JEE Advanced 2005S

LEVELJEE Main

Animated Solution for Mathematics - Three Dimensional Geometry: A variable plane at a distance of the one unit from the origin cuts the coordinates axes at $A, B$ and $C$ . If the centroid $D (x, y, z)$ of triangle $A B C$ satisfies the relation $\frac{1}{x ^{2}} + \frac{1}{y ^{2}} + \frac{1}{z ^{2}} = k$ , then the value $k$ is

Q: Why is the distance from the origin to the plane $\frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}}$?

Great question! The standard form is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} - 1 = 0$. Here, the coefficients are $A = \frac{1}{a}$, $B = \frac{1}{b}$, $C = \frac{1}{c}$, and $D = -1$. Plugging these into the distance formula $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$ gives exactly that expression. It is a beautiful result of the intercept form!

Q: Why do we multiply by 3 for the centroid coordinates?

Remember, the centroid of a triangle with vertices $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, and $(x_3, y_3, z_3)$ is $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3})$. Since our vertices are $(a,0,0)$, $(0,b,0)$, and $(0,0,c)$, the math simplifies to $(\frac{a}{3}, \frac{b}{3}, \frac{c}{3})$. It is just the average!

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Visualized Solution

Visualizing the 3D Coordinate System

We start by setting up a three-dimensional Cartesian coordinate system.
Let the three mutually perpendicular axes be the $x$ -axis, $y$ -axis, and $z$ -axis, intersecting at the origin $O (0, 0, 0)$ .

Defining the Intercepts $A$ , $B$ , and $C$

Let the variable plane intersect the coordinate axes at points $A$ , $B$ , and $C$ .
We can define these points as $A (a, 0, 0)$ on the $x$ -axis, $B (0, b, 0)$ on the $y$ -axis, and $C (0, 0, c)$ on the $z$ -axis.
Here, $a$ , $b$ , and $c$ represent the non-zero intercepts of the plane on the respective axes.

Intercept Form of the Plane

The equation of a plane making intercepts $a$ , $b$ , and $c$ on the coordinate axes is given by:
$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$
This is known as the intercept form of a plane.

Perpendicular Distance Formula

The perpendicular distance $d$ from a point $(x_{0}, y_{0}, z_{0})$ to a plane $A x + B y + C z + D = 0$ is:
$d = \frac{∣ A x _{0} + B y _{0} + C z _{0} + D ∣}{A ^{2} + B ^{2} + C ^{2}}$
In our case, the point is the origin $(0, 0, 0)$ and the distance is $1$ unit.

Setting Up the Distance Equation

Let's rewrite the plane equation as: $\frac{1}{a} x + \frac{1}{b} y + \frac{1}{c} z - 1 = 0$
Substituting the origin $(0, 0, 0)$ into the distance formula:
$d = \frac{∣ \frac{1}{a} ( 0 ) + \frac{1}{b} ( 0 ) + \frac{1}{c} ( 0 ) - 1∣}{( \frac{1}{a} ) ^{2} + ( \frac{1}{b} ) ^{2} + ( \frac{1}{c} ) ^{2}} = 1$

Simplifying the Relation

This simplifies to: $\frac{1}{\frac{1}{a ^{2}} + \frac{1}{b ^{2}} + \frac{1}{c ^{2}}} = 1$
Squaring both sides and taking the reciprocal, we get:
$\frac{1}{a ^{2}} + \frac{1}{b ^{2}} + \frac{1}{c ^{2}} = 1$

Centroid of Triangle $A B C$

The centroid $D (x, y, z)$ of a triangle with vertices $A (a, 0, 0)$ , $B (0, b, 0)$ , and $C (0, 0, c)$ is given by:
$(x, y, z) = (\frac{a + 0 + 0}{3}, \frac{0 + b + 0}{3}, \frac{0 + 0 + c}{3})$
Therefore, the coordinates of the centroid are: $x = \frac{a}{3}$ , $y = \frac{b}{3}$ , $z = \frac{c}{3}$

Expressing Intercepts in terms of Centroid

From the centroid coordinates, we can express $a$ , $b$ , and $c$ in terms of $x$ , $y$ , and $z$ :
$a = 3 x$
$b = 3 y$
$c = 3 z$

