JEE Advanced 2013

LEVELJEE Advanced

Animated Solution for Mathematics - Three Dimensional Geometry: Consider the lines $L_{1} : \frac{x - 1}{2} = \frac{y}{- 1} = \frac{z + 3}{1}, L_{2} : \frac{x - 4}{1} = \frac{y + 3}{1} = \frac{z + 3}{2}$ and the planes $P_{1} : 7 x + y + 2 z = 3, P_{2} : 3 x + 5 y - 6 z = 4$ . Let $a x + b y + cz = d$ be the equation of the plane passing through the point of intersection of lines $L_{1}$ and $L_{2}$ , and perpendicular to planes $P_{1}$ and $P_{2}$ . Match List I with List II:

Q: Does it matter if I simplify the normal vector by dividing by $-16$?

Not at all! The equation of a plane $ax + by + cz = d$ is defined by the *direction* of the normal vector $(a, b, c)$. Any scalar multiple of this vector (like dividing by $-16$) represents the same orientation. Simplifying makes your algebra much cleaner and less prone to errors, so I highly encourage it!

List-I

(P)

(Q)

(R)

(S)

List-II

(1)

(2)

-3

(3)

(4)

-2

Select Matching Pairs:

* Multiple Allowed

PMatches

QMatches

RMatches

SMatches

Progressive Hints

Hint 1

Hint 1 (The Spark): To find the point of intersection, visualize the lines as paths in 3D space. Since they intersect at a single point, there must be a unique set of parameters (let's call them

λ

and

μ

) that yield the same

(x, y, z)

coordinates for both lines. Start by expressing the general coordinates of any point on

L_{1}

and

L_{2}

in terms of these parameters.

Hint 2

Hint 2 (The Arsenal): Once you have the intersection point, you need the equation of the plane. The plane is perpendicular to two given planes,

P_{1}

and

P_{2}

. Recall that the normal vector of a plane is perpendicular to the plane itself. If our required plane is perpendicular to

P_{1}

and

P_{2}

, its normal vector must be perpendicular to the normal vectors of

P_{1}

and

P_{2}

. What mathematical operation finds a vector perpendicular to two other vectors?

Hint 3

Hint 3 (The Execution): Use the cross product of the normal vectors of

P_{1}

and

P_{2}

to find the normal vector

n = (a, b, c)

of your required plane. Once you have the normal vector

(a, b, c)

and the intersection point

(x_{0}, y_{0}, z_{0})

, use the point-normal form of the plane equation:

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

Visualized Solution

Visualizing the Lines $L_{1}$ and $L_{2}$

We are given two lines in 3D space:
$L_{1} : \frac{x - 1}{2} = \frac{y}{- 1} = \frac{z + 3}{1}$
$L_{2} : \frac{x - 4}{1} = \frac{y + 3}{1} = \frac{z + 3}{2}$
Our first objective is to find their point of intersection, $P$ .

Parametric Representation of $L_{1}$ and $L_{2}$

Let's express any general point on $L_{1}$ by equating it to parameter $λ$ :
$\frac{x - 1}{2} = \frac{y}{- 1} = \frac{z + 3}{1} = λ \Rightarrow P_{1} (2 λ + 1, - λ, λ - 3)$
Similarly, for $L_{2}$ , using parameter $μ$ :
$\frac{x - 4}{1} = \frac{y + 3}{1} = \frac{z + 3}{2} = μ \Rightarrow P_{2} (μ + 4, μ - 3, 2 μ - 3)$

Setting up the System of Equations

If the lines intersect, there must exist a unique pair of $(λ, μ)$ where $P_{1} = P_{2}$ .
Equating the $x$ -coordinates:
$2 λ + 1 = μ + 4 \Rightarrow 2 λ - μ = 3$ (Equation 1)
Equating the $y$ -coordinates:
$- λ = μ - 3 \Rightarrow λ + μ = 3$ (Equation 2)

Solving for $λ$ and $μ$

Adding Equation 1 and Equation 2:
$(2 λ - μ) + (λ + μ) = 3 + 3 \Rightarrow 3 λ = 6 \Rightarrow λ = 2$
Substitute $λ = 2$ into Equation 2:
$2 + μ = 3 \Rightarrow μ = 1$

