Equation of a Line in Space

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JEE Main 2011

LEVELJEE Main

Statement-1: The point $A (1, 0, 7)$ is the mirror image of the point $B (1, 6, 3)$ in the line: $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ . Statement-2: The line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ bisects the line segment joining $A (1, 0, 7)$ and $B (1, 6, 3)$ .

(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.

JEE Main 2008

LEVELJEE Main

The line passing through the points $(5, 1, a)$ and $(3, b, 1)$ crosses the $y z$ -plane at the point $(0, \frac{17}{2}, - \frac{13}{2})$ . Then

(A)

a = 2, b = 8

(B)

a = 4, b = 6

(C)

a = 6, b = 4

(D)

a = 8, b = 2

JEE Main 2007

LEVELJEE Main

If $(2, 3, 5)$ is one end of a diameter of the sphere $x^{2} + y^{2} + z^{2} - 6 x - 12 y - 2 z + 20 = 0$ , then the coordinates of the other end of the diameter are

(A)

(4, 3, 5)

(B)

(4, 3, - 3)

(C)

(4, 9, - 3)

(D)

(4, - 3, 3)

JEE Main 2007

LEVELJEE Main

Let $L$ be the line of intersection of the planes $2 x + 3 y + z = 1$ and $x + 3 y + 2 z = 2$ . If $L$ makes an angle $α$ with the positive $x$ -axis, then $cos α$ equals

(A)

(B)

\frac{1}{2}

(C)

\frac{1}{3}

(D)

\frac{1}{2}

JEE Main 2004

LEVELJEE Main

A line with direction cosines proportional to 2, 1, 2 meets each of the lines $x = y + a = z$ and $x + a = 2 y = 2 z$ . The co-ordinates of each of the points of intersection are given by

(A)

(2 a, 3 a, 3 a), (2 a, a, a)

(B)

(3 a, 2 a, 3 a), (a, a, a)

(C)

(3 a, 2 a, 3 a), (a, a, 2 a)

(D)

(a, 3 a, 3 a), (a, a, a)

JEE Main 2004

LEVELJEE Main

If the straight lines $x = 1 + s, y = - 3 - λ s, z = 1 + λ s$ and $x = \frac{t}{2}, y = 1 + t, z = 2 - t$ , with parameters $s$ and $t$ respectively, are co-planar, then $λ$ equals.

In $R^{3}$ , let $L$ be a straight line passing through the origin. Suppose that all the points on $L$ are at a constant distance from the two planes $P_{1} : x + 2 y - z + 1 = 0$ and $P_{2} : 2 x - y + z - 1 = 0$ . Let $M$ be the locus of the feet of the perpendiculars drawn from the points on $L$ to the plane $P_{1}$ . Which of the following points lie(s) on $M$ ?

* Multiple Correct Options

(A)

(0, - \frac{5}{6}, - \frac{2}{3})

(B)

(- \frac{1}{6}, - \frac{1}{6}, \frac{1}{6})

(C)

(- \frac{5}{6}, 0, \frac{1}{6})

(D)

(- \frac{1}{3}, 0, \frac{2}{3})

JEE Advanced 2014

LEVELJEE Advanced

From a point $P (λ, λ, λ)$ , perpendicular $P Q$ and $P R$ are drawn respectively on the lines $y = x, z = 1$ and $y = - x, z = - 1$ . If $P$ is such that $∠ QP R$ is a right angle, then the possible value(s) of $λ$ is/(are)

(A)

2

- 2

JEE Advanced 2013

LEVELJEE Advanced

A line $l$ passing through the origin is perpendicular to the lines $l_{1} : (3 + t) \hat{i} + (- 1 + 2 t) \hat{j} + (4 + 2 t) \hat{k}, - \infty < t < \infty$ and $l_{2} : (3 + 2 s) \hat{i} + (3 + 2 s) \hat{j} + (2 + s) \hat{k}, - \infty < s < \infty$ . Then, the coordinate(s) of the point(s) on $l_{2}$ at a distance of $17$ from the point of intersection of $l$ and $l_{1}$ is (are)

