Why is the line joining the two points the normal vector?

The problem states the plane is perpendicular to the line. In 3D geometry, if a line is perpendicular to a plane, its direction vector is parallel to the plane's normal vector. Thus, the vector connecting the two points on the line is exactly the direction we need for our normal vector $\vec{n}$.

Q: Does the order of subtraction for the normal vector matter?

Great question! If you subtract point $A$ from $B$ instead of $B$ from $A$, you get the negative of the normal vector. However, when you plug this into the plane equation, the signs will cancel out, and the final equation will be identical. So, don't worry about the order!

Q: Why do we use the absolute value in the distance formula?

Distance is a physical quantity and must always be non-negative. The expression $ax + by + cz + d$ can be negative depending on which side of the plane the point lies. The absolute value ensures we get the magnitude of the displacement, which is the physical distance.

Question

Let $P$ be the plane, passing through the point $(1, -1, -5)$ and perpendicular to the line joining the points $(4, 1, -3)$ and $(2, 4, 3)$. Then the distance of $P$ from the point $(3, -2, 2)$ is

Options:

(A) 6

(B) 4

(C) 5

(D) 7

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, explorers of the third dimension! Today, we are going to demystify the concept of a plane in space. Imagine a flat sheet of paper floating in the vastness of 3D coordinates.

It is not just sitting there; it is oriented in a specific way, held in place by a line that pierces through it at a perfect right angle. This is the essence of our problem: we have a plane $P$ passing through a point $P_0(1, -1, -5)$ and standing perpendicular to a line defined by two points, $A(4, 1, -3)$ and $B(2, 4, 3)$.

Our goal is to find the distance from this plane to a point $Q(3, -2, 2)$. Let us embark on this journey.

### Phase 1: Finding the Normal Vector

In the world of planes, the most important piece of information is the normal vector, $\vec{n}$. Think of this vector as the 'compass' of the plane—it tells us exactly which way the plane is facing.

Because our plane is perpendicular to the line joining $A$ and $B$, the direction of this line *is* the normal vector. To find it, we calculate the vector $\vec{BA}$.

We subtract the coordinates of $B$ from $A$:
$$\vec{n} = (4 - 2, 1 - 4, -3 - 3) = (2, -3, -6)$$

This vector $(2, -3, -6)$ is the backbone of our plane equation.

### Phase 2: Constructing the Plane Equation

Now that we have our normal vector $\vec{n} = (2, -3, -6)$ and a point on the plane $P_0(1, -1, -5)$, we can use the point-normal form of the plane equation:
$$a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$$

Substituting our values, we get:
$$2(x - 1) - 3(y - (-1)) - 6(z - (-5)) = 0$$

Expanding this, we carefully handle the signs:
$$2x - 2 - 3(y + 1) - 6(z + 5) = 0$$

Simplifying further:
$$2x - 2 - 3y - 3 - 6z - 30 = 0$$

Combining the constants, we arrive at the standard form:
$$2x - 3y - 6z - 35 = 0$$

This equation is the mathematical signature of our plane.

### Phase 3: The Final Leap

We are almost there! We need to find the distance from point $Q(3, -2, 2)$ to our plane. The formula for the perpendicular distance from a point $(x_1, y_1, z_1)$ to a plane $ax + by + cz + d = 0$ is:
$$d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}$$

Let us plug in our values. The numerator becomes:
$$|2(3) - 3(-2) - 6(2) - 35| = |6 + 6 - 12 - 35| = |-35| = 35$$

The denominator is the magnitude of the normal vector:
$$\sqrt{2^2 + (-3)^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$$

Finally, we calculate the distance:
$$d = \frac{35}{7} = 5$$

The distance is **5 units**. You have successfully navigated the 3D space and conquered the problem!

Animated Solution for Mathematics - Three Dimensional Geometry: Let $P$ be the plane, passing through the point $(1, - 1, - 5)$ and perpendicular to the line joining the points $(4, 1, - 3)$ and $(2, 4, 3)$ . Then the distance of $P$ from the point $(3, - 2, 2)$ is

Select Answer:

Visualized Solution

Visualizing the Geometry

Finding the Normal Vector $n$

Calculating Direction Ratios

The Point-Normal Form

Substituting the Values

Expanding the Equation

Simplifying to Standard Form

Distance from a Point

Plugging in the Coordinates

Calculating the Numerator

Calculating the Denominator

Final Result

The Sigma Insight: Equation of a Plane

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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