Consider the lines L1 : (x-1)/2 =...

MathematicsEquation of a PlaneJEE Advanced 2013Moderate

View in:English Hindi

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:

List-I

(P)

(Q)

(R)

(S)

List-II

(1)

(2)

-3

(3)

(4)

-2

Answer:P → 3, Q → 2, R → 4, S → 1

Visualized Solution (English)

Visualizing the Lines .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L_{1} and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L_{2}

Given lines: $L_{1} : \frac{x - 1}{2} = \frac{y}{- 1} = \frac{z + 3}{1}$
Given lines: $L_{2} : \frac{x - 4}{1} = \frac{y + 3}{1} = \frac{z + 3}{2}$
Goal: Find the intersection point $P$ of $L_{1}$ and $L_{2}$ .

Defining General Points

General point on $L_{1}$ : $P_{1} (2 λ + 1, - λ, λ - 3)$
General point on $L_{2}$ : $P_{2} (μ + 4, μ - 3, 2 μ - 3)$

Equating the Coordinates

Equating $x$ and $y$ coordinates:
1) $2 λ + 1 = μ + 4 \Rightarrow 2 λ - μ = 3$
2) $- λ = μ - 3 \Rightarrow λ + μ = 3$

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

Adding (1) and (2): $3 λ = 6 \Rightarrow λ = 2$
Substituting $λ = 2$ in (2): $2 + μ = 3 \Rightarrow μ = 1$
Check $z$ : $λ - 3 = 2 - 3 = - 1$ and $2 μ - 3 = 2 (1) - 3 = - 1$ .
Consistency confirmed.

Finding the Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Intersection point $P$ at $λ = 2$ :
$x = 2 (2) + 1 = 5$
$y = - (2) = - 2$
$z = 2 - 3 = - 1$
Point $P = (5, - 2, - 1)$

The Logic of the Normal Vector

Required plane is perpendicular to $P_{1} : 7 x + y + 2 z = 3$ and $P_{2} : 3 x + 5 y - 6 z = 4$ .
Normal vectors: $n_{1} = (7, 1, 2)$ and $n_{2} = (3, 5, - 6)$ .
The required normal $n$ is parallel to $n_{1} \times n_{2}$ .

Setting up the Cross Product

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

Calculating the Normal Vector

Expanding the determinant:
$n = \hat{i} (- 6 - 10) - \hat{j} (- 42 - 6) + \hat{k} (35 - 3)$
$n = - 16 \hat{i} + 48 \hat{j} + 32 \hat{k}$

Simplifying Direction Ratios

Direction ratios of the normal $n$ :
Dividing by $- 16$ : $(1, - 3, - 2)$
So, $a : b : c = 1 : - 3 : - 2$

The Equation of the Plane

Equation of a plane: $A (x - x_{1}) + B (y - y_{1}) + C (z - z_{1}) = 0$
Substitute point $(5, - 2, - 1)$ and normal $(1, - 3, - 2)$ .

Substitution and Expansion

Substituting values:
$1 (x - 5) - 3 (y + 2) - 2 (z + 1) = 0$
Expanding:
$x - 5 - 3 y - 6 - 2 z - 2 = 0$

The Final Plane Equation

Simplifying the terms:
$x - 3 y - 2 z - 13 = 0$
Standard form: $x - 3 y - 2 z = 13$
Comparing with $a x + b y + cz = d$ :
$a = 1, b = - 3, c = - 2, d = 13$

Matching the Results

Final Matching:
a = 1 $\to$ (3)
b = -3 $\to$ (2)
c = -2 $\to$ (4)
d = 13 $\to$ (1)
Matching Sequence: P-3, Q-2, R-4, S-1
Correct Option: (a)

00:00 / 00:00

Conceptually Similar Problems

MathematicsIntersection of a Line and a PlaneJEE Advanced 2012Moderate

View in:English Hindi

The equation of a plane passing through the line of intersection of the planes $x + 2 y + 3 z = 2$ and $x - y + z = 3$ and at a distance $\frac{2}{3}$ from the point $(3, 1, - 1)$ is

(A)

5 x - 11 y + z = 17

(B)