* Multiple Correct Options

(A)

(\frac{7}{3}, \frac{7}{3}, \frac{5}{3})

(B)

(- 1, - 1, 0)

(C)

(1, 1, 1)

(D)

(\frac{7}{9}, \frac{7}{9}, \frac{8}{9})

JEE Advanced 2013

LEVELJEE Main

Two lines $L_{1} : x = 5, \frac{y}{3 - α} = \frac{z}{- 2}$ and $L_{2} : x = α, \frac{y}{- 1} = \frac{z}{2 - α}$ are coplanar. Then $α$ can take value(s)

* Multiple Correct Options

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.

JEE Advanced 1996

LEVELJEE Advanced

The position vectors of the vertices $A, B$ and $C$ of a tetrahedron $A B C D$ are $\hat{i} + \hat{j} + \hat{k}, \hat{i}$ and $3 \hat{i}$ , respectively. The altitude from vertex $D$ to the opposite face $A B C$ meets the median line through $A$ of the triangle $A B C$ at a point $E$ . If the length of the side $A D$ is $4$ and the volume of the tetrahedron is $\frac{2 2}{3}$ , find the position vector of the point $E$ for all its possible positions.

JEE Advanced 2006

LEVELJEE Advanced

Match the following

List-I

(P)

\sum_{i = 1}^{\infty} tan^{- 1} (\frac{1}{2 i ^{2}}) = t

, then

tan t =

(Q)

Sides

a, b, c

of a triangle

A B C

are in AP and

cos θ_{1} = \frac{a}{b + c}

cos θ_{2} = \frac{b}{a + c}

cos θ_{3} = \frac{c}{a + b}

, then

tan^{2} (\frac{θ _{1}}{2}) + tan^{2} (\frac{θ _{3}}{2}) =

(R)

A line is perpendicular to

x + 2 y + 2 z = 0

and passes through

(0, 1, 0)

. The perpendicular distance of this line from the origin is

List-II

(1)

(2)

\frac{5}{3}

(3)

\frac{2}{3}

JEE Main 2019 (9 January)

LEVELJEE Advanced

The equation of the line passing through $(- 4, 3, 1)$ , parallel to the plane $x + 2 y - z - 5 = 0$ and intersecting the line $\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z - 2}{- 1}$ is:

(A)

\frac{x + 4}{- 1} = \frac{y - 3}{1} = \frac{z - 1}{1}

(B)

\frac{x + 4}{3} = \frac{y - 3}{- 1} = \frac{z - 1}{1}

(C)

\frac{x + 4}{1} = \frac{y - 3}{1} = \frac{z - 1}{3}

(D)

\frac{x - 4}{2} = \frac{y + 3}{1} = \frac{z + 1}{4}

JEE Main 2019 (9 January)

LEVELJEE Main

If the lines $x = a y + b, z = cy + d$ and $x = a^{'} z + b^{'}, y = c^{'} z + d^{'}$ are perpendicular, then:

(A)

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

(B)

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

(C)

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

(D)

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

JEE Main 2019 (8 April Shift 1)

LEVELJEE Main

The length of the perpendicular from the point $(2, - 1, 4)$ on the straight line, $\frac{x + 3}{10} = \frac{y - 2}{- 7} = \frac{z}{1}$ is :

(A)

less than 2

(B)

greater than 3 but less than 4

(C)

greater than 4

(D)

greater than 2 but less than 3

JEE Main 2019 (08 April Shift 2)

LEVELJEE Main

If a point R(4,y,z) lies on the line segment joining the points P(2,-3,4) and Q(8,0,10), then the distance of R from the origin is :

(A)

214

(B)

(C)

53

(D)

221

JEE Main 2019 (9 April)

LEVELJEE Main

The vertices B and C of a $Δ A B C$ lie on the line, $\frac{x + 2}{3} = \frac{y - 1}{0} = \frac{z}{4}$ such that $B C = 5$ units. Then the area (in sq. units) of this triangle, given that the point $A (1, - 1, 2)$ , is :-

