Distance of a point from a plane and the line of intersection of two planes

What you will be able to do

• find the distance of a point from a plane, the foot of the perpendicular, and the distance between parallel planes;
• find the line in which two planes meet, in vector and Cartesian form.

How high is the drone above the roof?

A drone hovers above a sloping roof. The shortest distance to the roof is along the line at right angles to it, which is the direction of the normal. We find it by projecting a vector from the roof onto the normal.

The distance formula

Let the plane be $\vec{r} \cdot \vec{n} = k$ and let $P$ have position vector $\vec{p}$. The vector $\vec{p}$ has component $\dfrac{\vec{p} \cdot \vec{n}}{\lvert \vec{n} \rvert}$ along the unit normal. A point of the plane has component $\dfrac{k}{\lvert \vec{n} \rvert}$. The difference is the signed distance:

The distance of the origin from the plane is $\dfrac{\lvert D \rvert}{\sqrt{A^2 + B^2 + C^2}}$. Two points are on the same side of the plane when the values $Ax + By + Cz + D$ at them have the same sign.

The foot and the parallel planes. The foot of the perpendicular from $P$ is $F = \vec{p} - \dfrac{\vec{p} \cdot \vec{n} - k}{\lvert \vec{n} \rvert^2}\,\vec{n}$. For parallel planes $Ax + By + Cz + D_1 = 0$ and $Ax + By + Cz + D_2 = 0$, with the same coefficients, the distance is $\dfrac{\lvert D_1 - D_2 \rvert}{\sqrt{A^2 + B^2 + C^2}}$. Divide through first if the coefficients differ.

The line where two planes meet

Two planes that are not parallel meet in a line. A point on both planes satisfies both equations. A direction along the line is perpendicular to both normals, so it is $\vec{n}_1 \times \vec{n}_2$. To find a point, give one coordinate a convenient value, such as $z = 0$, and solve for the other two.

A worked example

Distance and foot. Find the distance of $P(3, -1, 2)$ from $2x + 3y + 6z = 1$.

1. The value at $P$. $6 - 3 + 12 = 15$, and $15 - 1 = 14$.
2. The length of the normal. $\sqrt{4 + 9 + 36} = 7$.
3. The distance. $\dfrac{14}{7} = 2$ units.
4. The foot. $F = P - \dfrac{14}{49}(2, 3, 6) = \left(\dfrac{17}{7},\ -\dfrac{13}{7},\ \dfrac{2}{7}\right)$. Check: $\dfrac{34 - 39 + 12}{7} = 1$.

The line of two planes. Find the line in which $x + y + z = 6$ and $2x - y + z = 3$ meet.

1. The direction. $(1, 1, 1) \times (2, -1, 1) = (2, 1, -3)$.
2. A point. Put $z = 0$. Then $x + y = 6$ and $2x - y = 3$ give $x = 3$ and $y = 3$.
3. The line. $\dfrac{x - 3}{2} = \dfrac{y - 3}{1} = \dfrac{z}{-3}$.
4. A check. At $\lambda = 1$ the point is $(5, 4, -3)$. Then $5 + 4 - 3 = 6$ and $10 - 4 - 3 = 3$. Both planes are satisfied.

Practise with new numbers

Try it in Python

The program finds the distance and the foot for the worked example. It then finds the line in which two planes meet.

Common mistakes

• Forgetting the modulus. A distance is never negative. The sign only says which side of the plane the point is on.
• Not dividing by the length of the normal. The value of $Ax + By + Cz + D$ at a point is a number, not a distance.
• Comparing planes with different coefficients. Make the coefficients the same before using $\lvert D_1 - D_2 \rvert$.
• Setting a coordinate to $0$ when the line cannot pass through it. If the equations for the other two give no solution, set another coordinate to $0$.

Remember these

• $d = \dfrac{\lvert Ax_1 + By_1 + Cz_1 + D \rvert}{\sqrt{A^2 + B^2 + C^2}}$.
• Parallel planes with equal coefficients: $d = \dfrac{\lvert D_1 - D_2 \rvert}{\sqrt{A^2 + B^2 + C^2}}$.
• The foot of the perpendicular is $P$ minus the signed distance times the unit normal.
• Two planes meet in a line with direction $\vec{n}_1 \times \vec{n}_2$.

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 $(1, 2, 3)$ from the plane $x + 2y + 2z = 3$.

2. Find the distance of the origin from the plane $2x - 3y + 6z = 21$.

3. Find the distance between the planes $x + 2y - 2z = 3$ and $x + 2y - 2z = 12$.

4. Find the direction of the line in which the planes $x + y = 2$ and $y + z = 3$ meet.

5. The points $(1, 1, 1)$ and $(3, 3, 3)$ are tried against the plane $x + y + z = 6$. Are they on the same side?

References

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