Bisector Planes

Stand at the corner where two walls of a room meet. Hold a flat sheet of cardboard so it fits perfectly into the corner, making equal angles with both walls. That cardboard is lying along a bisector plane — a plane that splits the dihedral angle between the two walls into two equal halves.

But there is a second position for the cardboard. Instead of fitting into the acute corner, tilt it the other way so it bisects the obtuse opening between the walls on the outside. Two walls, two bisector planes — one for each pair of vertical angles formed at the line of intersection.

In 2D, when two lines cross, the angle bisectors are lines. In 3D, when two planes cross, the angle bisectors are planes. The idea is the same — equidistance — but the algebra lives in three dimensions. The payoff is a clean formula that handles every case.

The key idea: equidistance

Here is the principle that makes everything work. A point lies on the bisector of two planes if and only if its perpendicular distance from one plane equals its perpendicular distance from the other.

Think about why this is true. If a point P is equidistant from two planes P_1 and P_2, then the perpendicular feet from P to both planes are at the same distance. That means P lies on a plane that makes equal angles with both P_1 and P_2 — which is exactly the definition of an angle bisector. Conversely, if P lies on the angle bisector, the symmetry forces equal distances to both planes. Equidistance and angle-bisecting are the same condition, just described differently.

This is the same principle behind angle bisectors of two lines in 2D. The only difference is that "distance from a line" becomes "distance from a plane."

Recall: the perpendicular distance from a point (x_1, y_1, z_1) to the plane ax + by + cz + d = 0 is

The absolute value in the numerator means the distance is always non-negative. But for bisector planes, you will need to work with the signed version — the expression without the absolute value — because the sign tells you which side of the plane the point sits on. A point on one side of the plane gives a positive numerator; a point on the other side gives a negative numerator. The magnitude is the same in both cases — only the sign changes. This sign information is what separates the two bisector planes.

Deriving the equation of bisector planes

Take two planes:

A point (x, y, z) lies on a bisector plane exactly when its perpendicular distances from P_1 and P_2 are equal. Using the signed distance formula:

The \pm is doing real work. There are two cases:

Case 1 (positive sign):

This says the signed distances are equal — the point is on the same side of both planes (or on the same-side pair). This gives one bisector plane.

Case 2 (negative sign):

This says the signed distances are equal in magnitude but opposite in sign — the point is on opposite sides of the two planes. This gives the other bisector plane.

Together, these two equations give you both bisector planes. Write them compactly as:

Notice what this formula is really saying. Every point on the first bisector plane is equidistant from both original planes and lies on the same side of both. Every point on the second bisector plane is also equidistant from both, but it lies between the two planes — on opposite sides.

Which bisector contains the origin?

In many problems — especially in JEE — you need to identify which of the two bisector planes contains the origin. There is a clean test.

Step 1. Arrange both plane equations so that the constant terms d_1 and d_2 have the same sign (both positive or both negative). If they don't, multiply one equation through by -1.

Step 2. After this arrangement, the bisector plane given by the positive sign contains the origin.

Why does this work? The origin (0, 0, 0) has signed distance d_1 / \sqrt{a_1^2 + b_1^2 + c_1^2} from P_1 and d_2 / \sqrt{a_2^2 + b_2^2 + c_2^2} from P_2. If d_1 and d_2 have the same sign, both signed distances have the same sign, so the origin satisfies the positive-sign equation. If they had opposite signs, the origin would satisfy the negative-sign equation — but you fixed that by flipping one equation first.

This is a standard trick. Make the constants agree in sign, and then the positive bisector is the one the origin lives on.

Acute vs. obtuse bisector

The two bisector planes split the dihedral angle into two pairs of vertical angles — one pair acute, one pair obtuse (unless the original planes are perpendicular, in which case both bisectors make 45° angles and the distinction vanishes).

To decide which bisector is the acute-angle bisector, use the following test.

Step 1. Compute the angle \theta between the normals of the two original planes:

Step 2. Pick any point on one of the bisector planes (a convenient one — often you can set two coordinates to zero and solve for the third). Check the angle this bisector makes with one of the original planes.

There is also a more direct algebraic test. After ensuring d_1 and d_2 have the same sign:

• Compute a_1 a_2 + b_1 b_2 + c_1 c_2.
• If this quantity is negative, the positive-sign bisector (the one containing the origin) is the acute angle bisector.
• If this quantity is positive, the positive-sign bisector is the obtuse angle bisector.

The logic behind this rule requires careful reasoning. The dot product of the normals is |\mathbf{n}_1||\mathbf{n}_2|\cos\theta, where \theta is the angle between the normal vectors. The dihedral angle between the planes is either \theta or \pi - \theta, depending on the normal orientations.

After ensuring both d_1 and d_2 are positive, the normals \mathbf{n}_1 = (a_1, b_1, c_1) and \mathbf{n}_2 = (a_2, b_2, c_2) both point toward the side of their respective planes that contains the origin (since the origin makes both plane expressions positive). The origin sits in the region where both normals "agree" — the region bounded by both planes on the same side. When the dot product a_1a_2 + b_1b_2 + c_1c_2 is negative, the normals point in roughly opposite directions, meaning the dihedral angle on the origin's side is acute. So the origin-containing (positive-sign) bisector is the acute bisector. When the dot product is positive, the normals point in roughly the same direction, the dihedral angle on the origin's side is obtuse, and the origin-containing bisector is the obtuse bisector.

