Does it matter if I use $\vec{AB}$ or $\vec{BA}$ as the normal vector?

Not at all! The normal vector defines the orientation of the plane. $\vec{BA}$ is just $-\vec{AB}$. The equation will just be multiplied by $-1$, which represents the exact same plane. You will get the same result either way.

JEE Main 2020 - 3 Sep (Evening)

LEVELJEE Main

Animated Solution for Mathematics - Three Dimensional Geometry: The plane which bisects the line joining, the points $(4, - 2, 3)$ and $(2, 4, - 1)$ at right angles also passes through the point:

Q: Wait, why can we just use the vector $\vec{AB}$ as the normal vector?

Because the plane is perpendicular to the line segment $AB$ by definition. Any vector lying on the line $AB$ is perpendicular to the plane, making it a perfect normal vector. It is the most direct way to define the plane's orientation.

Select Answer:

Visualized Solution

Visualizing the Geometry

Given points: $A (4, - 2, 3)$ and $B (2, 4, - 1)$ .
The plane bisects segment $A B$ at right angles.

Conditions for the Plane

1. Passes through the midpoint of $A B$ .
2. Normal vector is parallel to $A B$ .

Setting up the Midpoint

Midpoint formula: $M = (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2}, \frac{z _{1} + z _{2}}{2})$
Substitute coordinates of $A$ and $B$ :
$M = (\frac{4 + 2}{2}, \frac{- 2 + 4}{2}, \frac{3 - 1}{2})$

Calculating Midpoint Coordinates

$x$ -coordinate: $\frac{6}{2} = 3$
$y$ -coordinate: $\frac{2}{2} = 1$
$z$ -coordinate: $\frac{2}{2} = 1$
Midpoint $M = (3, 1, 1)$

Setting up the Normal Vector $n$

Normal vector $n = A B = ⟨ x_{2} - x_{1}, y_{2} - y_{1}, z_{2} - z_{1} ⟩$
Substitute coordinates:
$n = ⟨ 2 - 4, 4 - (- 2), - 1 - 3 ⟩$

Calculating the Normal Vector

$n = ⟨ - 2, 6, - 4 ⟩$

Simplifying the Normal Vector

Direction ratios: $⟨ - 2, 6, - 4 ⟩$
Divide by $- 2$ to simplify:
Simplified D.R.s: $⟨ 1, - 3, 2 ⟩$

Constructing the Plane Equation

Equation of plane: $a (x - x_{1}) + b (y - y_{1}) + c (z - z_{1}) = 0$
Using $M (3, 1, 1)$ and $n = ⟨ 1, - 3, 2 ⟩$ :
$1 (x - 3) - 3 (y - 1) + 2 (z - 1) = 0$

Expanding the Equation

Expand the brackets:
$x - 3 - 3 y + 3 + 2 z - 2 = 0$

Final Plane Equation

Group variables and constants:
$x - 3 y + 2 z + (- 3 + 3 - 2) = 0$
$x - 3 y + 2 z - 2 = 0$
Final Equation: $x - 3 y + 2 z = 2$

Verifying the Options

Check point $(4, 0, - 1)$ in $x - 3 y + 2 z = 2$ :
Substitute $x = 4, y = 0, z = - 1$ into L.H.S.
L.H.S. $= 4 - 3 (0) + 2 (- 1)$

Final Conclusion

L.H.S. $= 4 - 0 - 2 = 2$
L.H.S. $=$ R.H.S.
The point $(4, 0, - 1)$ satisfies the equation.

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The Sigma Insight: Equation of a Plane

Solution Diagram

Analyzing the Setup

To find the perpendicular bisecting plane of the segment connecting points

A (4, - 2, 3)

and

B (2, 4, - 1)

, we require two specific components: a point on the plane and a normal vector.

The plane passes through the midpoint

M

of the segment

A B

. Furthermore, because the plane is perpendicular to the line segment

A B

, the vector

A B

acts as the normal vector

n

for the plane.

The Algebra of the Plane

First, we calculate the midpoint

M

using the formula:

M = (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2}, \frac{z _{1} + z _{2}}{2})

Substituting the coordinates of

A

and

B

M = (\frac{4 + 2}{2}, \frac{- 2 + 4}{2}, \frac{3 - 1}{2}) = (3, 1, 1)

Next, we determine the normal vector

n = A B

n = ⟨ 2 - 4, 4 - (- 2), - 1 - 3 ⟩ = ⟨ - 2, 6, - 4 ⟩

To simplify our calculations, we can scale this vector by dividing by

- 2

, yielding the simplified normal vector

n = ⟨ 1, - 3, 2 ⟩

Constructing the Equation

We use the standard point-normal form of a plane equation,

a (x - x_{0}) + b (y - y_{0}) + c (z - z_{0}) = 0

. Substituting our anchor point

M (3, 1, 1)

and normal vector

⟨ 1, - 3, 2 ⟩

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

Expanding the terms, we obtain:

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

This simplifies to the final plane equation:

x - 3 y + 2 z = 2

The Verification

To verify if a point such as

(4, 0, - 1)

lies on this plane, we substitute the coordinates into the left-hand side of our derived equation:

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

Since the result matches the right-hand side of the equation, we confirm that the point

(4, 0, - 1)

lies on the plane. The geometry of the 3D space is thus satisfied.

Common Traps & FAQs

Wait, why can we just use the vector

A B

as the normal vector?

Because the plane is perpendicular to the line segment

A B

by definition. Any vector lying on the line

A B

is perpendicular to the plane, making it a perfect normal vector. It is the most direct way to define the plane's orientation.