2 x + y = 32 - 1

(C)

x + y + z = 3

(D)

x - 2 y = 1 - 2

Answer:A

MathematicsEquation of a PlaneJEE Advanced 2010Moderate

View in:English Hindi

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)

x + 2 y - 2 z = 0

(B)

3 x + 2 y - 2 z = 0

(C)

x - 2 y + z = 0

(D)

5 x + 2 y - 4 z = 0

Answer:C

MathematicsEquation of a PlaneJEE Main 2002Moderate

View in:English Hindi

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

(A)

x - y + z = 1

(B)

x + y + z = 5

(C)

x + 2 y - z = 1

(D)

2 x - y + z = 5

Answer:A

MathematicsEquation of a PlaneJEE Main 2015Moderate

View in:English Hindi

The equation of the plane containing the line $2 x - 5 y + z = 3$ ; $x + y + 4 z = 5$ , and parallel to the plane, $x + 3 y + 6 z = 1$ , is:

(A)

x + 3 y + 6 z = 7

(B)

2 x + 6 y + 12 z = - 13

(C)

2 x + 6 y + 12 z = 13

(D)

x + 3 y + 6 z = - 7

Answer:C

MathematicsEquation of a PlaneJEE Advanced 2012Moderate

View in:English Hindi

If the straight lines $\frac{x - 1}{2} = \frac{y + 1}{k} = \frac{z}{2}$ and $\frac{x + 1}{5} = \frac{y + 1}{2} = \frac{z}{k}$ are coplanar, then the plane(s) containing these two lines is (are)

* Multiple Correct Options

(A)

y + 2 z = - 1

(B)

y + z = - 1

(C)

y - z = - 1

(D)

y - 2 z = - 1

Answer:B, C

MathematicsShortest Distance Between Two Skew LinesJEE Advanced 2008Moderate

View in:English Hindi

Comprehension Passage

Consider the lines

L_{1} : \frac{x + 1}{3} = \frac{y + 2}{1} = \frac{z + 1}{2}

and

L_{2} : \frac{x - 2}{1} = \frac{y + 2}{2} = \frac{z - 3}{3}

Question 1:

The unit vector perpendicular to both $L_{1}$ and $L_{2}$ is

(A)

\frac{- i ^ + 7 j ^ + 7 k ^}{99}

(B)

\frac{- i ^ - 7 j ^ + 5 k ^}{5 3}

(C)

\frac{- i ^ + 7 j ^ + 5 k ^}{5 3}

(D)

\frac{7 i ^ - 7 j ^ - k ^}{99}

Answer:B

Question 2:

The shortest distance between $L_{1}$ and $L_{2}$ is

(A)

(B)

\frac{17}{3}

(C)

\frac{41}{5 3}

(D)

\frac{17}{5 3}

Answer:D

Question 3:

The distance of the point (1, 1, 1) from the plane passing through the point (-1, -2, -1) and whose normal is perpendicular to both the lines $L_{1}$ and $L_{2}$ is

(A)

\frac{2}{75}

(B)

\frac{7}{75}

(C)

\frac{13}{75}

(D)

\frac{23}{75}

Answer:C

MathematicsEquation of a PlaneJEE Advanced 2015Moderate

View in:English Hindi

In $R^{3}$ , consider the planes $P_{1} : y = 0$ and $P_{2} : x + z = 1$ . Let $P_{3}$ be the plane, different from $P_{1}$ and $P_{2}$ , which passes through the intersection of $P_{1}$ and $P_{2}$ . If the distance of the point $(0, 1, 0)$ from $P_{3}$ is 1 and the distance of a point $(α, β, γ)$ from $P_{3}$ is 2, then which of the following relations is (are) true?

* Multiple Correct Options

(A)

2 α + β + 2 γ + 2 = 0

(B)

2 α - β + 2 γ + 4 = 0

(C)

2 α + β - 2 γ - 10 = 0

(D)

2 α - β + 2 γ - 8 = 0

Answer:B, D

MathematicsEquation of a PlaneJEE Advanced 2008Moderate

View in:English Hindi

Consider three planes $P_{1} : x - y + z = 1$ , $P_{2} : x + y - z = 1$ , $P_{3} : x - 3 y + 3 z = 2$ . Let $L_{1}, L_{2}, L_{3}$ be the lines of intersection of the planes $P_{2}$ and $P_{3}$ , $P_{3}$ and $P_{1}$ , $P_{1}$ and $P_{2}$ , respectively. STATEMENT-1 : At least two of the lines $L_{1}, L_{2}$ and $L_{3}$ are non-parallel and STATEMENT-2 : The three planes do not have a common point.

(A)

STATEMENT - 1 is True, STATEMENT - 2 is True; STATEMENT - 2 is a correct explanation for STATEMENT - 1

(B)

STATEMENT - 1 is True, STATEMENT - 2 is True; STATEMENT - 2 is NOT a correct explanation for STATEMENT - 1

(C)

STATEMENT - 1 is True, STATEMENT - 2 is False

(D)

STATEMENT - 1 is False, STATEMENT - 2 is True

Answer:D

MathematicsEquation of a Line in SpaceJEE Advanced 2007Moderate

View in:English Hindi

Consider the planes $3 x - 6 y - 2 z = 15$ and $2 x + y - 2 z = 5$ . STATEMENT-1 : The parametric equations of the line of intersection of the given planes are $x = 3 + 14 t, y = 1 + 2 t, z = 15 t$ . because STATEMENT-2 : The vector $14 \hat{i} + 2 \hat{j} + 15 \hat{k}$ is parallel to the line of intersection of given planes.

(A)

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

(B)

Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

(C)

Statement-1 is True, Statement-2 is False

(D)

Statement-1 is False, Statement-2 is True.

Answer:D

MathematicsEquation of a PlaneJEE Advanced 2005Moderate

View in:English Hindi

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 $(2, 1, - 1)$ .

Answer:

62 x + 29 y + 19 z - 105 = 0

List-I

List-II

Visualized Solution (English)

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

Defining General Points

Equating the Coordinates

Solving for $λ$ and $μ$

Finding the Intersection Point $P$

The Logic of the Normal Vector

Setting up the Cross Product

Calculating the Normal Vector

Simplifying Direction Ratios

The Equation of the Plane

Substitution and Expansion

The Final Plane Equation

Matching the Results

Conceptually Similar Problems

The equation of a plane passing through the line of intersection of the planes $x + 2 y + 3 z = 2$ and $x - y + z = 3$ and at a distance $\frac{2}{3}$ from the point $(3, 1, - 1)$ 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

The equation of the plane containing the line $2 x - 5 y + z = 3$ ; $x + y + 4 z = 5$ , and parallel to the plane, $x + 3 y + 6 z = 1$ , is:

If the straight lines $\frac{x - 1}{2} = \frac{y + 1}{k} = \frac{z}{2}$ and $\frac{x + 1}{5} = \frac{y + 1}{2} = \frac{z}{k}$ are coplanar, then the plane(s) containing these two lines is (are)

Comprehension Passage

The unit vector perpendicular to both $L_{1}$ and $L_{2}$ is

The shortest distance between $L_{1}$ and $L_{2}$ is

The distance of the point (1, 1, 1) from the plane passing through the point (-1, -2, -1) and whose normal is perpendicular to both the lines $L_{1}$ and $L_{2}$ is

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 $(2, 1, - 1)$ .

List-I

List-II

The equation of a plane passing through the line of intersection of the planes $x + 2 y + 3 z = 2$ and $x - y + z = 3$ and at a distance $\frac{2}{3}$ from the point $(3, 1, - 1)$ 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

The equation of the plane containing the line $2 x - 5 y + z = 3$ ; $x + y + 4 z = 5$ , and parallel to the plane, $x + 3 y + 6 z = 1$ , is:

If the straight lines $\frac{x - 1}{2} = \frac{y + 1}{k} = \frac{z}{2}$ and $\frac{x + 1}{5} = \frac{y + 1}{2} = \frac{z}{k}$ are coplanar, then the plane(s) containing these two lines is (are)

Comprehension Passage

The unit vector perpendicular to both $L_{1}$ and $L_{2}$ is

The shortest distance between $L_{1}$ and $L_{2}$ is

The distance of the point (1, 1, 1) from the plane passing through the point (-1, -2, -1) and whose normal is perpendicular to both the lines $L_{1}$ and $L_{2}$ is

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 $(2, 1, - 1)$ .

List-I

List-II

Visualized Solution (English)

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​

Defining General Points

Equating the Coordinates

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

Finding the Intersection Point .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

The Equation of the Plane

Substitution and Expansion

The Final Plane Equation

Matching the Results

Conceptually Similar Problems

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of a plane passing through the line of intersection of the planes x+2y+3z=2 and x−y+z=3 and at a distance 3​2​ from the point (3,1,−1) is

.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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of the plane containing the line 2x−5y+z=3; x+y+4z=5, and parallel to the plane, x+3y+6z=1, is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the straight lines 2x−1​=ky+1​=2z​ and 5x+1​=2y+1​=kz​ are coplanar, then the plane(s) containing these two lines is (are)

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The unit vector perpendicular to both L1​ and L2​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The shortest distance between L1​ and L2​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The distance of the point (1, 1, 1) from the plane passing through the point (-1, -2, -1) and whose normal is perpendicular to both the lines L1​ and L2​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find the equation of the plane containing the line 2x−y+z−3=0,3x+y+z=5 and at a distance of 6​1​ from the point (2,1,−1).

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of a plane passing through the line of intersection of the planes x+2y+3z=2 and x−y+z=3 and at a distance 3​2​ from the point (3,1,−1) is

.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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of the plane containing the line 2x−5y+z=3; x+y+4z=5, and parallel to the plane, x+3y+6z=1, is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the straight lines 2x−1​=ky+1​=2z​ and 5x+1​=2y+1​=kz​ are coplanar, then the plane(s) containing these two lines is (are)

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The unit vector perpendicular to both L1​ and L2​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The shortest distance between L1​ and L2​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The distance of the point (1, 1, 1) from the plane passing through the point (-1, -2, -1) and whose normal is perpendicular to both the lines L1​ and L2​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find the equation of the plane containing the line 2x−y+z−3=0,3x+y+z=5 and at a distance of 6​1​ from the point (2,1,−1).

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

Solving for $λ$ and $μ$

Finding the Intersection Point $P$

The equation of a plane passing through the line of intersection of the planes $x + 2 y + 3 z = 2$ and $x - y + z = 3$ and at a distance $\frac{2}{3}$ from the point $(3, 1, - 1)$ 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

The equation of the plane containing the line $2 x - 5 y + z = 3$ ; $x + y + 4 z = 5$ , and parallel to the plane, $x + 3 y + 6 z = 1$ , is:

If the straight lines $\frac{x - 1}{2} = \frac{y + 1}{k} = \frac{z}{2}$ and $\frac{x + 1}{5} = \frac{y + 1}{2} = \frac{z}{k}$ are coplanar, then the plane(s) containing these two lines is (are)

The unit vector perpendicular to both $L_{1}$ and $L_{2}$ is

The shortest distance between $L_{1}$ and $L_{2}$ is

The distance of the point (1, 1, 1) from the plane passing through the point (-1, -2, -1) and whose normal is perpendicular to both the lines $L_{1}$ and $L_{2}$ is

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 a plane passing through the line of intersection of the planes $x + 2 y + 3 z = 2$ and $x - y + z = 3$ and at a distance $\frac{2}{3}$ from the point $(3, 1, - 1)$ 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

The equation of the plane containing the line $2 x - 5 y + z = 3$ ; $x + y + 4 z = 5$ , and parallel to the plane, $x + 3 y + 6 z = 1$ , is:

If the straight lines $\frac{x - 1}{2} = \frac{y + 1}{k} = \frac{z}{2}$ and $\frac{x + 1}{5} = \frac{y + 1}{2} = \frac{z}{k}$ are coplanar, then the plane(s) containing these two lines is (are)

The unit vector perpendicular to both $L_{1}$ and $L_{2}$ is

The shortest distance between $L_{1}$ and $L_{2}$ is

The distance of the point (1, 1, 1) from the plane passing through the point (-1, -2, -1) and whose normal is perpendicular to both the lines $L_{1}$ and $L_{2}$ is

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 $(2, 1, - 1)$ .