Equation of a Plane

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JEE Main 2015

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:

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

JEE Main 2013

LEVELBoard

Distance between two parallel planes $2 x + y + 2 z = 8$ and $4 x + 2 y + 4 z + 5 = 0$ is

(A)

\frac{3}{2}

(B)

\frac{5}{2}

(C)

\frac{7}{2}

(D)

\frac{9}{2}

JEE Main 2012

LEVELJEE Main

A equation of a plane parallel to the plane $x - 2 y + 2 z - 5 = 0$ and at a unit distance from the origin is :

(A)

x - 2 y + 2 z - 3 = 0

(B)

x - 2 y + 2 z + 1 = 0

(C)

x - 2 y + 2 z - 1 = 0

(D)

x - 2 y + 2 z + 5 = 0

JEE Main 2006

LEVELJEE Main

The image of the point $(- 1, 3, 4)$ in the plane $x - 2 y = 0$ is

(A)

(15, 11, 4)

(B)

(- \frac{17}{3}, - \frac{19}{3}, 4)

(C)

( rac{17}{3}, -\frac{19}{3}, 1)

If the plane $2 a x - 3 a y + 4 a z + 6 = 0$ passes through the midpoint of the line joining the centres of the spheres $x^{2} + y^{2} + z^{2} + 6 x - 8 y - 2 z = 13$ and $x^{2} + y^{2} + z^{2} - 10 x + 4 y - 2 z = 8$ then $a$ equals

(A)

- 1

(B)

(C)

- 2

(D)

JEE Main 2005

LEVELJEE Main

The plane $x + 2 y - z = 4$ cuts the sphere $x^{2} + y^{2} + z^{2} - x + z - 2 = 0$ in a circle of radius

2

JEE Main 2004

LEVELJEE Main

Distance between two parallel planes $2 x + y + 2 z = 8$ and $4 x + 2 y + 4 z + 5 = 0$ is

The intersection of the spheres $x^{2} + y^{2} + z^{2} + 7 x - 2 y - z = 13$ and $x^{2} + y^{2} + z^{2} - 3 x + 3 y + 4 z = 8$ is the same as the intersection of one of the sphere and the plane

(A)

2 x - y - z = 1

(B)

x - 2 y - z = 1

(C)

x - y - 2 z = 1

(D)

x - y - z = 1

JEE Main 2003

LEVELJEE Main

The shortest distance from the plane $12 x + 4 y + 3 z = 327$ to the sphere $x^{2} + y^{2} + z^{2} + 4 x - 2 y - 6 z = 155$ is

(A)

(B)

(C)

11 \frac{4}{13}

(D)

JEE Main 2003

LEVELJEE Main

The radius of the circle in which the sphere $x^{2} + y^{2} + z^{2} + 2 x - 2 y - 4 z - 19 = 0$ is cut by the plane $x + 2 y + 2 z + 7 = 0$ is

Two system of rectangular axes have the same origin. If a plane cuts them at distances $a, b, c$ and $a^{'}, b^{'}, c^{'}$ from the origin then

(A)

\frac{1}{a ^{2}} + \frac{1}{b ^{2}} + \frac{1}{c ^{2}} - \frac{1}{a ^{'2}} - \frac{1}{b ^{'2}} - \frac{1}{c ^{'2}} = 0

(B)

\frac{1}{a ^{2}} + \frac{1}{b ^{2}} + \frac{1}{c ^{2}} + \frac{1}{a ^{'2}} + \frac{1}{b ^{'2}} + \frac{1}{c ^{'2}} = 0

(C)

\frac{1}{a ^{2}} + \frac{1}{b ^{2}} + \frac{1}{c ^{2}} - \frac{1}{a ^{'2}} - \frac{1}{b ^{'2}} - \frac{1}{c ^{'2}} = 0

(D)

\frac{1}{a ^{2}} - \frac{1}{b ^{2}} - \frac{1}{c ^{2}} + \frac{1}{a ^{'2}} + \frac{1}{b ^{'2}} + \frac{1}{c ^{'2}} = 0

