JEE Advanced 2004

LEVELJEE Main

Animated Solution for Mathematics - Three Dimensional Geometry: 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.

Q: Wait, why can we just scale the normal vector to $(1, 1, 1)$? Does that change the plane?

That is a fantastic question! The normal vector $\vec{n} = (a, b, c)$ defines the *orientation* of the plane, not its position. In the equation $ax + by + cz = d$, the coefficients $a, b, c$ represent the direction ratios of the normal. If you multiply the entire equation by a constant, the plane remains identical. So, using $(1, 1, 1)$ instead of $(-1, -1, -1)$ is perfectly valid and makes the arithmetic much cleaner.

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the plane as a flat sheet in 3D space. If it is parallel to two lines, it must be oriented such that it never touches them. This implies that the normal vector—the line segment perpendicular to the plane—must be orthogonal to the direction vectors of both lines. How do you find a vector orthogonal to two others?

Hint 2

Hint 2 (The Arsenal): You have the direction ratios of the two lines, say

L_{1} = (1, 0, - 1)

and

L_{2} = (1, - 1, 0)

. Recall that the cross product of two vectors yields a third vector perpendicular to both. Use the determinant method to calculate

n = L_{1} \times L_{2}

Hint 3

Hint 3 (The Execution): Once you have the normal vector

n = (a, b, c)

and the point

P (1, 1, 1)

that lies on the plane, use the point-normal form equation:

a (x - x_{1}) + b (y - y_{1}) + c (z - z_{1}) = 0

. Substitute your values to find the equation of the plane.

Visualized Solution

Visualizing the Geometry

Given anchor point: $P (1, 1, 1)$
The plane is parallel to two lines, $L_{1}$ and $L_{2}$ .
Direction ratios of $L_{1}$ : $(1, 0, - 1)$
Direction ratios of $L_{2}$ : $(1, - 1, 0)$

The Normal Vector Concept

A plane is uniquely defined by its normal vector $n$ .
Since the plane is parallel to $L_{1}$ and $L_{2}$ , $n$ must be perpendicular to both.
Therefore, $n = L_{1} \times L_{2}$

Setting up the Cross Product

$n = \hat{i} 11 \hat{j} 0 - 1 \hat{k} - 1 0$

Calculating $n$

$n = \hat{i} (0 - 1) - \hat{j} (0 - (- 1)) + \hat{k} (- 1 - 0)$
$n = - \hat{i} - \hat{j} - \hat{k}$
Direction ratios of normal: $(- 1, - 1, - 1)$
For simplicity, we can scale this to $(1, 1, 1)$

Point-Normal Form of Plane

Formula: $a (x - x_{1}) + b (y - y_{1}) + c (z - z_{1}) = 0$
Where $(a, b, c)$ are direction ratios of the normal.
And $(x_{1}, y_{1}, z_{1})$ is the given point on the plane.

Substituting the Values

Substitute $(a, b, c) = (1, 1, 1)$
Substitute $(x_{1}, y_{1}, z_{1}) = (1, 1, 1)$
$1 (x - 1) + 1 (y - 1) + 1 (z - 1) = 0$

Simplifying the Equation

$x - 1 + y - 1 + z - 1 = 0$
$x + y + z - 3 = 0$
Final Plane Equation: $x + y + z = 3$

Finding the Intercepts

Convert to Intercept Form: $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$
Divide $x + y + z = 3$ by $3$ :
$\frac{x}{3} + \frac{y}{3} + \frac{z}{3} = 1$
Intercepts: $A (3, 0, 0)$ , $B (0, 3, 0)$ , $C (0, 0, 3)$

Visualizing the Tetrahedron

The plane intersects the axes at $A, B, C$ .
The origin $O (0, 0, 0)$ and points $A, B, C$ form a tetrahedron.
This is a solid 3D figure with four triangular faces.

Volume of the Tetrahedron

For a tetrahedron formed by origin and intercepts $a, b, c$ :
Volume $V = \frac{1}{6} ∣ a \cdot b \cdot c ∣$
Here, $a = 3$ , $b = 3$ , $c = 3$ .

Calculating the Final Volume

Substitute the values: $V = \frac{1}{6} (3 \times 3 \times 3)$
$V = \frac{27}{6}$
Simplifying: $V = \frac{9}{2}$ cubic units.

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The Sigma Insight: Equation of a Plane

Solution Diagram

Analyzing the Setup

To define a plane in 3D space, we require a point on the plane and a normal vector

n

perpendicular to its surface. We are given the point

P (1, 1, 1)

and the condition that the plane must be parallel to two lines,

L_{1}

and

L_{2}

Since the plane is parallel to both lines, its normal vector must be perpendicular to the direction vectors of both lines. We find this normal vector by calculating the cross product of the direction vectors

L_{1}

and

L_{2}

The Master Equation

We set up the determinant to find the normal vector

n

n = \hat{i} 11 \hat{j} 0 - 1 \hat{k} - 1 0

Expanding this determinant, we obtain:

n = \hat{i} (0 - 1) - \hat{j} (0 - (- 1)) + \hat{k} (- 1 - 0) = - \hat{i} - \hat{j} - \hat{k}

