Distance of a point from a line and the foot of the perpendicular

What you will be able to do

• find the foot of the perpendicular from a point to a line;
• find the distance of a point from a line in two ways;
• find the points of a line at a given distance from a point.

A drone and a power line

A drone hovers at a point $P$ near a straight power line. The nearest point of the line is straight below, in the sense that the segment from the drone to it meets the line at a right angle. That point is the foot of the perpendicular, $F$. The distance $PF$ is the distance of the point from the line.

Method 1: the foot first

Let the line be $\vec{r} = \vec{a} + t\vec{b}$. Any point of the line is $F = \vec{a} + t\vec{b}$. We want $PF$ to be at right angles to the line, so

Put this $t$ back to get $F$. Then $PF$ is the distance. The method works in every book and also gives the foot, which the formula in Method 2 does not.

Method 2: the cross product

The vector $P - \vec{a}$ makes an angle $\theta$ with $\vec{b}$. The perpendicular distance is $\lvert P - \vec{a} \rvert \sin\theta$. Since $\lvert \vec{u} \times \vec{b} \rvert = \lvert \vec{u} \rvert \lvert \vec{b} \rvert \sin\theta$, this gives

For the coordinate axes the answer is quick. The distance of $(x, y, z)$ from the $x$-axis is $\sqrt{y^2 + z^2}$, from the $y$-axis it is $\sqrt{x^2 + z^2}$, and from the $z$-axis it is $\sqrt{x^2 + y^2}$.

A worked example

Find the foot of the perpendicular from $P(5, 1, 4)$ to the line $\dfrac{x - 1}{2} = \dfrac{y - 2}{1} = \dfrac{z - 3}{2}$, and the distance.

1. The vectors. $\vec{a} = (1, 2, 3)$ and $\vec{b} = (2, 1, 2)$. Then $P - \vec{a} = (4, -1, 1)$.
2. The parameter. $(4, -1, 1) \cdot (2, 1, 2) = 8 - 1 + 2 = 9$ and $\lvert \vec{b} \rvert^2 = 9$. So $t = 1$.
3. The foot. $F = (1 + 2, 2 + 1, 3 + 2) = (3, 3, 5)$.
4. The distance. $PF = \lvert (2, -2, -1) \rvert = \sqrt{4 + 4 + 1} = 3$ units.
5. The check by Method 2. $(4, -1, 1) \times (2, 1, 2) = (-3, -6, 6)$ with length $9$. Then $d = \dfrac{9}{3} = 3$. Both methods agree.

A second example. The distance of the point $(7, 5, 12)$ from the $x$-axis is $\sqrt{25 + 144} = 13$ units.

Practise with new numbers

Try it in Python

The program finds the foot and the distance of the worked example by Method 1. It then checks the answer by Method 2.

Common mistakes

• Dividing by $\lvert \vec{b} \rvert$ in $t$. The parameter divides by $\lvert \vec{b} \rvert^2$, because $\vec{b}$ appears twice in $t\vec{b}$.
• Using the distance to $\vec{a}$. The distance from $P$ to the point $A$ of the line is longer than the distance to the line itself.
• Forgetting that the foot is on the line. Check that $F = \vec{a} + t\vec{b}$ really satisfies the equation of the line.
• Mixing $P - \vec{a}$ and $\vec{a} - P$. The sign of $t$ changes. Keep one order.

Remember these

• The foot $F = \vec{a} + t\vec{b}$ with $t = \dfrac{(P - \vec{a}) \cdot \vec{b}}{\lvert \vec{b} \rvert^2}$.
• The distance is $\lvert PF \rvert$, or $\dfrac{\lvert (P - \vec{a}) \times \vec{b} \rvert}{\lvert \vec{b} \rvert}$.
• The distance from the $x$-axis is $\sqrt{y^2 + z^2}$.
• $PF \perp \vec{b}$ is the equation that fixes the foot.

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 $(7, 5, 12)$ from the $y$-axis.

2. Find the foot of the perpendicular from $(4, 5, 2)$ to the line through the origin with direction ratios $(1, 2, 2)$.

3. Find the distance of $(4, 5, 2)$ from the same line.

4. Why is $PF$ the shortest distance from $P$ to the line?

5. If the distance of $P$ from a line is $5$, how far is $P$ from a point of the line that is $12$ units from the foot?

References

• OpenStax, Calculus Volume 3, Chapter 2, sections on the dot product, the cross product and equations of lines in space (distance from a point to a line).
• NCERT, Mathematics Part II for Class XII, Chapter 11, section on the equation of a line in space.