Verifying Consistency and Finding $P$

Let's verify these values with the $z$ -coordinates:
For $L_{1}$ : $z = λ - 3 = 2 - 3 = - 1$
For $L_{2}$ : $z = 2 μ - 3 = 2 (1) - 3 = - 1$
Since both yield $z = - 1$ , the lines intersect at:
$P (5, - 2, - 1)$

The Logic of the Normal Vector

The required plane is perpendicular to planes:
$P_{1} : 7 x + y + 2 z = 3 \Rightarrow n_{1} = (7, 1, 2)$
$P_{2} : 3 x + 5 y - 6 z = 4 \Rightarrow n_{2} = (3, 5, - 6)$
Therefore, the normal vector $n$ of our plane must be perpendicular to both $n_{1}$ and $n_{2}$ .
This means $n$ is parallel to the cross product $n_{1} \times n_{2}$ .

Setting up the Cross Product

Let's set up the vector cross product using a determinant:
$n = n_{1} \times n_{2} = \hat{i} 73 \hat{j} 15 \hat{k} 2 - 6$

Calculating the Normal Vector

Expanding along the first row:
$n = \hat{i} (1 (- 6) - 2 (5)) - \hat{j} (7 (- 6) - 2 (3)) + \hat{k} (7 (5) - 1 (3))$
$n = \hat{i} (- 6 - 10) - \hat{j} (- 42 - 6) + \hat{k} (35 - 3)$
$n = - 16 \hat{i} + 48 \hat{j} + 32 \hat{k}$

Simplifying Direction Ratios

The direction ratios of the normal vector are $(- 16, 48, 32)$ .
We can simplify these ratios by dividing each component by $- 16$ :
$(a, b, c) = (1, - 3, - 2)$

Writing the Equation of the Plane

The equation of a plane passing through $(x_{1}, y_{1}, z_{1})$ with normal $(a, b, c)$ is:
$a (x - x_{1}) + b (y - y_{1}) + c (z - z_{1}) = 0$
Substituting $P (5, - 2, - 1)$ and $(a, b, c) = (1, - 3, - 2)$ :
$1 (x - 5) - 3 (y - (- 2)) - 2 (z - (- 1)) = 0$

Final Equation and Matching

Expanding the equation:
$x - 5 - 3 y - 6 - 2 z - 2 = 0 \Rightarrow x - 3 y - 2 z = 13$
Comparing with $a x + b y + cz = d$ :
$a = 1, b = - 3, c = - 2, d = 13$
Matching with List II:
$a = 1 \to (3)$ , $b = - 3 \to (2)$ , $c = - 2 \to (4)$ , $d = 13 \to (1)$

00:00 / 00:00

The Sigma Insight: Equation of a Plane

Solution Diagram

Analyzing the Intersection

To determine if the lines

L_{1}

and

L_{2}

intersect, we parameterize them. By setting the symmetric equations of

L_{1}

equal to

λ

, any point on

L_{1}

is defined as

P_{1} (2 λ + 1, - λ, λ - 3)

Similarly, for

L_{2}

, we introduce the parameter

μ

to define any point as

P_{2} (μ + 4, μ - 3, 2 μ - 3)

. If these lines intersect, there must exist a unique pair

(λ, μ)

such that

P_{1} = P_{2}

This yields the following system of linear equations:

2 λ - μ = 3

λ + μ = 3

Adding these equations yields

3 λ = 6

, which gives

λ = 2

. Substituting this back, we find

μ = 1

. A quick check of the

z

-coordinates confirms that both lines pass through

z = - 1

at this intersection point,

P (5, - 2, - 1)

The Power of the Normal Vector

Now that we have the point of intersection, we must determine the orientation of the plane. We are given that the plane is perpendicular to two other planes with normal vectors

n_{1} = (7, 1, 2)

and

n_{2} = (3, 5, - 6)

The normal vector of our required plane,

n

, must be perpendicular to both

n_{1}

and

n_{2}

. We find this vector using the cross product:

n = n_{1} \times n_{2} = \hat{i} 73 \hat{j} 15 \hat{k} 2 - 6

Expanding the determinant, we obtain:

n = \hat{i} (1 (- 6) - 2 (5)) - \hat{j} (7 (- 6) - 2 (3)) + \hat{k} (7 (5) - 1 (3))

n = - 16 \hat{i} + 48 \hat{j} + 32 \hat{k}

The Final Assembly

The direction ratios of our normal vector are

(- 16, 48, 32)