(A)

234

(B)

34

(C)

6

(D)

517

JEE Main 2019 (10 April Shift 1)

LEVELJEE Main

If the length of the perpendicular from the point $(β, 0, β) (β \neq = 0)$ to the line $\frac{x - 1}{1} = \frac{y - 0}{0} = \frac{z + 1}{- 1}$ is $3/2$ , then $β$ is equal to :

JEE Main 2019 (10 April Shift 2)

LEVELJEE Main

The distance of the point having position vector $- \hat{i} + 2 \hat{j} + 6 \hat{k}$ from the straight line passing through the point $(2, 3, - 4)$ and parallel to the vector, $6 \hat{i} + 3 \hat{j} - 4 \hat{k}$ is :

JEE Main 11 Jan 2019 (Evening)

LEVELJEE Main

Two lines $\frac{x - 3}{1} = \frac{y + 1}{3} = \frac{z - 6}{- 1}$ and $\frac{x + 5}{7} = \frac{y - 2}{- 6} = \frac{z - 3}{4}$ intersect at the point $R$ . The reflection of $R$ in the xy-plane has coordinates:

(A)

(2, - 4, - 7)

(B)

(2, - 4, 7)

(C)

(- 2, 4, 7)

(D)

(2, 4, 7)

JEE Main 2020 (7 January Shift 1)

LEVELJEE Main

If the foot of perpendicular drawn from the point $(1, 0, 3)$ on a line passing through $(α, 7, 1)$ is $(\frac{5}{3}, \frac{7}{3}, \frac{17}{3})$ , then $α$ is equal to ________.

JEE Main 2020 - 7 Jan (Evening)

LEVELJEE Main

If the foot of the perpendicular drawn from the point $(1, 0, 3)$ on a line passing through $(α, 7, 1)$ is $(5/3, 7/3, 17/3)$ , then $α$ is equal to .....

JEE Main 2020 - 5 Sep (Morning)

LEVELJEE Main

If $(a, b, c)$ is the image of the point $(1, 2, - 3)$ in the line $\frac{x + 1}{2} = \frac{y - 3}{- 2} = \frac{z}{- 1}$ , then $a + b + c$ is equal to:

(A)

2

(B)

1

(C)

3

(D)

- 1

JEE Main 2021 (01 Sep Shift 2)

LEVELJEE Advanced

The distance of line $3 y - 2 z - 1 = 0 = 3 x - z + 4$ from the point $(2, - 1, 6)$ is :

(A)

26

(B)

25

(C)

26

(D)

42

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The line passing through the points (5,1,a) and (3,b,1) crosses the yz-plane at the point (0,217​,−213​). Then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If (2,3,5) is one end of a diameter of the sphere x2+y2+z2−6x−12y−2z+20=0, then the coordinates of the other end of the diameter are

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let L be the line of intersection of the planes 2x+3y+z=1 and x+3y+2z=2. If L makes an angle α with the positive x-axis, then cosα equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A line with direction cosines proportional to 2, 1, 2 meets each of the lines x=y+a=z and x+a=2y=2z. The co-ordinates of each of the points of intersection are given by

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the straight lines x=1+s,y=−3−λs,z=1+λs and x=2t​,y=1+t,z=2−t, with parameters s and t respectively, are co-planar, then λ equals.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Two lines L1​:x=5,3−αy​=−2z​ and L2​:x=α,−1y​=2−αz​ are coplanar. Then α can take value(s)

Match the following

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 the line passing through (−4,3,1), parallel to the plane x+2y−z−5=0 and intersecting the line −3x+1​=2y−3​=−1z−2​ is:

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The length of the perpendicular from the point (2,−1,4) on the straight line, 10x+3​=−7y−2​=1z​ is :

If a point R(4,y,z) lies on the line segment joining the points P(2,-3,4) and Q(8,0,10), then the distance of R from the origin is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the length of the perpendicular from the point (β,0,β)(β=0) to the line 1x−1​=0y−0​=−1z+1​ is 3/2​, then β is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Two lines 1x−3​=3y+1​=−1z−6​ and 7x+5​=−6y−2​=4z−3​ intersect at the point R. The reflection of R in the xy-plane has coordinates:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the foot of perpendicular drawn from the point (1,0,3) on a line passing through (α,7,1) is (35​,37​,317​), then α is equal to ________.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the foot of the perpendicular drawn from the point (1,0,3) on a line passing through (α,7,1) is (5/3,7/3,17/3), then α is equal to .....