Substitution into the Relation

Substitute $a = 3 x$ , $b = 3 y$ , and $c = 3 z$ into the distance relation:
$\frac{1}{( 3 x ) ^{2}} + \frac{1}{( 3 y ) ^{2}} + \frac{1}{( 3 z ) ^{2}} = 1$
This gives: $\frac{1}{9 x ^{2}} + \frac{1}{9 y ^{2}} + \frac{1}{9 z ^{2}} = 1$

Finding the Value of $k$

Factor out $\frac{1}{9}$ from the left-hand side:
$\frac{1}{9} (\frac{1}{x ^{2}} + \frac{1}{y ^{2}} + \frac{1}{z ^{2}}) = 1$
Multiplying both sides by $9$ :
$\frac{1}{x ^{2}} + \frac{1}{y ^{2}} + \frac{1}{z ^{2}} = 9$
Comparing this with the given relation $\frac{1}{x ^{2}} + \frac{1}{y ^{2}} + \frac{1}{z ^{2}} = k$ , we get:
$k = 9$

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The Sigma Insight: Equation of a Plane

Solution Diagram

Analyzing the Setup

Imagine standing at the origin of a three-dimensional coordinate system. You are looking at a plane that slices through the

x

y

, and

z

axes.

This is a geometric dance where the plane's orientation is constrained by its distance from the origin. We aim to uncover the relationship between the plane's intercepts and the centroid of the triangle it forms.

The Plane's Identity

We define the plane using the 'intercept form'. When a plane cuts the axes at points

A (a, 0, 0)

B (0, b, 0)

, and

C (0, 0, c)

, the equation is:

\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1

This equation represents the "DNA" of our plane. If we know the values of

a

b

, and

c

, the entire surface is uniquely determined.

The Distance Constraint

The problem states that the plane is exactly

1

unit away from the origin. We utilize the perpendicular distance formula for a plane

A x + B y + C z + D = 0

, which is given by:

d = \frac{∣ D ∣}{A ^{2} + B ^{2} + C ^{2}}

Rewriting our intercept equation as

\frac{1}{a} x + \frac{1}{b} y + \frac{1}{c} z - 1 = 0

, we identify

A = \frac{1}{a}

B = \frac{1}{b}

C = \frac{1}{c}

, and

D = - 1

. Substituting these into the distance formula with

d = 1

1 = \frac{∣ - 1∣}{( \frac{1}{a} ) ^{2} + ( \frac{1}{b} ) ^{2} + ( \frac{1}{c} ) ^{2}}

Squaring both sides and taking the reciprocal, we arrive at the fundamental geometric relation:

\frac{1}{a ^{2}} + \frac{1}{b ^{2}} + \frac{1}{c ^{2}} = 1

The Centroid Connection

Next, we consider the centroid

G (x, y, z)

of the triangle

A B C

. The centroid is the arithmetic mean of the vertices

(a, 0, 0)

(0, b, 0)

, and

(0, 0, c)

This yields the coordinates:

x = \frac{a}{3}, y = \frac{b}{3}, z = \frac{c}{3}

Consequently, we can express the intercepts in terms of the centroid coordinates as

a = 3 x

b = 3 y

, and

c = 3 z

Final Synthesis

We substitute these expressions for

a

b

, and

c

into our distance relation:

\frac{1}{( 3 x ) ^{2}} + \frac{1}{( 3 y ) ^{2}} + \frac{1}{( 3 z ) ^{2}} = 1

Simplifying the denominators, we obtain:

\frac{1}{9 x ^{2}} + \frac{1}{9 y ^{2}} + \frac{1}{9 z ^{2}} = 1

Factoring out

\frac{1}{9}

and multiplying both sides by

9

, we reach the final result:

\frac{1}{x ^{2}} + \frac{1}{y ^{2}} + \frac{1}{z ^{2}} = 9

Comparing this to the form

\frac{1}{x ^{2}} + \frac{1}{y ^{2}} + \frac{1}{z ^{2}} = k

, we conclude that $k = 9$ .

Common Traps & FAQs

Why is the distance from the origin to the plane

\frac{1}{\frac{1}{a ^{2}} + \frac{1}{b ^{2}} + \frac{1}{c ^{2}}}

Great question! The standard form is

\frac{x}{a} + \frac{y}{b} + \frac{z}{c} - 1 = 0

. Here, the coefficients are

A = \frac{1}{a}

B = \frac{1}{b}

C = \frac{1}{c}

, and

D = - 1

. Plugging these into the distance formula

d = \frac{∣ D ∣}{A ^{2} + B ^{2} + C ^{2}}

gives exactly that expression. It is a beautiful result of the intercept form!

