A plane bisects the line segment joining the points (1, 2, 3) and (-3, 4, 5) at right angles. Then t

Q: Wait, why is the vector $\vec{AB}$ the normal vector of the plane? I thought the normal vector had to be perpendicular to the plane, not the line segment.

That is a fantastic question! Remember the definition of a perpendicular bisector. If a line is perpendicular to a plane, then the line itself is parallel to the normal vector of that plane. Since our segment $AB$ is perpendicular to the plane, the vector $\vec{AB}$ acts as the direction of the normal vector. You are essentially using the line segment to define the 'tilt' or orientation of the plane.

Q: Can I use the distance formula to solve this? Like, finding the locus of points equidistant from $A$ and $B$?

You absolutely can! The locus of points equidistant from two points $A$ and $B$ is indeed the perpendicular bisector plane. You would set $PA^2 = PB^2$, where $P$ is $(x, y, z)$. While this is a mathematically sound approach, it involves squaring terms and expanding quadratics, which is prone to calculation errors. The point-normal form is much more elegant and faster for JEE exams.

Q: Does it matter if I use $\vec{AB}$ or $\vec{BA}$ as my normal vector? Will it change the final equation?

It will not change the final result! If you use $\vec{BA}$, your direction ratios will be the negative of those from $\vec{AB}$. This will result in an equation where every term has the opposite sign (e.g., $-2x + y + z - 9 = 0$ vs $2x - y - z + 9 = 0$). Since both equations represent the same plane, they will yield the same result when you test the points. Just be consistent with your signs!

Question

A plane bisects the line segment joining the points (1, 2, 3) and (-3, 4, 5) at right angles. Then this plane also passes through the point :-

Options:

(A) (1, 2, -3)

(B) (-1, 2, 3)

(C) (-3, 2, 1)

(D) (3, 2, 1)

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of Symmetry

Imagine you are standing in a vast 3D coordinate space with two points, $A(1, 2, 3)$ and $B(-3, 4, 5)$. You are tasked with finding the equation of the plane that acts as the perpendicular bisector of the segment $AB$.

In the context of JEE Advanced, this problem is a study of symmetry. We must visualize the geometric soul of the problem to ensure our algebraic steps remain grounded in spatial logic.

### Phase 1: Finding the Heart of the Segment

Every plane requires an anchor point. Since our plane is a bisector, it must pass through the midpoint $M$ of segment $AB$.

We calculate this by averaging the coordinates:

$$M = \left( \frac{1-3}{2}, \frac{2+4}{2}, \frac{3+5}{2} \right)$$

Performing the arithmetic, we find the anchor point:

$$M = (-1, 3, 4)$$

Any equation we derive for this plane must satisfy these coordinates.

### Phase 2: The Orientation of the Plane

To define the tilt of the plane, we identify its normal vector $\vec{n}$. Because the plane is perpendicular to the segment $AB$, the vector $\vec{AB}$ serves as the normal vector.

We calculate $\vec{AB}$ by subtracting the position vector of $A$ from $B$:

$$\vec{n} = \vec{AB} = (-3-1)\hat{i} + (4-2)\hat{j} + (5-3)\hat{k} = -4\hat{i} + 2\hat{j} + 2\hat{k}$$

A crucial exam strategy is to recognize that direction ratios are proportional. We can simplify the vector $(-4, 2, 2)$ by dividing by $2$ to obtain the reduced normal vector:

$$\vec{n} = (-2, 1, 1)$$

Using this simplified vector makes our subsequent algebra cleaner and significantly reduces the probability of arithmetic errors.

### Phase 3: Constructing the Equation

We now possess the point $M(-1, 3, 4)$ and the normal vector $\vec{n} = (-2, 1, 1)$. We apply the point-normal form of a plane:

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

Substituting our specific values into the formula:

$$-2(x - (-1)) + 1(y - 3) + 1(z - 4) = 0$$

Expanding this expression, we obtain:

$$-2(x + 1) + (y - 3) + (z - 4) = 0$$

$$-2x - 2 + y - 3 + z - 4 = 0$$

Grouping the constants, we arrive at $-2x + y + z - 9 = 0$. Multiplying by $-1$ yields the standard form:

**$$2x - y - z + 9 = 0$$**

### Phase 4: The Final Verification

To verify our result, we test a point such as $P(-3, 2, 1)$ to see if it lies on the plane. Substituting these coordinates into our derived equation:

$$2(-3) - (2) - (1) + 9 = -6 - 2 - 1 + 9 = 0$$

Since the equation holds true, we have successfully navigated the geometry. Remember, in JEE, mathematics is the tool, but visualization is the key to mastery.

Animated Solution for Mathematics - Three Dimensional Geometry: A plane bisects the line segment joining the points (1, 2, 3) and (-3, 4, 5) at right angles. Then this plane also passes through the point :-

Select Answer:

Visualized Solution

Visualizing the Segment $A B$

The Concept of a Bisecting Plane

Calculating the Midpoint $M$

Midpoint Result: $M (- 1, 3, 4)$

Identifying the Normal Vector

Calculating Vector $A B$

Simplifying the Normal Vector

The Point-Normal Equation Form

Substituting Values into the Equation

Expanding the Plane Equation

Final Equation of the Plane

Testing the Given Options

Verifying Point $P (- 3, 2, 1)$

Conclusion

The Sigma Insight: Equation of a Plane

Analyzing the Geometry of Symmetry

Phase 1

Phase 2

Phase 3

Phase 4

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