For simplicity, we scale this vector to

n = (1, 1, 1)

. Using the point-normal form

a (x - x_{1}) + b (y - y_{1}) + c (z - z_{1}) = 0

, we substitute our point

P (1, 1, 1)

and normal vector

(1, 1, 1)

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

Simplifying the expression, we arrive at the equation of the plane:

$x + y + z = 3$

The Tetrahedron

A Pyramid in Space

The plane

x + y + z = 3

intersects the coordinate axes. To identify these intersection points, we convert the equation to the intercept form:

\frac{x}{3} + \frac{y}{3} + \frac{z}{3} = 1

This reveals that the plane intersects the axes at

A (3, 0, 0)

B (0, 3, 0)

, and

C (0, 0, 3)

. These points, together with the origin

O (0, 0, 0)

, form a tetrahedron.

Final Calculation

The volume

V

of a tetrahedron formed by the coordinate planes and a plane with intercepts

a, b, c

is given by:

V = \frac{1}{6} ∣ a \cdot b \cdot c ∣

Substituting the intercepts

a = 3, b = 3, c = 3

into the formula:

V = \frac{1}{6} (3 \times 3 \times 3) = \frac{27}{6} = 4.5

The volume of the tetrahedron is $4.5$ cubic units.

Common Traps & FAQs

Wait, why can we just scale the normal vector to

(1, 1, 1)

? Does that change the plane?

That is a fantastic question! The normal vector

n = (a, b, c)

defines the *orientation* of the plane, not its position. In the equation

a x + b y + cz = d

, the coefficients

a, b, c

represent the direction ratios of the normal. If you multiply the entire equation by a constant, the plane remains identical. So, using

(1, 1, 1)

instead of

(- 1, - 1, - 1)

is perfectly valid and makes the arithmetic much cleaner.

Why is the volume of the tetrahedron

\frac{1}{6} ab c

Think of the tetrahedron as a pyramid. The volume of any pyramid is

\frac{1}{3} \times Base Area \times Height

. Here, the base is a right-angled triangle in the

x y

-plane with vertices at

(0, 0, 0)

(a, 0, 0)

, and

(0, b, 0)

. Its area is

\frac{1}{2} ab

. The height of the tetrahedron along the

z

-axis is

c

. Thus, the volume is

\frac{1}{3} \times (\frac{1}{2} ab) \times c = \frac{1}{6} ab c

. It is a beautiful, time-saving shortcut!

Can I use the scalar triple product to find the volume?

Yes, you absolutely can! The volume of a tetrahedron with vertices at the origin and points

A, B, C

\frac{1}{6} ∣ A \cdot (B \times C) ∣

. However, since your points are simply the intercepts on the axes, the

\frac{1}{6} ab c

formula is essentially the result of that scalar triple product. Using the intercept formula is just a more direct path to the same destination.

Visualized Solution

Visualizing the Geometry

The Normal Vector Concept

Setting up the Cross Product

Calculating $n$

Point-Normal Form of Plane

Substituting the Values

Simplifying the Equation

Finding the Intercepts

Visualizing the Tetrahedron

Volume of the Tetrahedron

Calculating the Final Volume

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Master Equation

The Tetrahedron

Final Calculation

Similar Questions

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 equation of the plane passing through the line of intersection of the planes $r \cdot (\hat{i} + \hat{j} + \hat{k}) = 1$ and $r \cdot (2 \hat{i} + 3 \hat{j} - \hat{k}) + 4 = 0$ and parallel to the $x$ -axis 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

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

Find the equation of the plane containing the line $2 x - y + z - 3 = 0, 3 x + y + z = 5$ and at a distance of $\frac{1}{6}$ from the point $(2, 1, - 1)$ .

The 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:

Equation of a plane at a distance $\frac{2}{21}$ from the origin, which contains the line of intersection of the planes $x - y - z - 1 = 0$ and $2 x + y - 3 z + 4 = 0$ 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.

Visualized Solution

Visualizing the Geometry

The Normal Vector Concept

Setting up the Cross Product

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

Point-Normal Form of Plane

Substituting the Values

Simplifying the Equation

Finding the Intercepts

Visualizing the Tetrahedron

Volume of the Tetrahedron

Calculating the Final Volume

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Master Equation

The Tetrahedron

Final Calculation

Similar Questions

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Calculating $n$

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 equation of the plane passing through the line of intersection of the planes $r \cdot (\hat{i} + \hat{j} + \hat{k}) = 1$ and $r \cdot (2 \hat{i} + 3 \hat{j} - \hat{k}) + 4 = 0$ and parallel to the $x$ -axis 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

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

Find the equation of the plane containing the line $2 x - y + z - 3 = 0, 3 x + y + z = 5$ and at a distance of $\frac{1}{6}$ from the point $(2, 1, - 1)$ .

Equation of a plane at a distance $\frac{2}{21}$ from the origin, which contains the line of intersection of the planes $x - y - z - 1 = 0$ and $2 x + y - 3 z + 4 = 0$ is :