. To simplify, we divide by

- 16

to obtain the reduced normal vector

n = (1, - 3, - 2)

We now apply the point-normal form of a plane equation,

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

, using point

P (5, - 2, - 1)

1 (x - 5) - 3 (y + 2) - 2 (z + 1) = 0

Expanding this expression:

x - 5 - 3 y - 6 - 2 z - 2 = 0

x - 3 y - 2 z = 13

Comparing this to the standard form

a x + b y + cz = d

, we identify the coefficients as

a = 1

b = - 3

c = - 2

, and

d = 13

. The final equation of the plane is $x - 3 y - 2 z = 13$ .

Common Traps & FAQs

Wait, why do we use parameters

λ

and

μ

instead of just solving the equations directly?

Great question! In 3D space, a line is not just a single point; it is an infinite set of points. By setting the symmetric equations equal to a parameter (like

λ

), we are essentially creating a 'map' of every point on that line. This allows us to test if there is a specific value of

λ

and

μ

where the coordinates of the two lines coincide, which is exactly what we need to find the intersection point.

Why is the normal vector of the new plane the cross product of the normals of the given planes?

Think of it geometrically. If a plane is perpendicular to another plane, its normal vector must lie within the other plane (or be parallel to it). If our plane is perpendicular to *both*

P_{1}

and

P_{2}

, its normal vector must be perpendicular to the normal vectors of both

P_{1}

and

P_{2}

. The cross product is the perfect tool for this, as it produces a vector that is mutually perpendicular to the two input vectors.

Does it matter if I simplify the normal vector by dividing by

- 16

Not at all! The equation of a plane

a x + b y + cz = d

is defined by the *direction* of the normal vector

(a, b, c)

. Any scalar multiple of this vector (like dividing by

- 16

) represents the same orientation. Simplifying makes your algebra much cleaner and less prone to errors, so I highly encourage it!

List-I

List-II

Select Matching Pairs:

Visualized Solution

Visualizing the Lines $L_{1}$ and $L_{2}$

Parametric Representation of $L_{1}$ and $L_{2}$

Setting up the System of Equations

Solving for $λ$ and $μ$

Verifying Consistency and Finding $P$

The Logic of the Normal Vector

Setting up the Cross Product

Calculating the Normal Vector

Simplifying Direction Ratios

Writing the Equation of the Plane

Final Equation and Matching

The Sigma Insight: Equation of a Plane

Analyzing the Intersection

The Power of the Normal Vector

The Final Assembly

Similar Questions

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

The equation of the plane passing through the point $(1, 2, - 3)$ and perpendicular to the planes $3 x + y - 2 z = 5$ and $2 x - 5 y - z = 7$ , 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:

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 :

Let the equation of the plane, that passes through the point $(1, 4, - 3)$ and contains the line of intersection of the planes $3 x - 2 y + 4 z - 7 = 0$ and $x + 5 y - 2 z + 9 = 0$ , be $α x + β y + γ z + 3 = 0$ , then $α + β + γ$ is equal to :

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

List-I

List-II

Select Matching Pairs:

Visualized Solution

Visualizing the Lines .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Parametric Representation of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Setting up the System of Equations

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } μ

Verifying Consistency and Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

The Logic of the Normal Vector

Setting up the Cross Product

Calculating the Normal Vector

Simplifying Direction Ratios

Writing the Equation of the Plane

Final Equation and Matching

The Sigma Insight: Equation of a Plane

Analyzing the Intersection

The Power of the Normal Vector

The Final Assembly

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of the plane passing through the point (1,2,−3) and perpendicular to the planes 3x+y−2z=5 and 2x−5y−z=7, 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:

.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; } Equation of the plane containing the straight line 2x​=3y​=4z​ and perpendicular to the plane containing the straight lines 3x​=4y​=2z​ and 4x​=2y​=3z​ 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

Visualizing the Lines $L_{1}$ and $L_{2}$

Parametric Representation of $L_{1}$ and $L_{2}$

Solving for $λ$ and $μ$

Verifying Consistency and Finding $P$

The equation of the plane passing through the point $(1, 2, - 3)$ and perpendicular to the planes $3 x + y - 2 z = 5$ and $2 x - 5 y - z = 7$ , 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:

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 :

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