The plane which bisects the line segment joining the points (-3, -3, 4) and (3, 7, 6) at right angle

Q: Wait, why do we use the vector $\vec{AB}$ as the normal vector? Shouldn't it be something else?

That is a fantastic question! In 3D geometry, the 'normal' is simply any vector that is perpendicular to the plane. Since the problem explicitly states that our plane bisects the segment $AB$ at 'right angles,' the segment $AB$ itself is perpendicular to the plane. Therefore, the vector $\vec{AB}$ is the most natural and direct choice for our normal vector. It is the 'spine' of the plane, if you will.

Q: Why did we divide the direction ratios by 2? Does that change the plane?

Not at all! The direction ratios $(a, b, c)$ of a normal vector only define the *orientation* of the plane, not its position. Any vector proportional to $(a, b, c)$, such as $(ka, kb, kc)$, will represent the same orientation. Dividing by 2 just simplifies our arithmetic, making the final equation cleaner and less prone to calculation errors. It is a professional shortcut, not a mathematical change.

Q: Can I just use the distance formula to check the options instead of finding the plane equation?

You absolutely could! You could calculate the distance from each option point to $A$ and to $B$ and see if they are equal. However, that would require calculating the distance formula six times (twice for each of the three options). Finding the equation of the plane is a one-time investment that makes checking the options trivial. In JEE, efficiency is just as important as accuracy!

Question

The plane which bisects the line segment joining the points $(-3, -3, 4)$ and $(3, 7, 6)$ at right angles, passes through which one of the following points ?

Options:

(A) (4, -1, 7)

(B) (4, 1, -2)

(C) (-2, 3, 5)

(D) (2, 1, 3)

Sigma Percentile · Accepted Answer

### The Geometry of Balance: Mastering the Perpendicular Bisector Plane

Welcome, future engineers. Today, we are not just solving a coordinate geometry problem; we are exploring the concept of symmetry in three-dimensional space.

When we talk about a plane that bisects a line segment at right angles, we are talking about a boundary of perfect equilibrium. Imagine a line segment $AB$ floating in the vastness of 3D space. Our goal is to find the 'mirror' that sits exactly in the middle, slicing through it with absolute precision.

### Phase 1: Finding the Anchor Point

Every plane needs an anchor—a point that we know for certain lies on its surface. The problem gives us the points $A(-3, -3, 4)$ and $B(3, 7, 6)$.

Because our plane is a **bisector**, it must pass through the midpoint of $AB$. Think of this as the center of gravity of our segment.

To find this midpoint $M$, we use the arithmetic mean of the coordinates:

$$M = \left( \frac{-3+3}{2}, \frac{-3+7}{2}, \frac{4+6}{2} \right)$$

Calculating this, we get $M(0, 2, 5)$. This point is our anchor and the heart of our plane. If we know the orientation of the plane, this point will allow us to lock it into its correct position in space.

### Phase 2: Defining the Orientation

Now, how do we define the 'tilt' or orientation of this plane? The problem tells us the plane is perpendicular to the segment $AB$.

In the language of vectors, this means the vector $\vec{AB}$ is the normal vector $\vec{n}$ to our plane. Let us calculate the components of $\vec{AB}$ by subtracting the coordinates of $A$ from $B$:

$$\vec{n} = (3 - (-3))\hat{i} + (7 - (-3))\hat{j} + (6 - 4)\hat{k}$$

$$\vec{n} = 6\hat{i} + 10\hat{j} + 2\hat{k}$$

Here is a pro-tip for your JEE preparation: whenever you have a normal vector, you are only interested in its **direction ratios**. We can scale this vector by any non-zero constant without changing the plane's orientation.

Dividing by $2$, we get the simplified direction ratios $(3, 5, 1)$. This makes our algebra significantly lighter and much more elegant.

### Phase 3: Synthesizing the Equation

We have our anchor point $M(0, 2, 5)$ and our normal vector $\vec{n}(3, 5, 1)$. The general equation of a plane passing through $(x_0, y_0, z_0)$ with normal $(a, b, c)$ is given by:

$$a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$$

Substituting our values, we get:

$$3(x - 0) + 5(y - 2) + 1(z - 5) = 0$$

Now, let us expand this carefully. There is no need to rush:

$$3x + 5y - 10 + z - 5 = 0$$

Combining the constants, we arrive at the final, beautiful equation of our plane:

$$3x + 5y + z = 15$$

### Phase 4: The Final Verification

We have constructed the plane. Now, we must identify which of the given points lies on it. This is the moment of truth.

We take the coordinates of the options and test them against our equation. Let us test the point $(4, 1, -2)$:

$$3(4) + 5(1) + (-2) = 12 + 5 - 2 = 15$$

Since $15 = 15$, the point satisfies the equation perfectly. We have found our answer.

Remember, in JEE Advanced, the math is rarely just about calculation; it is about visualizing the physical reality behind the numbers. When you see a perpendicular bisector, see the symmetry. When you see a normal vector, see the orientation. Keep practicing, keep visualizing, and the geometry will start to speak to you.

Animated Solution for Mathematics - Three Dimensional Geometry: The plane which bisects the line segment joining the points $(- 3, - 3, 4)$ and $(3, 7, 6)$ at right angles, passes through which one of the following points ?

Select Answer:

Visualized Solution

Visualizing the Given Points

The Perpendicular Bisector Plane

Condition 1: The Midpoint

Substituting into Midpoint Formula

Calculating the Midpoint $M$

Condition 2: The Normal Vector

Setting up the Normal Vector $n$

Calculating Direction Ratios

The Plane Equation Formula

Substituting into the Plane Equation

Expanding the Equation

Final Equation of the Plane

Testing the Options

Verifying Point $(4, 1, - 2)$

Final Conclusion

The Sigma Insight: Equation of a Plane

The Geometry of Balance

Phase 1

Phase 2

Phase 3

Phase 4

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