Angle Between Two Lines

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JEE Main 2013

LEVELJEE Main

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, then $k$ can have

The two lines $x = a y + b, z = cy + d$ ; and $x = a^{'} y + b^{'}, z = c^{'} y + d^{'}$ are perpendicular to each other if

(A)

a a^{'} + c c^{'} = - 1

(B)

a a^{'} + c c^{'} = 1

(C)

\frac{a}{a ^{'}} + \frac{c}{c ^{'}} = - 1

(D)

\frac{a}{a ^{'}} + \frac{c}{c ^{'}} = 1

JEE Main 2005

LEVELJEE Main

The angle between the lines $2 x = 3 y = - z$ and $6 x = - y = - 4 z$ is

(A)

0^{\circ}

(B)

9 0^{\circ}

(C)

4 5^{\circ}

(D)

3 0^{\circ}

JEE Main 2003

LEVELJEE Main

The two lines $x = a y + b, z = cy + d$ and $x = a^{'} y + b^{'}, z = c^{'} y + d^{'}$ will be perpendicular, if and only if

(A)

a a^{'} + c c^{'} + 1 = 0

(B)

a a^{'} + b b^{'} + c c^{'} + 1 = 0

(C)

a a^{'} + b b^{'} + c c^{'} = 0

(D)

(a + a^{'}) (b + b^{'}) + (c + c^{'}) = 0

JEE Main 2018 (15 April Evening)

LEVELJEE Main

An angle between the lines whose direction cosines are given by the equations, $l + 3 m + 5 n = 0$ and $5 l m - 2 mn + 6 n l = 0$ , is :-

(A)

cos^{- 1} (\frac{1}{8})

(B)

cos^{- 1} (\frac{1}{3})

(C)

cos^{- 1} (\frac{1}{4})

(D)

cos^{- 1} (\frac{1}{6})

JEE Main 2018 (16 April Shift 1)

LEVELJEE Main

If the angle between the lines, $\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$ and $\frac{5 - x}{- 2} = \frac{7 y - 14}{p} = \frac{z - 3}{4}$ is $cos^{- 1} (\frac{2}{3})$ , then $p$ is equal to :

(A)

- \frac{4}{7}

(B)

\frac{7}{2}

(C)

- \frac{7}{4}

(D)

\frac{2}{7}

JEE Main 2021 (25 July Shift 1)

LEVELJEE Advanced

Let the foot of perpendicular from a point $P (1, 2, - 1)$ to the straight line $L : \frac{x}{1} = \frac{y}{0} = \frac{z}{- 1}$ be $N$ . Let a line be drawn from $P$ parallel to the plane $x + y + 2 z = 0$ which meets $L$ at point $Q$ . If $α$ is the acute angle between the lines $P N$ and $P Q$ , then $cos α$ is equal to

(A)

\frac{1}{5}

(B)

\frac{3}{2}

(C)

\frac{1}{3}

(D)

\frac{1}{2 3}

JEE Main 2022 (26 June Shift 1)

LEVELJEE Main

If the two lines $l_{1} : \frac{x - 2}{3} = \frac{y + 1}{- 2}, z = 2$ and $l_{2} : \frac{x - 1}{1} = \frac{2 y + 3}{α} = \frac{z + 5}{2}$ are perpendicular, then an angle between the lines $l_{2}$ and $l_{3} : \frac{1 - x}{3} = \frac{2 y - 1}{- 4} = \frac{z}{4}$ is:

(A)

cos^{- 1} (\frac{29}{4})

(B)

sec^{- 1} (\frac{29}{4})

(C)

cos^{- 1} (\frac{2}{29})

(D)

cos^{- 1} (\frac{2}{29})

JEE Main 2024 (05 Apr Shift 1)

LEVELJEE Main

If the line $\frac{2 - x}{3} = \frac{3 y - 2}{4 λ + 1} = 4 - z$ makes a right angle with the line $\frac{x + 3}{3 μ} = \frac{1 - 2 y}{6} = \frac{5 - z}{7}$ , then $4 λ + 9 μ$ is equal to :

JEE Main 2024 (29 Jan Shift 2)

