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JEE • MATHEMATICS • 3D PLANES

3d planes practice · Medium

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 31-59 of 59 questions

31 Medium

31 of 59

Application +4 | -1

The equation of the line passing through

(1,2,3)

and parallel to the planes

x-y+2 z=5

and

3 x+y+z=6

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because the direction ratios of the line are found by taking the cross product of the normal vectors of the planes, and the line passes through the given point.

32 Medium

32 of 59

Analyzing +4 | -1

A ray of light coming along the line

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

, strikes the plane mirror kept along the plane through points

(2,1,1),(3,0,2)

and

(2,-1,-1)

. Then the equation of reflected ray is:

View answer and explanation

Correct option: B

\frac{x-3}{1}=\frac{y-3}{5}=\frac{z-2}{10}

The correct answer is A (

\frac{x-3}{1}=\frac{y-3}{5}=\frac{z-2}{10}

) because it represents the equation of the reflected ray after finding the reflection point and the intersection point.

33 Medium

33 of 59

Analyzing +4 | -1

Let the image of the point

P(1,2,6)

in the plane passing through the points

A(1,2,0)

and

B(1,4,1)

C(0,5,1)

Q(\alpha, \beta, \gamma)

. Then

\left(\alpha^2+\beta^2+\gamma^2\right)

is equal to

View answer and explanation

Correct option: A

Option A (65) is correct because after finding the normal vector and equation of the plane, the image point Q is calculated, and then the sum of squares of its coordinates is found to be 65.

34 Medium

34 of 59

Analyzing +4 | -1

If the equation of the plane passing through the line of intersection of the planes

2 x-y+z=3,4 x-3 y

+5 z+9=0

and parallel to the line

\frac{x+1}{-2}=\frac{y+3}{4}=\frac{z-2}{5}

a x+b y+c z+6=0

, then

a

+b+c

is equal to

View answer and explanation

Correct option: B

Option B (14) is correct because it is the sum of the coefficients a, b, and c of the plane equation after finding the plane parallel to the given line and passing through the intersection of the two planes.

35 Medium

35 of 59

Analyzing +4 | -1

Plane is drawn from the point

P(1,1,1)

and perpendicular to the line with direction ratios

(1,1,1)

to intersect the plane

x+2 y+3 z=4

Q

. The locus of point

\mathrm{Q}

View answer and explanation

Correct option: A

\frac{x}{1}=\frac{y-5}{-2}=\frac{z+2}{1}

Option A (

\frac{x}{1}=\frac{y-5}{-2}=\frac{z+2}{1}

) is correct because it represents the line of intersection between the plane perpendicular to (1,1,1) passing through P(1,1,1) and the plane x+2y+3z=4.

36 Medium

36 of 59

Application +4 | -1

The volume of the tetrahedron included between the plane

3 x+4 y-5 z-60=0

and the co-ordinate planes is

View answer and explanation

Correct option: B

600

Option B (600) is the correct answer because it represents the volume of the tetrahedron formed by the plane and the coordinate planes, calculated using the intercepts.

37 Medium

37 of 59

Analyzing +4 | -1

For

a, b \in Z

and

|a-b| \leq 10

, let the angle between the plane

P: a x+y-z=b

and the line

I: x-1=

a-y=z+1

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

. If the distance of the point

(6,-6,4)

from the plane

P

3 \sqrt{6}

, then

a^{4}+b^{2}

is equal to

View answer and explanation

Correct option: A

Option A (32) is correct. By using the given angle between the plane and the line, we find 'a'. Then, using the distance formula, we find 'b'. Finally, we calculate a⁴ + b².

38 Medium

38 of 59

Application +4 | -1

The equation of the plane passing through the intersection of the planes

2 x-5 y+z=3

and

x+y+4 z=5

and parallel to the plane

x+3 y+6 z=1

x+3 y+6 z=k

, where

k

View answer and explanation

Correct option: A

Option A (7) is correct because finding the equation of the plane passing through the intersection and parallel to the given plane yields k=7.

39 Medium

39 of 59

Application +4 | -1

If a variable plane forms a tetrahedron of constant volume

64 \mathrm{k}^3

with the coordinate planes then the locus of the centroid of the tetrahedron is

\mathrm{xyz}=\mathrm{uk}^3

. Find

\mathrm{u}

View answer and explanation

Correct option: A

Option A (6) is correct because the locus of the centroid is found by relating the centroid coordinates to the intercepts and using the volume condition.

