Let the acute angle bisector of the two planes x - 2y - 2z + 1 = 0 and 2x - 3y - 6z + 1 = 0 be the p

Q: Wait, why do we have a $\pm$ sign in the bisector equation?

Great question! When two planes intersect, they create four regions. The angle bisectors divide these regions. There are always two bisector planes: one for the acute angle and one for the obtuse angle. The $\pm$ sign allows us to represent both possibilities algebraically.

Q: Does the order of the planes matter when checking the sign?

Yes, it is absolutely critical. You must first ensure that the constant terms $D_1$ and $D_2$ in your plane equations are positive. If they are not, multiply the entire equation by $-1$ before proceeding. Only then does the dot product rule $A_1A_2 + B_1B_2 + C_1C_2$ correctly identify the acute bisector.

Q: Why can't we just use energy conservation to solve this?

Energy conservation is a powerful tool in physics, but this is a pure 3D geometry problem. We rely on the geometric definition of an angle bisector as the locus of points equidistant from both planes. Stick to the distance formula; it is the most reliable path to the solution.

Question

Let the acute angle bisector of the two planes $x - 2y - 2z + 1 = 0$ and $2x - 3y - 6z + 1 = 0$ be the plane $P$. Then which of the following points lies on $P$?

Options:

(A) $(3, 1, -\frac{1}{2})$

(B) $(-2, 0, -\frac{1}{2})$

(C) $(0, 2, -4)$

(D) $(4, 0, -2)$

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of Intersection

Imagine standing in a room where two walls meet at an angle. You are looking for the plane that perfectly slices through the middle of that angle. This is the essence of our problem.

We have two planes defined by:
$$P_1: x - 2y - 2z + 1 = 0$$
$$P_2: 2x - 3y - 6z + 1 = 0$$

Our goal is to find the equation of the plane $P$ that bisects the acute angle between them. This is not just about plugging numbers into a formula; it is about understanding the symmetry of space.

### The Distance Equality

The fundamental property of any angle bisector is that every point on it is equidistant from the two intersecting planes. If we take an arbitrary point $(x, y, z)$ on the bisector, its perpendicular distance to $P_1$ must equal its perpendicular distance to $P_2$.

The distance from a point to a plane $Ax + By + Cz + D = 0$ is given by:
$$\frac{|Ax + By + Cz + D|}{\sqrt{A^2 + B^2 + C^2}}$$

By equating these distances, we get:
$$\frac{x - 2y - 2z + 1}{\sqrt{1^2 + (-2)^2 + (-2)^2}} = \pm \frac{2x - 3y - 6z + 1}{\sqrt{2^2 + (-3)^2 + (-6)^2}}$$

Calculating the denominators, we find $\sqrt{1 + 4 + 4} = 3$ and $\sqrt{4 + 9 + 36} = 7$. This simplifies our equation to:
$$\frac{x - 2y - 2z + 1}{3} = \pm \frac{2x - 3y - 6z + 1}{7}$$

### The Acute vs. Obtuse Mystery

Now, we face the $\pm$ sign. To determine which one gives us the acute bisector, we use the dot product of the normal vectors $\vec{n_1} = (1, -2, -2)$ and $\vec{n_2} = (2, -3, -6)$.

We calculate the expression $A_1A_2 + B_1B_2 + C_1C_2$:
$$A_1A_2 + B_1B_2 + C_1C_2 = (1)(2) + (-2)(-3) + (-2)(-6) = 2 + 6 + 12 = 20$$

Since $20 > 0$ and our constants $D_1, D_2$ are both positive, the rule states that the negative sign corresponds to the acute bisector. We discard the positive sign and proceed with the negative one.

### The Algebraic Resolution

We now solve the following equation:
$$\frac{x - 2y - 2z + 1}{3} = - \frac{2x - 3y - 6z + 1}{7}$$

Cross-multiplying gives:
$$7(x - 2y - 2z + 1) = -3(2x - 3y - 6z + 1)$$

Expanding this, we get:
$$7x - 14y - 14z + 7 = -6x + 9y + 18z - 3$$

Rearranging all terms to one side, we arrive at the final equation of our plane $P$:
**$$13x - 23y - 32z + 10 = 0$$**

### Verification

Finally, we test the point $(-2, 0, -\frac{1}{2})$. Substituting these values into our equation:
$$13(-2) - 23(0) - 32\left(-\frac{1}{2}\right) + 10 = -26 - 0 + 16 + 10 = 0$$

The point satisfies the equation, confirming it lies on the plane. You have navigated the geometry, mastered the sign rule, and verified the result.

Animated Solution for Mathematics - Three Dimensional Geometry: Let the acute angle bisector of the two planes $x - 2 y - 2 z + 1 = 0$ and $2 x - 3 y - 6 z + 1 = 0$ be the plane $P$ . Then which of the following points lies on $P$ ?

Select Answer:

Visualized Solution

Visualizing the Intersecting Planes

The Angle Bisector Formula

Calculating the Denominators

Setting up the General Equation

The Acute vs Obtuse Rule

Evaluating the Condition

Selecting the Correct Sign

Cross-Multiplication Setup

Expanding the Equation

The Final Equation of Plane P

Testing the Options

Final Verification

Conclusion and Key Takeaway

The Sigma Insight: Angle Between Two Planes

Analyzing the Geometry of Intersection

The Distance Equality

The Acute vs

The Algebraic Resolution

Verification

Similar Questions

A plane which bisects the angle between the two given planes $2 x - y + 2 z - 4 = 0$ and $x + 2 y + 2 z - 2 = 0$ , passes through the point:

Let $P_{1} : 2 x + y - z = 3$ and $P_{2} : x + 2 y + z = 2$ be two planes. Then, which of the following statement(s) is (are) TRUE ?

Let the points on the plane $P$ be equidistant from the points $(- 4, 2, 1)$ and $(2, - 2, 3)$ . Then the acute angle between the plane $P$ and the plane $2 x + y + 3 z = 1$ is

The acute angle between the planes $P_{1}$ and $P_{2}$ , when $P_{1}$ and $P_{2}$ are the planes passing through the intersection of the planes $5 x + 8 y + 13 z - 29 = 0$ and $8 x - 7 y + z - 20 = 0$ and the points $(2, 1, 3)$ and $(0, 1, 2)$ , respectively, is

The d.r. of normal to the plane through $(1, 0, 0), (0, 1, 0)$ which makes an angle $π /4$ with plane $x + y = 3$ are

A tetrahedron has vertices $P (1, 2, 1)$ , $Q (2, 1, 3)$ , $R (- 1, 1, 2)$ and $O (0, 0, 0)$ . The angle between the faces $O P Q$ and $P QR$ is :

A tetrahedron has vertices at $O (0, 0, 0), A (1, 2, 1), B (2, 1, 3)$ and $C (- 1, 1, 2)$ . Then the angle between the faces OAB and ABC will be