LEVELBoard

Let $P (3, 2, 3), Q (4, 6, 2)$ and $R (7, 3, 2)$ be the vertices of $△ P QR$ . Then, the angle $∠ QP R$ is

(A)

\frac{π}{6}

(B)

cos^{- 1} (\frac{7}{18})

(C)

cos^{- 1} (\frac{1}{18})

(D)

\frac{π}{3}

JEE Main 2025 April

LEVELJEE Main

Let $A$ be the point of intersection of the lines $L_{1} : \frac{x - 7}{1} = \frac{y - 5}{0} = \frac{z - 3}{- 1}$ and $L_{2} : \frac{x - 1}{3} = \frac{y + 3}{4} = \frac{z + 7}{5}$ . Let $B$ and $C$ be the point on the lines $L_{1}$ and $L_{2}$ respectively such that $A B = A C = 15$ . Then the square of the area of the triangle $A B C$ is :

(A)

54

(B)

63

(C)

57

(D)

60

JEE Advanced 2020

LEVELJEE Main

Let $L_{1}$ and $L_{2}$ be the following straight line. $L_{1} : \frac{x - 1}{1} = \frac{y}{- 1} = \frac{z - 1}{3}$ and $L_{2} : \frac{x - 1}{- 3} = \frac{y}{- 1} = \frac{z - 1}{1}$ . Suppose the straight line $L : \frac{x - α}{l} = \frac{y - 1}{m} = \frac{z - γ}{- 2}$ lies in the plane containing $L_{1}$ and $L_{2}$ , and passes through the point of intersection of $L_{1}$ and $L_{2}$ . If the line $L$ bisects the acute angle between the lines $L_{1}$ and $L_{2}$ , then which of the following statements is/are TRUE?

* Multiple Correct Options

(A)

α - γ = 3

(B)

l + m = 2

(C)

α - γ = 1

(D)

l + m = 0

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The two lines x=ay+b,z=cy+d; and x=a′y+b′,z=c′y+d′ are perpendicular to each other if

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The angle between the lines 2x=3y=−z and 6x=−y=−4z is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The two lines x=ay+b,z=cy+d and x=a′y+b′,z=c′y+d′ will be perpendicular, if and only if

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } An angle between the lines whose direction cosines are given by the equations, l+3m+5n=0 and 5lm−2mn+6nl=0, is :-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the angle between the lines, 2x​=2y​=1z​ and −25−x​=p7y−14​=4z−3​ is cos−1(32​), then p is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the line 32−x​=4λ+13y−2​=4−z makes a right angle with the line 3μx+3​=61−2y​=75−z​, then 4λ+9μ is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let P(3,2,3),Q(4,6,2) and R(7,3,2) be the vertices of △PQR. Then, the angle ∠QPR is

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, then $k$ can have

The two lines $x = a y + b, z = cy + d$ ; and $x = a^{'} y + b^{'}, z = c^{'} y + d^{'}$ are perpendicular to each other if

The angle between the lines $2 x = 3 y = - z$ and $6 x = - y = - 4 z$ is

The two lines $x = a y + b, z = cy + d$ and $x = a^{'} y + b^{'}, z = c^{'} y + d^{'}$ will be perpendicular, if and only if

An angle between the lines whose direction cosines are given by the equations, $l + 3 m + 5 n = 0$ and $5 l m - 2 mn + 6 n l = 0$ , is :-

If the angle between the lines, $\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$ and $\frac{5 - x}{- 2} = \frac{7 y - 14}{p} = \frac{z - 3}{4}$ is $cos^{- 1} (\frac{2}{3})$ , then $p$ is equal to :

If the line $\frac{2 - x}{3} = \frac{3 y - 2}{4 λ + 1} = 4 - z$ makes a right angle with the line $\frac{x + 3}{3 μ} = \frac{1 - 2 y}{6} = \frac{5 - z}{7}$ , then $4 λ + 9 μ$ is equal to :

Let $P (3, 2, 3), Q (4, 6, 2)$ and $R (7, 3, 2)$ be the vertices of $△ P QR$ . Then, the angle $∠ QP R$ is