JEE Main 2002

LEVELJEE Main

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

JEE Advanced 2016

LEVELJEE Main

Let $P$ be the image of the point $(3, 1, 7)$ with respect to the plane $x - y + z = 3$ . Then the equation of the plane passing through $P$ and containing the straight line $\frac{x}{1} = \frac{y}{2} = \frac{z}{1}$ is

(A)

x + y - 3 z = 0

(B)

3 x + z = 0

(C)

x - 4 y + 7 z = 0

(D)

2 x - y = 0

JEE Advanced 2016

LEVELJEE Advanced

Consider a pyramid $O P QR S$ located in the first octant ( $x \geq 0, y \geq 0, z \geq 0$ ) with $O$ as origin, and $O P$ and $O R$ along the x-axis and the y-axis, respectively. The base $O P QR$ of the pyramid is a square with $O P = 3$ . The point $S$ is directly above the mid-point, $T$ of diagonal $O Q$ such that $T S = 3$ . Then

* Multiple Correct Options

(A)

the acute angle between

O Q

and

O S

\frac{π}{3}

(B)

the equation of the plane containing the triangle

O QS

x - y = 0

(C)

the length of the perpendicular from

P

to the plane containing the triangle

O QS

\frac{3}{2}

(D)

the perpendicular distance from

O

to the straight line containing

R S

\frac{15}{2}

JEE Advanced 2015

LEVELJEE Advanced

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

JEE Advanced 2012

LEVELJEE Main

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

JEE Advanced 2010

LEVELJEE Advanced

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

JEE Advanced 2008

LEVELJEE Advanced

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

JEE Advanced 2006

LEVELJEE Main

A plane which is perpendicular to two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ , passes through $(1, - 2, 1)$ . The distance of the plane from the point $(1, 2, 2)$ is

(A)

0

(B)

1

(C)

2

(D)

22

JEE Advanced 2006

LEVELJEE Main

Let $A$ be vector parallel to line of intersection of planes $P_{1}$ and $P_{2}$ . Plane $P_{1}$ is parallel to the vectors $2 \hat{j} + 3 \hat{k}$ and $4 \hat{j} - 3 \hat{k}$ and that $P_{2}$ is parallel to $\hat{j} - \hat{k}$ and $3 \hat{i} + 3 \hat{j}$ , then the angle between vector $A$ and a given vector $2 \hat{i} + \hat{j} - 2 \hat{k}$ is

* Multiple Correct Options

(A)

\frac{π}{2}

(B)

\frac{π}{4}

(C)

\frac{π}{6}

(D)

\frac{3 π}{4}

JEE Advanced 2005S

LEVELJEE Main

A variable plane at a distance of the one unit from the origin cuts the coordinates axes at $A, B$ and $C$ . If the centroid $D (x, y, z)$ of triangle $A B C$ satisfies the relation $\frac{1}{x ^{2}} + \frac{1}{y ^{2}} + \frac{1}{z ^{2}} = k$ , then the value $k$ is

(A)

3

(B)

1

(C)

1/3

(D)

9

JEE Advanced 2010

LEVELJEE Advanced

If the distance between the plane $A x - 2 y + z = d$ and the plane containing the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ is $6$ , then find $∣ d ∣$ .

JEE Advanced 2005

LEVELJEE Advanced

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 $(2, 1, - 1)$ .

JEE Advanced 2004

LEVELJEE Main

Find the equation of plane passing through $(1, 1, 1)$ & parallel to the lines $L_{1}, L_{2}$ having direction ratios $(1, 0, - 1), (1, - 1, 0)$ . Find the volume of tetrahedron formed by origin and the points where these planes intersect the coordinate axes.