40 Medium

40 of 59

Application +4 | -1

The plane through the intersection of the planes

x + y + z = 1

and

2x + 3y - z + 4 = 0

and parallel to Y-axis also passes through the point

View answer and explanation

Correct option: C

(3, 2, 1)

Option C (3, 2, 1) is correct because it satisfies the equation of the plane formed by the intersection of the given planes and parallel to the Y-axis.

41 Medium

41 of 59

Application +4 | -1

The distance of the point of intersection of the line

\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}

and the plane

x-y+z=5

from the point

(-1,-5,-10)

View answer and explanation

Correct option: A

Option A (13) is correct because the distance is calculated by finding the point of intersection of the line and the plane, and then calculating the distance between that point and the given point.

42 Medium

42 of 59

Application +4 | -1

The equation of the plane through the point

(2,5,-3)

perpendicular to the planes

x+2 y+2 z=1

and

x-2 y+3 z=4

is -

View answer and explanation

Correct option: D

10 x-y-4 z=27

Option D (

10 x-y-4 z=27

) is correct because it represents the equation of a plane passing through the given point and perpendicular to the two given planes.

43 Medium

43 of 59

Analyzing +4 | -1

A variable plane makes intercepts on

X, Y

and

Z

-axes and it makes a tetrahedron of volume

64 \mathrm{cu}

. u. The locus of foot of perpendicular from origin on this plane is

View answer and explanation

Correct option: A

\left(x^2+y^2+z^2\right)^3=384 x y z

Option A,

\left(x^2+y^2+z^2\right)^3=384 x y z

, is the correct locus of the foot of the perpendicular from the origin on the plane, derived using the volume of the tetrahedron and the equation of the plane in intercept form.

44 Medium

44 of 59

Analyzing +4 | -1

If the planes

x-y-z=0, x-y+\sin \alpha \cdot z=0

and

x+\sin \alpha \cdot y-z=0

pass through a straight line, then number of values of

\alpha

[0,2021 \pi]

is/are

View answer and explanation

Correct option: A

1010

Option A (1010) is correct because solving the determinant of the coefficient matrix leads to sin α = -1, yielding 1010 solutions for α in the given interval.

45 Medium

45 of 59

Application +4 | -1

The direction ratios of normal to the plane through the points (0, - 1, 0) and (0, 0, 1) and making an angle

\frac{\pi }{4}

with the plane y – z + 5 = 0 are

View answer and explanation

Correct option: D

2\sqrt 3 ,1, - 1

Option D (

2\sqrt 3 ,1, - 1

) is correct because these direction ratios satisfy the conditions of the plane passing through the given points and making the specified angle.

Explore Other Chapters in MATHEMATICS

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SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs

46 Medium

46 of 59

Analyzing +4 | -1

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

\alpha

with the positive

X

-axis, then

\cos \alpha

equals

View answer and explanation

Correct option: A

1 / \sqrt{3}

Option A (

1 / \sqrt{3}

) is correct because the direction cosines of the line of intersection are found by solving the system of equations derived from the normal vectors of the planes.

47 Medium

47 of 59

Application +4 | -1

Equation of plane containing

(1,1,2)

and which is at a maximum distance from

(2,2,1)

View answer and explanation

Correct option: D

x+y-z=0

Option D (

x+y-z=0

) is correct because the normal vector to the plane is parallel to the vector connecting the point on the plane and the external point.

48 Medium

48 of 59

Application +4 | -1

Let

N

be the foot of perpendicular from the point

P(1,-2,3)

on the line passing through the points

(4,5,8)

and

(1,-7,5)

. Then the distance of

N

from the plane

2 x-2 y+z+5=0

View answer and explanation

Correct option: D

Option D (7) is correct because after finding the coordinates of point N and calculating its distance from the given plane, the result is 7.

49 Medium

49 of 59

Application +4 | -1

If the line

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

lies in the plane

lx + my - z = 9

, then

{l^2} + {m^2}

is equal to

View answer and explanation

Correct option: D

Option D (2) is correct because it is the value of

{l^2} + {m^2}

calculated from the given conditions that the line lies in the plane.

50 Medium

50 of 59

Application +4 | -1

If equation of the plane through the straight line

\frac{x-1}{2}=\frac{y+2}{-3}=\frac{z}{5}

and perpendicular to the plane

\mathrm{x}-\mathrm{y}+\mathrm{z}+2=0

ax - by + c z+4=0

, then find the value of

\frac{10^3 a+10^2 \mathrm{~b}+10 \mathrm{c}}{342}

View answer and explanation

Correct option: A

Option A (5) is correct because solving the system of equations derived from the line and plane conditions yields the correct values for a, b, and c, and thus the correct answer.

