A plane which bisects the angle between the two given planes 2x - y + 2z - 4 = 0 and x + 2y + 2z - 2

Q: Wait, why do we have two bisector planes? Shouldn't there be just one?

That is a fantastic question! Think about two lines intersecting in 2D; they create two pairs of vertically opposite angles, and each pair has its own bisector. Similarly, in 3D, two intersecting planes divide space into four regions. The angle bisector planes are the two planes that bisect these regions. So, yes, there are always two!

Q: Can I just ignore the absolute value bars and solve it as one equation?

If you ignore the absolute value, you are only finding one of the two bisectors. The absolute value is essential because it accounts for the two sides of the plane. By setting the expressions equal with a $\pm$ sign, you are mathematically capturing both possible bisector planes. Always remember to consider both cases!

Q: What if the denominators of the distance formulas were not equal? Would the method change?

The core logic remains identical! You would still equate the two distance expressions. If the denominators were different, you would simply cross-multiply them to clear the fractions before solving for the two cases. The beauty of this specific problem is that the denominators cancel out, but the underlying principle of equidistance is universal.

Question

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

Options:

(A) (2, 4, 1)

(B) (2, -4, 1)

(C) (1, 4, -1)

(D) (1, -4, 1)

Sigma Percentile · Accepted Answer

### The Geometry of Intersection

Imagine you are standing in a vast, empty room. Suddenly, two massive, infinite sheets of glass appear, slicing through the space and intersecting along a single, sharp line.

This is the visual reality of our problem. We are given two planes, $P_1: 2x - y + 2z - 4 = 0$ and $P_2: x + 2y + 2z - 2 = 0$.

Our mission is to find a point that lies on the plane which bisects the angle between these two.

### The Locus of Equidistance

What defines an angle bisector plane? It is not just a random slice; it is a special locus.

Geometrically, any point $P(x, y, z)$ lying on this bisector plane must be at an equal perpendicular distance from both $P_1$ and $P_2$.

If we denote the distance to the first plane as $d_1$ and the distance to the second as $d_2$, our fundamental condition is simply $d_1 = d_2$. This is the heartbeat of the entire solution.

### The Arsenal: The Distance Formula

To translate this geometric intuition into algebra, we reach for our most reliable tool: the perpendicular distance formula from a point to a plane.

For any plane $ax + by + cz + d = 0$, the distance $d$ from a point $(x, y, z)$ is given by:

$$d = \frac{|ax + by + cz + d|}{\sqrt{a^2 + b^2 + c^2}}$$

Applying this to our two planes, we get:

$$d_1 = \frac{|2x - y + 2z - 4|}{\sqrt{2^2 + (-1)^2 + 2^2}}$$

$$d_2 = \frac{|x + 2y + 2z - 2|}{\sqrt{1^2 + 2^2 + 2^2}}$$

### The Moment of Mathematical Elegance

Now, look closely at the denominators. For $P_1$, the sum of squares is $2^2 + (-1)^2 + 2^2 = 4 + 1 + 4 = 9$, and $\sqrt{9} = 3$.

For $P_2$, it is $1^2 + 2^2 + 2^2 = 1 + 4 + 4 = 9$, and $\sqrt{9} = 3$. The denominators are identical!

They cancel out perfectly, leaving us with a much cleaner equation:

$$|2x - y + 2z - 4| = |x + 2y + 2z - 2|$$

### Unlocking the Two Worlds

Removing the absolute value bars is where we branch into two distinct possibilities. We must consider both the positive and negative cases.

**Case 1 (Positive):** $2x - y + 2z - 4 = +(x + 2y + 2z - 2)$

Rearranging the terms, we get $2x - x - y - 2y + 2z - 2z - 4 + 2 = 0$, which simplifies to $x - 3y - 2 = 0$. This is our first bisector plane, $B_1$.

**Case 2 (Negative):** $2x - y + 2z - 4 = -(x + 2y + 2z - 2)$

Expanding the right side gives $2x - y + 2z - 4 = -x - 2y - 2z + 2$. Bringing everything to the left, we get $3x + y + 4z - 6 = 0$. This is our second bisector plane, $B_2$.

### The Final Verification

We have two candidate planes. The question asks which of the given points lies on the bisector. We test the options against these equations.

Let us check the point $(2, -4, 1)$ against $B_2: 3x + y + 4z - 6 = 0$:

$$3(2) + (-4) + 4(1) - 6 = 6 - 4 + 4 - 6 = 0$$

Since the result is exactly $0$, the point $(2, -4, 1)$ satisfies the equation. We have successfully navigated the geometry and the algebra to find our answer.

Remember, in JEE Advanced, the beauty lies not just in the final result, but in the logical steps that lead you there. Keep practicing, and keep visualizing!

Animated Solution for Mathematics - Three Dimensional Geometry: 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:

Select Answer:

Visualized Solution

Given Planes

Concept of Angle Bisector

Equating Perpendicular Distances

Simplifying the Denominators

The Absolute Value Equation

Case 1: The Positive Bisector

Case 2: The Negative Bisector

Testing the Options

Substituting the Point

Evaluating the Expression

Conclusion

The Sigma Insight: Angle Between Two Planes

The Geometry of Intersection

The Locus of Equidistance

The Arsenal

The Moment of Mathematical Elegance

Unlocking the Two Worlds

The Final Verification

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