JEE Main 2005

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Animated Solution for Mathematics - Three Dimensional Geometry: If the plane $2 a x - 3 a y + 4 a z + 6 = 0$ passes through the midpoint of the line joining the centres of the spheres $x^{2} + y^{2} + z^{2} + 6 x - 8 y - 2 z = 13$ and $x^{2} + y^{2} + z^{2} - 10 x + 4 y - 2 z = 8$ then $a$ equals

Q: Why do we ignore the constant term $d$ when finding the center?

The constant term $d$ in the sphere equation $x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0$ affects the radius of the sphere, not its position. The center is determined solely by the linear coefficients of $x, y,$ and $z$. Since we only need the center to find the midpoint, the constant term is irrelevant for this specific calculation.

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Visualized Solution

Visualizing the Spheres and the Plane

We are given two spheres, let's call them $S_{1}$ and $S_{2}$ .
A plane passes through the midpoint of the line segment joining their centers.
Our goal is to find the value of the parameter $a$ .

Finding the Center of Sphere $S_{1}$

Equation of $S_{1} : x^{2} + y^{2} + z^{2} + 6 x - 8 y - 2 z - 13 = 0$
Compare with general form: $x^{2} + y^{2} + z^{2} + 2 ux + 2 v y + 2 w z + d = 0$
Here, $2 u = 6 ⟹ u = 3$ , $2 v = - 8 ⟹ v = - 4$ , $2 w = - 2 ⟹ w = - 1$
Center $C_{1} = (- u, - v, - w) = (- 3, 4, 1)$

Finding the Center of Sphere $S_{2}$

Equation of $S_{2} : x^{2} + y^{2} + z^{2} - 10 x + 4 y - 2 z - 8 = 0$
Compare with general form: $2 u = - 10 ⟹ u = - 5$ , $2 v = 4 ⟹ v = 2$ , $2 w = - 2 ⟹ w = - 1$
Center $C_{2} = (- u, - v, - w) = (5, - 2, 1)$

The Line Segment and Midpoint Formula

We have two points: $C_{1} (- 3, 4, 1)$ and $C_{2} (5, - 2, 1)$
The midpoint $M (x_{m}, y_{m}, z_{m})$ of the line joining $(x_{1}, y_{1}, z_{1})$ and $(x_{2}, y_{2}, z_{2})$ is:
$M = (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2}, \frac{z _{1} + z _{2}}{2})$

Calculating Coordinates of Midpoint $M$

Substitute $C_{1} (- 3, 4, 1)$ and $C_{2} (5, - 2, 1)$ :
$x_{m} = \frac{- 3 + 5}{2} = \frac{2}{2} = 1$
$y_{m} = \frac{4 + ( - 2 )}{2} = \frac{2}{2} = 1$
$z_{m} = \frac{1 + 1}{2} = \frac{2}{2} = 1$
Midpoint $M = (1, 1, 1)$

The Plane Passing Through Midpoint $M$

Equation of the plane: $2 a x - 3 a y + 4 a z + 6 = 0$
Since the plane passes through $M (1, 1, 1)$ , this point must satisfy the plane's equation.

Substituting $M (1, 1, 1)$ into the Plane Equation

Substitute $x = 1, y = 1, z = 1$ :
$2 a (1) - 3 a (1) + 4 a (1) + 6 = 0$
This simplifies to: $2 a - 3 a + 4 a + 6 = 0$

Solving the Linear Equation for $a$

Combine the terms of $a$ :
$(2 a - 3 a + 4 a) + 6 = 0 ⟹ 3 a + 6 = 0$
Subtract $6$ from both sides: $3 a = - 6$
Divide by $3$ : $a = - 2$

Summary and Final Answer

The centers of the spheres are $C_{1} (- 3, 4, 1)$ and $C_{2} (5, - 2, 1)$ .
The midpoint of $C_{1} C_{2}$ is $M (1, 1, 1)$ .
Substituting $M$ into the plane equation yields $a = - 2$ .
Therefore, the correct option is $- 2$ .

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The Sigma Insight: Equation of a Plane

Solution Diagram

Analyzing the Setup

The general equation of a sphere is given by

x^{2} + y^{2} + z^{2} + 2 ux + 2 v y + 2 w z + d = 0

. By comparing this to the given equations for spheres

S_{1}

and

S_{2}

, we extract the coordinates of their centers.

For sphere

S_{1}

, we identify the center as

C_{1} = (- 3, 4, 1)

. For sphere

S_{2}

, we identify the center as

C_{2} = (5, - 2, 1)

Finding the Midpoint

We now consider the line segment connecting

C_{1}

and

C_{2}

. To find the midpoint

M

of this segment, we use the standard midpoint formula:

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

Applying this to our coordinates, we calculate the components:

x_{m} = \frac{- 3 + 5}{2} = 1

y_{m} = \frac{4 - 2}{2} = 1

z_{m} = \frac{1 + 1}{2} = 1

Thus, the midpoint

M

(1, 1, 1)

Solving for the Parameter

The problem states that the plane

2 a x - 3 a y + 4 a z + 6 = 0

passes through the point

M

. This implies that the coordinates of

M

must satisfy the equation of the plane.

Substituting

x = 1

y = 1

, and

z = 1

into the plane equation, we obtain:

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

Simplifying the expression, we get:

2 a - 3 a + 4 a + 6 = 0

3 a + 6 = 0

Solving for

a

, we find

3 a = - 6

, which leads to the final result:

$a = - 2$

Common Traps & FAQs

Why do we ignore the constant term

d

when finding the center?