51 Medium

51 of 59

Application +4 | -1

If the distance of the point

(1,-2,3)

from the plane

x+2 y-3 z+10=0

measured parallel to the line

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

\sqrt{\frac{7}{2}}

then the value of

|m|=

View answer and explanation

Correct option: B

Option B (2) is correct because it is the absolute value of the parameter 'm' that satisfies the given distance condition between the point and the plane, measured parallel to the given line.

52 Medium

52 of 59

Application +4 | -1

If the lines

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

and

\frac{x-11}{4}=\frac{y-9}{2}=\frac{z-4}{3}

intersects at point

p

, then perpendicular distance of

p

from plane

3 x+2 y+6 z=10

View answer and explanation

Correct option: B

\frac{3}{7}

Option B (3/7) is correct because it is the perpendicular distance of the intersection point (3, 5, -2) from the plane 3x + 2y + 6z = 10, calculated using the point-to-plane distance formula.

53 Medium

53 of 59

Analyzing +4 | -1

Let the equation of plane passing through the line of intersection of the planes

x+2 y+a z=2

and

x-y+z=3

5 x-11 y+b z=6 a-1

. For

c \in \mathbb{Z}

, if the distance of this plane from the point

(a,-c, c)

\frac{2}{\sqrt{a}}

, then

\frac{a+b}{c}

is equal to

View answer and explanation

Correct option: C

-4

Option C (

-4

) is correct because after finding the values of a and b, and using the distance formula, the value of c is determined, leading to the correct ratio.

54 Medium

54 of 59

Application +4 | -1

The angle between the line

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

and the plane

3 x+6 y-2 z+5=0

View answer and explanation

Correct option: D

\sin ^{-1}\left(\frac{4}{21}\right)

The correct answer is D (

\sin ^{-1}(\frac{4}{21})

) because it is calculated using the formula for the angle between a line and a plane.

55 Medium

55 of 59

Application +4 | -1

If line

\frac{x-1}{2}=\frac{y+\alpha}{\alpha}=\frac{z-\beta}{2}

lies in plane

2 x+y+z=5

, then

\alpha+\beta

View answer and explanation

Correct option: D

-9

Option D (-9) is correct because by substituting the parametric form of the line into the plane equation, we can solve for α and β, and then find their sum.

56 Medium

56 of 59

Application +4 | -1

The plane

x-2 y+3 z=17

divides the line joining the points

(-2,4,7)

and

(3,-5,8)

in the ratio

View answer and explanation

Correct option: B

3: 10

Option B (

3: 10

) is correct because it represents the ratio in which the plane divides the line segment, calculated by substituting the section formula into the plane equation.

57 Medium

57 of 59

Application +4 | -1

The plane, passing through the points

(0,-1,2)

and

(-1,2,1)

and parallel to the line passing through

(5,1,-7)

and

(1,-1,-1)

, also passes through the point

View answer and explanation

Correct option: A

(-2,5,0)

Option A ((-2,5,0)) is correct because after finding the equation of the plane, substituting the coordinates of the point (-2,5,0) satisfies the equation.

58 Medium

58 of 59

Analyzing +4 | -1

Let the plane containing the line of intersection of the planes

P 1: x+(\lambda+4) y+z=1

and

P 2: 2 x+y

+z=2

pass through the points

(0,1,0)

and

(1,0, 1).

Then the distance of the point

(2 \lambda, \lambda,-\lambda)

from the plane

P 2

View answer and explanation

Correct option: B

3 \sqrt{6}

Option B (

3 \sqrt{{6}}

) is correct because after finding λ, the distance from the point to the plane P2 is calculated using the distance formula.

59 Medium

59 of 59

Application +4 | -1

The angle between the line

\frac{{x - 1}}{2} = \frac{y}{3} = \frac{{z - 3}}{6}

and the plane

10x+2y+11z=3

, is

View answer and explanation

Correct option: A

{\sin ^{ - 1}}\left( {\frac{{92}}{{105}}} \right)

Option A (

{\sin ^{ - 1}}\left( {\frac{{92}}{{105}}} \right)

) is correct because it is the angle between the line and the plane, calculated using the formula involving the dot product of the direction vector of the line and the normal vector of the plane.

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