How far apart are the points (2,0,0) and (-3,0,0)?

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(2,0,0)

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Concept understanding — Distance From Point To Line

Distance from a Point to a Line

The distance from a point to a line is the shortest distance — the length of the perpendicular dropped from the point onto the line. In 3D we compute it with vectors and the cross product.

Let the line be $\vec{r}=\vec{a}+\lambda\vec{b}$ (a point $A$ with position vector $\vec{a}$ , direction $\vec{b}$ ), and let $P$ be the given point with position vector $\vec{p}$ .

The idea

Look at the triangle formed by $A$ , $P$ and the foot of the perpendicular $M$ . The segment $\overrightarrow{AP}=\vec{p}-\vec{a}$ is the hypotenuse, and the perpendicular distance $d=PM$ is the side opposite the angle $\theta$ between $\overrightarrow{AP}$ and the line:

d = |\overrightarrow{AP}|\sin\theta.

But the cross product already contains $\sin\theta$ : $|\overrightarrow{AP}\times\vec{b}| = |\overrightarrow{AP}|\,|\vec{b}|\sin\theta$ . Dividing by $|\vec{b}|$ isolates the distance.

d = \frac{|(\vec{p}-\vec{a})\times\vec{b}|}{|\vec{b}|}

Example

Distance of $P(1,2,3)$ from the line $\vec{r}=(\hat{i}+\hat{j})+\lambda(2\hat{i}-\hat{j}+2\hat{k})$ .

Here $\vec{a}=(1,1,0)$ , $\vec{b}=(2,-1,2)$ , and $\overrightarrow{AP}=\vec{p}-\vec{a}=(0,1,3)$ .

$\overrightarrow{AP}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\0&1&3\\2&-1&2\end{vmatrix} = (2+3)\hat{i} - (0-6)\hat{j} + (0-2)\hat{k} = 5\hat{i}+6\hat{j}-2\hat{k}.$ …

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The distance between two points on a coordinate axis is the absolute difference of their coordinates; for $(2,0,0)$ and $(-3,0,0)$ , this distance is $\boxed{5}$ .

When we talk about the "distance" between two points in geometry, we are referring to the shortest possible path connecting them, which is always a straight line segment. In a 3D coordinate system, this distance can be visualized as the length of the hypotenuse of a right-angled triangle (or a series of them) formed by the differences in their coordinates.

Consider the given points: $P_1 = (2,0,0)$ and $P_2 = (-3,0,0)$ .

Notice that both points have their $y$ and $z$ coordinates equal to $0$ . This means both points lie directly on the $x$ -axis.

Intuition for Points on an Axis

Imagine a number line. If you have a point at $2$ and another at $-3$ , how do you find the distance between them? You simply find the absolute difference of their positions.

The distance between $2$ and $-3$ on a number line is $|2 - (-3)| = |2+3| = |5| = 5$ .

Alternatively, it's $|-3 - 2| = |-5| = 5$ .

This is a fundamental concept: distance is always a non-negative value.

The General 3D Distance Formula

While the intuition for points on an axis is simple, it's important to understand the general formula for the distance between any two points in 3D space. This formula is a direct extension of the Pythagorean theorem.

d

Let's apply this formula step-by-step to our specific problem.

Identify the coordinates of the two points. We have $P_1 = (2,0,0)$ and $P_2 = (-3,0,0)$ . So, we can assign: $x_1 = 2$ , $y_1 = 0$ , $z_1 = 0$ $x_2 = -3$ , $y_2 = 0$ , $z_2 = 0$ …

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Q.How far apart are the points (2,0,0)(2,0,0)(2,0,0) and (−3,0,0)(-3,0,0)(−3,0,0)?

Intuition for Points on an Axis

The General 3D Distance Formula

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Q.How far apart are the points $(2,0,0)$ and $(-3,0,0)$ ?

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