Does it matter if I use

A B

B A

as the normal vector?

Not at all! The normal vector defines the orientation of the plane.

B A

is just

- A B

. The equation will just be multiplied by

- 1

, which represents the exact same plane. You will get the same result either way.

Why did we simplify the normal vector from

⟨ - 2, 6, - 4 ⟩

⟨ 1, - 3, 2 ⟩

It is a mathematical shortcut! The direction ratios only need to be proportional. Using smaller integers makes the arithmetic cleaner and significantly reduces the chance of making a silly calculation error. It is always good practice to simplify your vectors.

Animated Solution for Mathematics - Three Dimensional Geometry: The plane which bisects the line joining, the points $(4, - 2, 3)$ and $(2, 4, - 1)$ at right angles also passes through the point:

Select Answer:

Visualized Solution

Visualizing the Geometry

Conditions for the Plane

Setting up the Midpoint

Calculating Midpoint Coordinates

Setting up the Normal Vector $n$

Calculating the Normal Vector

Simplifying the Normal Vector

Constructing the Plane Equation

Expanding the Equation

Final Plane Equation

Verifying the Options

Final Conclusion

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Algebra of the Plane

Constructing the Equation

The Verification

Similar Questions

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 :-

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 ?

The plane passing through the point (4, -1, 2) and parallel to the lines $\frac{x + 2}{3} = \frac{y - 2}{- 1} = \frac{z + 1}{2}$ and $\frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 4}{3}$ also passes through the point :

The plane passing through the points $(1, 2, 1)$ , $(2, 1, 2)$ and parallel to the line, $2 x = 3 y, z = 1$ also passes through the point:

The plane through the intersection of the planes $x + y + z = 1$ and $2 x + 3 y - z + 4 = 0$ and parallel to y-axis also passes through the point :

Let $P$ be the plane, passing through the point $(1, - 1, - 5)$ and perpendicular to the line joining the points $(4, 1, - 3)$ and $(2, 4, 3)$ . Then the distance of $P$ from the point $(3, - 2, 2)$ is

A plane passing through the points $(0, - 1, 0)$ and $(0, 0, 1)$ and making an angle $π /4$ with the plane $y - z + 5 = 0$ , also passes through the point

A plane which is perpendicular to two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ , passes through $(1, - 2, 1)$ . The distance of the plane from the point $(1, 2, 2)$ is

A plane which passes through the point $(3, 2, 0)$ and the line $\frac{x - 4}{1} = \frac{y - 7}{5} = \frac{z - 4}{4}$ is

Let the plane passing through the point $(- 1, 0, - 2)$ and perpendicular to each of the planes $2 x + y - z = 2$ and $x - y - z = 3$ be $a x + b y + cz + 8 = 0$ . Then the value of $a + b + c$ is equal to:

Select Answer:

Visualized Solution

Visualizing the Geometry

Conditions for the Plane

Setting up the Midpoint

Calculating Midpoint Coordinates

Setting up the Normal Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Calculating the Normal Vector

Simplifying the Normal Vector

Constructing the Plane Equation

Expanding the Equation

Final Plane Equation

Verifying the Options

Final Conclusion

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Algebra of the Plane

Constructing the Equation

The Verification

Similar Questions

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 :-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 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 ?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The plane passing through the point (4, -1, 2) and parallel to the lines 3x+2​=−1y−2​=2z+1​ and 1x−2​=2y−3​=3z−4​ also passes through the point :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The plane passing through the points (1,2,1), (2,1,2) and parallel to the line, 2x=3y,z=1 also passes through the point:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The plane through the intersection of the planes x+y+z=1 and 2x+3y−z+4=0 and parallel to y-axis also passes through the point :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A plane passing through the points (0,−1,0) and (0,0,1) and making an angle π/4 with the plane y−z+5=0, also passes through the point

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A plane which is perpendicular to two planes 2x−2y+z=0 and x−y+2z=4, passes through (1,−2,1). The distance of the plane from the point (1,2,2) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A plane which passes through the point (3,2,0) and the line 1x−4​=5y−7​=4z−4​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the plane passing through the point (−1,0,−2) and perpendicular to each of the planes 2x+y−z=2 and x−y−z=3 be ax+by+cz+8=0. Then the value of a+b+c is equal to:

Setting up the Normal Vector $n$

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 ?

The plane passing through the point (4, -1, 2) and parallel to the lines $\frac{x + 2}{3} = \frac{y - 2}{- 1} = \frac{z + 1}{2}$ and $\frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 4}{3}$ also passes through the point :

The plane passing through the points $(1, 2, 1)$ , $(2, 1, 2)$ and parallel to the line, $2 x = 3 y, z = 1$ also passes through the point:

The plane through the intersection of the planes $x + y + z = 1$ and $2 x + 3 y - z + 4 = 0$ and parallel to y-axis also passes through the point :

A plane passing through the points $(0, - 1, 0)$ and $(0, 0, 1)$ and making an angle $π /4$ with the plane $y - z + 5 = 0$ , also passes through the point

A plane which is perpendicular to two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ , passes through $(1, - 2, 1)$ . The distance of the plane from the point $(1, 2, 2)$ is

A plane which passes through the point $(3, 2, 0)$ and the line $\frac{x - 4}{1} = \frac{y - 7}{5} = \frac{z - 4}{4}$ is

Let the plane passing through the point $(- 1, 0, - 2)$ and perpendicular to each of the planes $2 x + y - z = 2$ and $x - y - z = 3$ be $a x + b y + cz + 8 = 0$ . Then the value of $a + b + c$ is equal to: