Consider the lines given by L1 : x + 3y...

MathematicsVarious Forms of Equations of a LineJEE Advanced 2008Moderate

Consider the lines given by $L_{1} : x + 3 y - 5 = 0$ ; $L_{2} : 3 x - k y - 1 = 0$ ; $L_{3} : 5 x + 2 y - 12 = 0$ . Match the Statements / Expressions in Column I with the Statements / Expressions in Column II and indicate your answer by darkening the appropriate bubbles in the $4 \times 4$ matrix given in the ORS.

List-I

(P)

L_{1}, L_{2}, L_{3}

are concurrent, if

(Q)

One of

L_{1}, L_{2}, L_{3}

is parallel to at least one of the other two, if

(R)

L_{1}, L_{2}, L_{3}

form a triangle, if

(S)

L_{1}, L_{2}, L_{3}

do not form a triangle, if

List-II

(1)

k = - 9

(2)

k = - \frac{6}{5}

(3)

k = \frac{5}{6}

(4)

k = 5

Answer:P → 4, P → 2, R → 3, S → NaN

Visualized Solution (Hindi)

Defining the Three Lines

Given lines:
Line $L_{1} : x + 3 y - 5 = 0$
Line $L_{2} : 3 x - k y - 1 = 0$
Line $L_{3} : 5 x + 2 y - 12 = 0$

Condition for Concurrency

Lines $L_{1}, L_{2}, L_{3}$ are concurrent if:
$135 3 - k 2 - 5 - 1 - 12 = 0$

Solving for Concurrency (.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k = 5)

Expanding the determinant:
$1 (12 k + 2) - 3 (- 36 + 5) - 5 (6 + 5 k) = 0$
$12 k + 2 + 93 - 30 - 25 k = 0$
$- 13 k + 65 = 0 \Rightarrow k = 5$
Thus, (A) matches with (s).

Checking for Parallelism (.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L_{1} ∥ L_{2})

For $L_{1} ∥ L_{2}$ :
$\frac{1}{3} = \frac{3}{- k} \Rightarrow k = - 9$
This matches with (p).

Checking for Parallelism (.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L_{2} ∥ L_{3})

For $L_{2} ∥ L_{3}$ :
$\frac{3}{5} = \frac{- k}{2} \Rightarrow k = - \frac{6}{5}$
This matches with (q).

Condition to Form a Triangle

$L_{1}, L_{2}, L_{3}$ form a triangle if:
1. No two lines are parallel ( $k \neq = - 9, - \frac{6}{5}$ )
2. Lines are not concurrent ( $k \neq = 5$ )
Possible value from Column II is $k = \frac{5}{6}$ .
Thus, (C) matches with (r).

Condition for Not Forming a Triangle

$L_{1}, L_{2}, L_{3}$ do not form a triangle if:
1. They are concurrent ( $k = 5$ )
2. At least two lines are parallel ( $k = - 9$ or $k = - \frac{6}{5}$ )
Thus, (D) matches with (p), (q), and (s).

Final Summary and Takeaway

Final Match:
(A) $\to$ (s)
(B) $\to$ (p), (q)
(C) $\to$ (r)
(D) $\to$ (p), (q), (s)
Key Takeaway: Triangle formation fails if lines are concurrent or parallel.

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Conceptually Similar Problems

MathematicsFamily of LinesJEE Advanced 1985Easy

View in:English Hindi

Three lines $p x + q y + r = 0, q x + r y + p = 0$ and $r x + p y + q = 0$ are concurrent if

* Multiple Correct Options

(A)

p + q + r = 0

(B)

p^{2} + q^{2} + r^{2} = q r + r p + pq

(C)

p^{3} + q^{3} + r^{3} = 3 pq r

(D)

none of these.

Answer:A, B, C

MathematicsSolution of System of Linear Equations (Matrix Method and Cramer's Rule)JEE Advanced 2007Moderate

View in:English Hindi

Consider the following linear equations $a x + b y + cz = 0; b x + cy + a z = 0; c x + a y + b z = 0$ Match the conditions/expressions in Column I with statements in Column II and indicate your answer by darkening the appropriate bubbles in the $4 \times 4$ matrix given in the ORS.

List-I

(P)

a + b + c \neq = 0

and

a^{2} + b^{2} + c^{2} = ab + b c + c a

(Q)

a + b + c = 0

and

a^{2} + b^{2} + c^{2} \neq = ab + b c + c a

(R)

a + b + c \neq = 0

and

a^{2} + b^{2} + c^{2} \neq = ab + b c + c a

(S)

a + b + c = 0

and

a^{2} + b^{2} + c^{2} = ab + b c + c a

List-II

(1)

the equations represent planes meeting only at a single point

(2)

the equations represent the line

x = y = z

(3)

the equations represent identical planes.

(4)

the equations represent the whole of the three dimensional space.

Answer:P → 3, Q → 2, R → 1, S → 4

MathematicsSolution of System of Linear Equations (Matrix Method and Cramer's Rule)JEE Advanced 2008Easy

View in:English Hindi

Consider the system of equations $x - 2 y + 3 z = - 1$ , $- x + y - 2 z = k$ , $x - 3 y + 4 z = 1$ . STATEMENT - 1 : The system of equations has no solution for $k \neq = 3$ and STATEMENT - 2 : The determinant $1 - 1 1 3 - 2 4 - 1 k 1 \neq = 0$ , for $k \neq = 3$ .

(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:A

MathematicsAngle Between Two LinesJEE Main 2011Moderate

View in:English Hindi

The lines $L_{1} : y - x = 0$ and $L_{2} : 2 x + y = 0$ intersect the line $L_{3} : y + 2 = 0$ at $P$ and $Q$ respectively. The bisector of the acute angle between $L_{1}$ and $L_{2}$ intersects $L_{3}$ at $R$ . Statement-1: The ratio $P R : R Q$ equals $22 : 5$ . Statement-2: In any triangle, bisector of an angle divides the triangle into two similar triangles.

(A)

Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

(B)

Statement-1 is true, Statement-2 is false.

(C)

Statement-1 is false, Statement-2 is true.

(D)

Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Answer:B

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 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

MathematicsShortest Distance Between Two Skew LinesJEE Main 2003Moderate

View in:English Hindi

The lines $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{- k}$ and $\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}$ are coplanar if

(A)

k = 3

- 2

(B)

k = 0

- 1

(C)

k = 1

- 1

(D)

k = 0

- 3

Answer:D

MathematicsShortest Distance Between Two Skew LinesJEE Advanced 2004SModerate

View in:English Hindi

If the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$ intersect, then the value of $k$ is

(A)

3/2

(B)

9/2

(C)

- 2/9

(D)

- 3/2

Answer:B

MathematicsAngle Between Two LinesJEE Advanced 2007Moderate

View in:English Hindi

Lines $L_{1} : y - x = 0$ and $L_{2} : 2 x + y = 0$ intersect the line $L_{3} : y + 2 = 0$ at $P$ and $Q$ , respectively. The bisector of the acute angle between $L_{1}$ and $L_{2}$ intersects $L_{3}$ at $R$ . STATEMENT-1 : The ratio $P R : R Q$ equals $22 : 5$ . because STATEMENT-2 : In any triangle, bisector of an angle divides the triangle into two similar triangles.

(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:C

MathematicsShortest Distance Between Two Skew LinesJEE Main 2012Moderate

View in:English Hindi

If the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$ intersect, then $k$ is equal to:

(A)

- 1

(B)

\frac{2}{9}

(C)

\frac{9}{2}

(D)

0

Answer:C

List-I

List-II

Visualized Solution (Hindi)

Defining the Three Lines

Condition for Concurrency

Solving for Concurrency ( $k = 5$ )

Checking for Parallelism ( $L_{1} ∥ L_{2}$ )

Checking for Parallelism ( $L_{2} ∥ L_{3}$ )

Condition to Form a Triangle

Condition for Not Forming a Triangle

Final Summary and Takeaway

Conceptually Similar Problems

Three lines $p x + q y + r = 0, q x + r y + p = 0$ and $r x + p y + q = 0$ are concurrent if

Consider the following linear equations $a x + b y + cz = 0; b x + cy + a z = 0; c x + a y + b z = 0$ Match the conditions/expressions in Column I with statements in Column II and indicate your answer by darkening the appropriate bubbles in the $4 \times 4$ matrix given in the ORS.

List-I

List-II

Consider the system of equations $x - 2 y + 3 z = - 1$ , $- x + y - 2 z = k$ , $x - 3 y + 4 z = 1$ . STATEMENT - 1 : The system of equations has no solution for $k \neq = 3$ and STATEMENT - 2 : The determinant $1 - 1 1 3 - 2 4 - 1 k 1 \neq = 0$ , for $k \neq = 3$ .

List-I

List-II

The lines $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{- k}$ and $\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}$ are coplanar if

If the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$ intersect, then the value of $k$ is

If the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$ intersect, then $k$ is equal to:

List-I

List-II

Three lines $p x + q y + r = 0, q x + r y + p = 0$ and $r x + p y + q = 0$ are concurrent if

Consider the following linear equations $a x + b y + cz = 0; b x + cy + a z = 0; c x + a y + b z = 0$ Match the conditions/expressions in Column I with statements in Column II and indicate your answer by darkening the appropriate bubbles in the $4 \times 4$ matrix given in the ORS.

List-I

List-II

Consider the system of equations $x - 2 y + 3 z = - 1$ , $- x + y - 2 z = k$ , $x - 3 y + 4 z = 1$ . STATEMENT - 1 : The system of equations has no solution for $k \neq = 3$ and STATEMENT - 2 : The determinant $1 - 1 1 3 - 2 4 - 1 k 1 \neq = 0$ , for $k \neq = 3$ .

List-I

List-II

The lines $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{- k}$ and $\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}$ are coplanar if

If the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$ intersect, then the value of $k$ is

If the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$ intersect, then $k$ is equal to:

List-I

List-II

Visualized Solution (Hindi)

Defining the Three Lines

Condition for Concurrency

Solving for Concurrency ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k=5)

Checking for Parallelism ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​∥L2​)

Checking for Parallelism ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​∥L3​)

Condition to Form a Triangle

Condition for Not Forming a Triangle

Final Summary and Takeaway

Conceptually Similar Problems

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Three lines px+qy+r=0,qx+ry+p=0 and rx+py+q=0 are concurrent if

List-I

List-II

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The lines 1x−2​=1y−3​=−kz−4​ and kx−1​=2y−4​=1z−5​ are coplanar if

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the lines 2x−1​=3y+1​=4z−1​ and 1x−3​=2y−k​=1z​ intersect, then the value of k is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the line 2x−1​=3y+1​=4z−1​ and 1x−3​=2y−k​=1z​ intersect, then k is equal to:

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Three lines px+qy+r=0,qx+ry+p=0 and rx+py+q=0 are concurrent if

List-I

List-II

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The lines 1x−2​=1y−3​=−kz−4​ and kx−1​=2y−4​=1z−5​ are coplanar if

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the lines 2x−1​=3y+1​=4z−1​ and 1x−3​=2y−k​=1z​ intersect, then the value of k is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the line 2x−1​=3y+1​=4z−1​ and 1x−3​=2y−k​=1z​ intersect, then k is equal to:

Solving for Concurrency ( $k = 5$ )

Checking for Parallelism ( $L_{1} ∥ L_{2}$ )

Checking for Parallelism ( $L_{2} ∥ L_{3}$ )

Three lines $p x + q y + r = 0, q x + r y + p = 0$ and $r x + p y + q = 0$ are concurrent if

The lines $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{- k}$ and $\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}$ are coplanar if

If the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$ intersect, then the value of $k$ is

If the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$ intersect, then $k$ is equal to:

Three lines $p x + q y + r = 0, q x + r y + p = 0$ and $r x + p y + q = 0$ are concurrent if

The lines $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{- k}$ and $\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}$ are coplanar if

If the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$ intersect, then the value of $k$ is

If the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$ intersect, then $k$ is equal to: