Distance of a point from the origin, an axis and a plane

What you will be able to do

• find the distance of a point from the origin;
• find the distance of a point from each axis;
• find the distance of a point from a coordinate plane and from a plane parallel to it.

The distance from the origin

How far from the corner of the room is the fly? Walk along the floor to the point $N$ below the fly, and then straight up.

Let $P(x, y, z)$ be the point. Its shadow on the $xy$-plane is $N(x, y, 0)$. Triangle $ONP$ has a right angle at $N$, because $NP$ is vertical and $ON$ lies on the floor. In the floor, $ON$ is the diagonal of a rectangle with sides $|x|$ and $|y|$. So

Pythagoras in triangle $ONP$ gives

So the distance of $P(x, y, z)$ from the origin is $\sqrt{x^2 + y^2 + z^2}$.

The distance from an axis

The foot of the perpendicular from $P$ to the $x$-axis is $(x, 0, 0)$. The distance is the length of the segment from $(x, 0, 0)$ to $P$. Only $y$ and $z$ differ, so

In the same way $d_y = \sqrt{x^2 + z^2}$ and $d_z = \sqrt{x^2 + y^2}$. The distance from an axis uses the two coordinates that are not along that axis.

The distance from a plane

The distance of $P$ from the $xy$-plane is the length of the perpendicular, which is $|z|$. The distance from the $yz$-plane is $|x|$, and from the $zx$-plane it is $|y|$.

A plane parallel to the $xy$-plane has the equation $z = k$. All its points are at height $k$. The distance from $P$ to it is $|z - k|$. In the same way the plane $x = k$ is at distance $|x - k|$ from $P$.

A worked example

Take $P(3, -4, 12)$.

1. From the origin. $\sqrt{9 + 16 + 144} = 13$.
2. From the $x$-axis. $\sqrt{16 + 144} = \sqrt{160} = 4\sqrt{10}$.
3. From the $z$-axis. $\sqrt{9 + 16} = 5$.
4. From the $xy$-plane. $|12| = 12$.
5. From the plane $z = 20$. $|12 - 20| = 8$.

The distance from the origin is the greatest of the distances from the axes, because it adds all three squares.

Practise with new numbers

Try it in Python

The program finds all the distances of one point. It also checks which axis is nearest.

Common mistakes

• Using all three squares for an axis. For an axis leave out the coordinate that lies along it.
• Forgetting the square root. $x^2 + y^2 + z^2$ is the square of the distance.
• Dropping the sign too early. A distance to a plane is $|z - k|$. If $z$ is less than $k$, take $k - z$.
• Mixing up an axis and a plane. The distance from the $z$-axis is $\sqrt{x^2 + y^2}$. The distance from the $xy$-plane is $|z|$.

Remember these

• Distance from the origin: $\sqrt{x^2 + y^2 + z^2}$.
• Distances from the axes: $\sqrt{y^2 + z^2}$, $\sqrt{x^2 + z^2}$, $\sqrt{x^2 + y^2}$.
• Distances from the planes: $|z|$, $|x|$, $|y|$. From the plane $z = k$: $|z - k|$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the distance of $(2, -3, 6)$ from the origin.

2. Find the distance of $(3, 4, 12)$ from the $y$-axis.

3. Find the distance of $(-5, 12, 8)$ from the $z$-axis.

4. Find the distances of $(4, -2, 9)$ from the $xy$-plane, the $yz$-plane and the $zx$-plane.

5. Find the distance of $(5, 2, -3)$ from the plane $z = 2$.

References

• NCERT, Mathematics Textbook for Class XI, the chapter Introduction to Three Dimensional Geometry, section on the distance between two points.
• OpenStax, Calculus Volume 3, Chapter 2, section 2.2 on vectors in three dimensions.