Summary of the decision procedure

Here is the full algorithm for finding and classifying bisector planes, laid out for reference.

1. Write both planes in the form a_i x + b_i y + c_i z + d_i = 0.

2. If d_1 and d_2 have opposite signs, multiply one equation by -1 so both constants are positive (or both negative).

3. The two bisector planes are obtained by setting:

   \frac{a_1x + b_1y + c_1z + d_1}{\sqrt{a_1^2 + b_1^2 + c_1^2}} = +\frac{a_2x + b_2y + c_2z + d_2}{\sqrt{a_2^2 + b_2^2 + c_2^2}} \quad \text{(contains origin)}
   \frac{a_1x + b_1y + c_1z + d_1}{\sqrt{a_1^2 + b_1^2 + c_1^2}} = -\frac{a_2x + b_2y + c_2z + d_2}{\sqrt{a_2^2 + b_2^2 + c_2^2}} \quad \text{(does not contain origin)}

4. Compute a_1 a_2 + b_1 b_2 + c_1 c_2. If negative, the origin-bisector is the acute bisector. If positive, it is the obtuse bisector.

Common confusions

• "There is only one bisector plane." There are always two. Just as two intersecting lines in 2D create two pairs of vertical angles and two perpendicular bisectors, two intersecting planes in 3D create two bisector planes. One bisects the acute dihedral angle, the other bisects the obtuse.

• "I can skip the sign adjustment for d_1 and d_2." If you skip it, the positive-sign equation might give you the bisector that does not contain the origin. The sign adjustment is a small step that prevents a wrong classification.

• "The acute bisector is always the one containing the origin." Not necessarily. The origin could be in the acute region or the obtuse region — it depends on where the origin sits relative to the two planes. You must check a_1 a_2 + b_1 b_2 + c_1 c_2 to decide.

• "Bisector planes are perpendicular to both original planes." Not true. They are perpendicular to each other (the two bisectors always meet at right angles), but they make equal angles with the original planes — not right angles.

• "This only works when the two planes are not parallel." Correct instinct. Parallel planes do not intersect, so there is no dihedral angle to bisect. The equidistance idea still works — the "bisector" of two parallel planes is a single plane parallel to both, sitting halfway between them — but it is not really a bisector-plane problem anymore.

Going deeper

Why the two bisector planes are perpendicular

This is a fact worth proving. Let \hat{\mathbf{n}}_1 and \hat{\mathbf{n}}_2 be the unit normals to the two original planes. The bisector planes have normals that are proportional to \hat{\mathbf{n}}_1 + \hat{\mathbf{n}}_2 and \hat{\mathbf{n}}_1 - \hat{\mathbf{n}}_2.

Compute their dot product:

The dot product is zero, so the two normals are perpendicular. Therefore the two bisector planes are perpendicular to each other. This is exactly analogous to the 2D result that the two angle bisectors of intersecting lines are perpendicular.

You can verify this with Example 2 above. The normals of the two bisectors are (3, -1, 0) and (1, 3, -4):

Connection to the 2D angle bisector formula

In 2D, the angle bisectors of two lines a_1 x + b_1 y + c_1 = 0 and a_2 x + b_2 y + c_2 = 0 are given by:

The bisector plane formula is exactly the same equation, with the z-terms added. The principle — equidistance — is identical in any number of dimensions. If you already know the 2D angle bisector formula well, the 3D version requires no new ideas at all.

Bisecting specific dihedral angles

Some textbook problems ask for the bisector of the angle that contains a specific point (not the origin). The method is the same: substitute the point into both signed distance expressions, check whether the signs match, and pick the corresponding bisector. If both signed distances have the same sign at the test point, the positive-sign bisector contains it. If they have opposite signs, the negative-sign bisector does.

This generalises the origin test: the origin is just one particular test point that happens to have coordinates (0, 0, 0), making the signed distances reduce to d_1/\sqrt{\cdots} and d_2/\sqrt{\cdots}.

Bisector of the angle containing a given point — worked through

Suppose you need the bisector of the angle between 2x - y + z + 3 = 0 and x + 2y - 2z + 1 = 0 that contains the point (1, 0, 0).

First, compute the signed distances from (1, 0, 0) to each plane:

Both signed distances are positive — the point is on the same side of both planes. So the bisector containing this point is the positive-sign bisector:

If the signed distances had opposite signs, you would use the negative-sign bisector instead. The method is completely general: compute signs at the test point, match to the bisector.

Locus interpretation

The bisector plane is a locus — the set of all points satisfying a geometric condition, in this case "equidistant from two given planes." This connects bisector planes to the broader idea of loci in 3D geometry. The perpendicular bisector plane of a line segment (the set of points equidistant from two given points) is a different locus that uses the same equidistance principle with a different distance formula.

The equidistance principle appears throughout Indian mathematical traditions. Brahmagupta's work on cyclic quadrilaterals used the idea that a circumcentre is equidistant from all four vertices — the same flavour of equidistance, applied to points rather than planes. The underlying principle is always the same: equal distances produce symmetric objects.

Where this leads next

• Plane: Basic Equations — the foundation: general form, normal form, and intercept form of a plane's equation.
• Distance Formulas for Planes — the perpendicular distance formula that powers the bisector plane derivation.
• Angle and Intersection of Planes — how to find the angle between two planes and the line where they meet.
• Angle Bisectors (2D) — the simpler, two-dimensional version of the same idea, for comparison.
• Plane: More Forms — additional ways to specify a plane, including the three-point form and the parametric form.