Can we just use the distance formula for a point to a plane directly?

Absolutely! That is exactly what we did. The key is to recognize that the 'intercept form' is just a specific case of the general plane equation. Once you see that, the distance formula becomes your most powerful tool.

Why do we multiply by 3 for the centroid coordinates?

Remember, the centroid of a triangle with vertices

(x_{1}, y_{1}, z_{1})

(x_{2}, y_{2}, z_{2})

, and

(x_{3}, y_{3}, z_{3})

(\frac{x _{1} + x _{2} + x _{3}}{3}, \frac{y _{1} + y _{2} + y _{3}}{3}, \frac{z _{1} + z _{2} + z _{3}}{3})

. Since our vertices are

(a, 0, 0)

(0, b, 0)

, and

(0, 0, c)

, the math simplifies to

(\frac{a}{3}, \frac{b}{3}, \frac{c}{3})

. It is just the average!

Select Answer:

Visualized Solution

Visualizing the 3D Coordinate System

Defining the Intercepts $A$ , $B$ , and $C$

Intercept Form of the Plane

Perpendicular Distance Formula

Setting Up the Distance Equation

Simplifying the Relation

Centroid of Triangle $A B C$

Expressing Intercepts in terms of Centroid

Substitution into the Relation

Finding the Value of $k$

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Plane's Identity

The Distance Constraint

The Centroid Connection

Final Synthesis

Similar Questions

A plane $P$ meets the coordinate axes at $A, B$ and $C$ respectively. The centroid of $Δ A B C$ is given to be $(1, 1, 2)$ . Then the equation of the line through this centroid and perpendicular to the plane $P$ is:

If a plane passes through the points $(- 1, k, 0)$ , $(2, k, - 1)$ , $(1, 1, 2)$ and is parallel to the line $\frac{x - 1}{1} = \frac{2 y + 1}{2} = \frac{z + 1}{- 1}$ , then the value of $\frac{k ^{2} + 1}{( k - 1 ) ( k - 2 )}$ 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

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Let the plane $x + 3 y - 2 z + 6 = 0$ meet the co-ordinate axes at the points $A, B, C$ . If the orthocenter of the triangle $A B C$ is $(α, β, \frac{6}{7})$ , then $98 (α + β)^{2}$ is equal to __________.

Let a plane $P$ contain two lines $r = \hat{i} + λ (\hat{i} + \hat{j}), λ \in R$ and $r = - \hat{j} + μ (\hat{j} - \hat{k}), μ \in R$ . If $Q (α, β, γ)$ is the foot of the perpendicular drawn from the point $M (1, 0, 1)$ to $P$ , then $3 (α + β + γ)$ equals

Find the equation of plane passing through $(1, 1, 1)$ & parallel to the lines $L_{1}, L_{2}$ having direction ratios $(1, 0, - 1), (1, - 1, 0)$ . Find the volume of tetrahedron formed by origin and the points where these planes intersect the coordinate axes.

Select Answer:

Visualized Solution

Visualizing the 3D Coordinate System

Intercept Form of the Plane

Perpendicular Distance Formula

Setting Up the Distance Equation

Simplifying the Relation

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

Expressing Intercepts in terms of Centroid

Substitution into the Relation

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

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Plane's Identity

The Distance Constraint

The Centroid Connection

Final Synthesis

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a plane passes through the points (−1,k,0), (2,k,−1), (1,1,2) and is parallel to the line 1x−1​=22y+1​=−1z+1​, then the value of (k−1)(k−2)k2+1​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the plane x+3y−2z+6=0 meet the co-ordinate axes at the points A,B,C. If the orthocenter of the triangle ABC is (α,β,76​), then 98(α+β)2 is equal to __________.

Centroid of Triangle $A B C$

Finding the Value of $k$

If a plane passes through the points $(- 1, k, 0)$ , $(2, k, - 1)$ , $(1, 1, 2)$ and is parallel to the line $\frac{x - 1}{1} = \frac{2 y + 1}{2} = \frac{z + 1}{- 1}$ , then the value of $\frac{k ^{2} + 1}{( k - 1 ) ( k - 2 )}$ is

Let the plane $x + 3 y - 2 z + 6 = 0$ meet the co-ordinate axes at the points $A, B, C$ . If the orthocenter of the triangle $A B C$ is $(α, β, \frac{6}{7})$ , then $98 (α + β)^{2}$ is equal to __________.