The constant term

d

in the sphere equation

x^{2} + y^{2} + z^{2} + 2 ux + 2 v y + 2 w z + d = 0

affects the radius of the sphere, not its position. The center is determined solely by the linear coefficients of

x, y,

and

z

. Since we only need the center to find the midpoint, the constant term is irrelevant for this specific calculation.

Is there a faster way to find the center than comparing coefficients?

You could complete the square for each variable, but comparing coefficients with the general form

x^{2} + y^{2} + z^{2} + 2 ux + 2 v y + 2 w z + d = 0

is generally faster and less prone to arithmetic errors. It is the standard, most reliable method for JEE problems.

What if the plane equation had a constant on the other side?

Always bring all terms to one side to form an equation of the type

f (x, y, z) = 0

before substituting the point. This ensures you do not make a sign error when evaluating the constant term.

Animated Solution for Mathematics - Three Dimensional Geometry: If the plane $2 a x - 3 a y + 4 a z + 6 = 0$ passes through the midpoint of the line joining the centres of the spheres $x^{2} + y^{2} + z^{2} + 6 x - 8 y - 2 z = 13$ and $x^{2} + y^{2} + z^{2} - 10 x + 4 y - 2 z = 8$ then $a$ equals

Select Answer:

Visualized Solution

Visualizing the Spheres and the Plane

Finding the Center of Sphere $S_{1}$

Finding the Center of Sphere $S_{2}$

The Line Segment and Midpoint Formula

Calculating Coordinates of Midpoint $M$

The Plane Passing Through Midpoint $M$

Substituting $M (1, 1, 1)$ into the Plane Equation

Solving the Linear Equation for $a$

Summary and Final Answer

The Sigma Insight: Equation of a Plane

Analyzing the Setup

Finding the Midpoint

Solving for the Parameter

Similar Questions

The intersection of the spheres $x^{2} + y^{2} + z^{2} + 7 x - 2 y - z = 13$ and $x^{2} + y^{2} + z^{2} - 3 x + 3 y + 4 z = 8$ is the same as the intersection of one of the sphere and the plane

If the equation of the plane passing through the line of intersection of the planes $2 x - y + z = 3, 4 x - 3 y + 5 z + 9 = 0$ and parallel to the line $\frac{x + 1}{- 2} = \frac{y + 3}{4} = \frac{z - 2}{5}$ is $a x + b y + cz + 6 = 0$ , then $a + b + c$ is equal to

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:

The plane $x + 2 y - z = 4$ cuts the sphere $x^{2} + y^{2} + z^{2} - x + z - 2 = 0$ in a circle of radius

The shortest distance from the plane $12 x + 4 y + 3 z = 327$ to the sphere $x^{2} + y^{2} + z^{2} + 4 x - 2 y - 6 z = 155$ is

If the equation of a plane $P$ , passing through the intersection of the planes $x + 4 y - z + 7 = 0$ and $3 x + y + 5 z = 8$ is $a x + b y + 6 z = 15$ for some $a, b \in R$ , then the distance of the point $(3, 2, - 1)$ from the plane $P$ is

(i) Find the equation of the plane passing through the points $(2, 1, 0), (5, 0, 1)$ and $(4, 1, 1)$ . (ii) If $P$ is the point $(2, 1, 6)$ then find the point $Q$ such that $P Q$ is perpendicular to the plane in (i) and the midpoint of $P Q$ lies on it.

Equation of the plane containing the straight line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$ is

If the distance between the plane, $23 x - 10 y - 2 z + 48 = 0$ and the plane containing the lines $\frac{x + 1}{2} = \frac{y - 3}{4} = \frac{z + 1}{3}$ and $\frac{x + 3}{2} = \frac{y + 2}{6} = \frac{z - 1}{λ}$ ( $λ \in R$ ) is equal to $\frac{k}{633}$ , then $k$ equal to

Select Answer:

Visualized Solution

Visualizing the Spheres and the Plane

Finding the Center of Sphere .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1​

Finding the Center of Sphere .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S2​

The Line Segment and Midpoint Formula

Calculating Coordinates of Midpoint .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M

The Plane Passing Through Midpoint .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M(1,1,1) into the Plane Equation

Solving the Linear Equation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Summary and Final Answer

The Sigma Insight: Equation of a Plane

Analyzing the Setup

Finding the Midpoint

Solving for the Parameter

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The shortest distance from the plane 12x+4y+3z=327 to the sphere x2+y2+z2+4x−2y−6z=155 is

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Finding the Center of Sphere $S_{1}$

Finding the Center of Sphere $S_{2}$

Calculating Coordinates of Midpoint $M$

The Plane Passing Through Midpoint $M$

Substituting $M (1, 1, 1)$ into the Plane Equation

Solving the Linear Equation for $a$

The intersection of the spheres $x^{2} + y^{2} + z^{2} + 7 x - 2 y - z = 13$ and $x^{2} + y^{2} + z^{2} - 3 x + 3 y + 4 z = 8$ is the same as the intersection of one of the sphere and the plane

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:

The plane $x + 2 y - z = 4$ cuts the sphere $x^{2} + y^{2} + z^{2} - x + z - 2 = 0$ in a circle of radius

The shortest distance from the plane $12 x + 4 y + 3 z = 327$ to the sphere $x^{2} + y^{2} + z^{2} + 4 x - 2 y - 6 z = 155$ is

Equation of the plane containing the straight line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$ is