JEE Advanced 2003

LEVELJEE Advanced

(i) Find the equation of the plane passing through the points $(2, 1, 0), (5, 0, 1)$ and $(4, 1, 1)$ . (ii) If $P$ is the point $(2, 1, 6)$ then find the point $Q$ such that $P Q$ is perpendicular to the plane in (i) and the midpoint of $P Q$ lies on it.

.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; } Distance between two parallel planes 2x+y+2z=8 and 4x+2y+4z+5=0 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A equation of a plane parallel to the plane x−2y+2z−5=0 and at a unit distance from the origin is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The image of the point (−1,3,4) in the plane x−2y=0 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the plane 2ax−3ay+4az+6=0 passes through the midpoint of the line joining the centres of the spheres x2+y2+z2+6x−8y−2z=13 and x2+y2+z2−10x+4y−2z=8 then a equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The plane x+2y−z=4 cuts the sphere x2+y2+z2−x+z−2=0 in a circle of radius

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Distance between two parallel planes 2x+y+2z=8 and 4x+2y+4z+5=0 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The intersection of the spheres x2+y2+z2+7x−2y−z=13 and x2+y2+z2−3x+3y+4z=8 is the same as the intersection of one of the sphere and the plane

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The shortest distance from the plane 12x+4y+3z=327 to the sphere x2+y2+z2+4x−2y−6z=155 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The radius of the circle in which the sphere x2+y2+z2+2x−2y−4z−19=0 is cut by the plane x+2y+2z+7=0 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Two system of rectangular axes have the same origin. If a plane cuts them at distances a,b,c and a′,b′,c′ from the origin then

.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; } 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)

.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 is perpendicular to two planes 2x−2y+z=0 and x−y+2z=4, passes through (1,−2,1). The distance of the plane from the point (1,2,2) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the distance between the plane Ax−2y+z=d and the plane containing the lines 2x−1​=3y−2​=4z−3​ and 3x−2​=4y−3​=5z−4​ is 6​, then find ∣d∣.

.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).

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:

Distance between two parallel planes $2 x + y + 2 z = 8$ and $4 x + 2 y + 4 z + 5 = 0$ is

A equation of a plane parallel to the plane $x - 2 y + 2 z - 5 = 0$ and at a unit distance from the origin is :

The image of the point $(- 1, 3, 4)$ in the plane $x - 2 y = 0$ is

If the plane $2 a x - 3 a y + 4 a z + 6 = 0$ passes through the midpoint of the line joining the centres of the spheres $x^{2} + y^{2} + z^{2} + 6 x - 8 y - 2 z = 13$ and $x^{2} + y^{2} + z^{2} - 10 x + 4 y - 2 z = 8$ then $a$ equals

The plane $x + 2 y - z = 4$ cuts the sphere $x^{2} + y^{2} + z^{2} - x + z - 2 = 0$ in a circle of radius

Distance between two parallel planes $2 x + y + 2 z = 8$ and $4 x + 2 y + 4 z + 5 = 0$ is

The intersection of the spheres $x^{2} + y^{2} + z^{2} + 7 x - 2 y - z = 13$ and $x^{2} + y^{2} + z^{2} - 3 x + 3 y + 4 z = 8$ is the same as the intersection of one of the sphere and the plane

The shortest distance from the plane $12 x + 4 y + 3 z = 327$ to the sphere $x^{2} + y^{2} + z^{2} + 4 x - 2 y - 6 z = 155$ is

The radius of the circle in which the sphere $x^{2} + y^{2} + z^{2} + 2 x - 2 y - 4 z - 19 = 0$ is cut by the plane $x + 2 y + 2 z + 7 = 0$ is

Two system of rectangular axes have the same origin. If a plane cuts them at distances $a, b, c$ and $a^{'}, b^{'}, c^{'}$ from the origin then

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 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)

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 is perpendicular to two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ , passes through $(1, - 2, 1)$ . The distance of the plane from the point $(1, 2, 2)$ is

If the distance between the plane $A x - 2 y + z = d$ and the plane containing the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ is $6$ , then find $∣ d ∣$ .

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 $(2, 1, - 1)$ .