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If (a,b,c) is the image of the point (1,2,−3) in the line 2x+1​=−2y−3​=−1z​, then 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 distance of line 3y−2z−1=0=3x−z+4 from the point (2,−1,6) is :

The line passing through the points $(5, 1, a)$ and $(3, b, 1)$ crosses the $y z$ -plane at the point $(0, \frac{17}{2}, - \frac{13}{2})$ . Then

If $(2, 3, 5)$ is one end of a diameter of the sphere $x^{2} + y^{2} + z^{2} - 6 x - 12 y - 2 z + 20 = 0$ , then the coordinates of the other end of the diameter are

Let $L$ be the line of intersection of the planes $2 x + 3 y + z = 1$ and $x + 3 y + 2 z = 2$ . If $L$ makes an angle $α$ with the positive $x$ -axis, then $cos α$ equals

A line with direction cosines proportional to 2, 1, 2 meets each of the lines $x = y + a = z$ and $x + a = 2 y = 2 z$ . The co-ordinates of each of the points of intersection are given by

If the straight lines $x = 1 + s, y = - 3 - λ s, z = 1 + λ s$ and $x = \frac{t}{2}, y = 1 + t, z = 2 - t$ , with parameters $s$ and $t$ respectively, are co-planar, then $λ$ equals.

Two lines $L_{1} : x = 5, \frac{y}{3 - α} = \frac{z}{- 2}$ and $L_{2} : x = α, \frac{y}{- 1} = \frac{z}{2 - α}$ are coplanar. Then $α$ can take value(s)

The equation of the line passing through $(- 4, 3, 1)$ , parallel to the plane $x + 2 y - z - 5 = 0$ and intersecting the line $\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z - 2}{- 1}$ is:

If the lines $x = a y + b, z = cy + d$ and $x = a^{'} z + b^{'}, y = c^{'} z + d^{'}$ are perpendicular, then:

The length of the perpendicular from the point $(2, - 1, 4)$ on the straight line, $\frac{x + 3}{10} = \frac{y - 2}{- 7} = \frac{z}{1}$ is :

If the length of the perpendicular from the point $(β, 0, β) (β \neq = 0)$ to the line $\frac{x - 1}{1} = \frac{y - 0}{0} = \frac{z + 1}{- 1}$ is $3/2$ , then $β$ is equal to :

Two lines $\frac{x - 3}{1} = \frac{y + 1}{3} = \frac{z - 6}{- 1}$ and $\frac{x + 5}{7} = \frac{y - 2}{- 6} = \frac{z - 3}{4}$ intersect at the point $R$ . The reflection of $R$ in the xy-plane has coordinates:

If the foot of perpendicular drawn from the point $(1, 0, 3)$ on a line passing through $(α, 7, 1)$ is $(\frac{5}{3}, \frac{7}{3}, \frac{17}{3})$ , then $α$ is equal to ________.

If the foot of the perpendicular drawn from the point $(1, 0, 3)$ on a line passing through $(α, 7, 1)$ is $(5/3, 7/3, 17/3)$ , then $α$ is equal to .....

If $(a, b, c)$ is the image of the point $(1, 2, - 3)$ in the line $\frac{x + 1}{2} = \frac{y - 3}{- 2} = \frac{z}{- 1}$ , then $a + b + c$ is equal to:

The distance of line $3 y - 2 z - 1 = 0 = 3 x - z + 4$ from the point $(2, - 1, 6)$ is :