JEE Exam  >  JEE Tests  >  Mathematics (Maths) Class 12  >  JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - JEE MCQ

JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - JEE MCQ


Test Description

30 Questions MCQ Test Mathematics (Maths) Class 12 - JEE Advanced Level Test: Three Dimensional 3D Geometry- 1

JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 for JEE 2024 is part of Mathematics (Maths) Class 12 preparation. The JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 questions and answers have been prepared according to the JEE exam syllabus.The JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 below.
Solutions of JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 questions in English are available as part of our Mathematics (Maths) Class 12 for JEE & JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 solutions in Hindi for Mathematics (Maths) Class 12 course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 | 30 questions in 60 minutes | Mock test for JEE preparation | Free important questions MCQ to study Mathematics (Maths) Class 12 for JEE Exam | Download free PDF with solutions
1 Crore+ students have signed up on EduRev. Have you? Download the App
JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 1

A line makes angles α,β,γ with the coordinates axes. If α+β = 90°, then (gamma) equal to

Detailed Solution for JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 1

If a line makes the angles α,β,γ,
Then,  cosα+cosβ+cos2 γ=1
It is given that, α+β=90°
⇒  α=β−90°
⇒  cosα=cos(90o−β) 
⇒  cosα=sinβ
⇒  cosα = sinβ=1−cos2 β
⇒  cos2α+cos2β=1
As,   cosα+cos2 β+cos^2γ=1
⇒  1+cos2 γ=1
⇒  cos2 γ=0
hence,  γ=π/2=90°

JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 2

The coordinates of the point A, B, C, D are (4, α, 2), (5, –3, 2), (β, 1, 1) & (3, 3, – 1). Line AB would be perpendicular to line CD when

JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 3

The locus represented by xy + yz = 0 is

JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 4

The equation of plane which passes through (2, –3, 1) & is normal to the line joining the points (3, 4, –1) & (2, – 1, 5) is given by

Detailed Solution for JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 4

A(3,4,−1) and  B(2,−1,5)

Vector AB is normal to the required plane 
⇒ Directions of normal (−1,−5,6)
∴ Equation ,−x−5y+6z=k
Point (2,−3,1) passes through the plane,
∴ −2+15+6=k⇒k=19
∴ −x−5y+6z = 19
x+5y−6z+19 = 0

JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 5

If the sum of the squares of the distances of a point from the three coordinate axes be 36, then its distance from the origin is

Detailed Solution for JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 5

Let (x,y,z) be the point.
Given sum of the squares of distance from point to the axes is 36. 
⇒(x2+y2)+(y2+z2)+(z2+x2)=36
⇒2(x2+y2+z2)=36⇒x2+y2+z2=18
So the distance of the point from the origin is =3(2)1/2

JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 6

The locus of a point P which moves such that PA2 – PB2 = 2k2 where A and B are (3, 4, 5) and (–1, 3, –7) respectively is

JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 7

The equation of the plane passing through the point (1, – 3, –2) and perpendicular to planes x + 2y + 2z = 5 and 3x + 3y + 2z = 8, is

Detailed Solution for JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 7

The normals to the planes x+2y+2z=5 and 3x+3y+2z=8 are their respective unit vectors ie


Since the required plane is perpendicular to the planes 
x+2y+2z=5 and 3x+3y+2z=8,
 its normal would be perpendicular to the normals to the planes
The cross product of  is normal to the required plane

The equation of the required plane would be

-2x+4y-3z=2+12-6 ie 2x-4y+3z-8 = 0

JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 8

A variable plane passes through a fixed point (1, 2, 3). The locus of the foot of the perpendicular drawn from origin to this plane is

Detailed Solution for JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 8

α(x − α) + β(y − β) + γ(z − γ) = 0
α(1 − α) + β (2 − β) + γ ( 3 − γ) = 0
α + 2β + 3γ = α+ β+ γ2
α+ β+ γ− α − 2β − 3γ = 0
x+ y+ z− x − 2y− 3z = 0

JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 9

The reflection of the point (2, –1, 3) in the plane 3x – 2y – z = 9 is

Detailed Solution for JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 9

line AB = (x-2)/3 = (y+1)/-2 = (z-3)/-1 = λ
(x,y,z)=(3λ+2,−2λ−1,−λ+3)
3x−2y−z=9
3(3λ+2)−2(−2λ−1)−3+λ=9
9λ+6+4λ+2−3+λ=9
14λ=4
λ=2/7
C(x,y,z)=(207, -117, 19/7)
A(2,-1,3)
C is MP of A and B
C= (A+B)/2,    B= (2C-A)
B=(26/7, −22/7+1, 38/7−3)
B=(26/7,−15/7,17/7)

JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 10

The distance of the point (–1, –5, –10) from the point of intersection of the line,  and the plane, x – y + z = 5, is

JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 11

The distance of the point (1, –2, 3) from the plane x – y + z = 5 measured parallel to the line, 

JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 12

The straight l ines and 

JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 13

If plane cuts off intercepts OA = a, OB = b, OC = c from the coordinate axes, then the area of the triangle ABC equal to

Detailed Solution for JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 13

 =AC = −ai^ + ck^
AB = −ai^ + bj^
 Area of △ABC= ½|AB × AC∣ 
|AB × AC∣ =

−(bc)i^− (ac)j^ − (ab)k^
∣AB × AC∣ = (b2c2 + a2c2 + a2b2)1/2
Area = 1/2(a2b2 + b2c2 + c2a2)1/2

JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 14

A point moves so that the sum of the squares of its distances from the six faces of a cube given by x = ± 1, y = ± 1, z = ± 1 is 10 units. The locus of the point is

Detailed Solution for JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 14

 Let P(x,y,z) be any point on the locus, then the distances from the six faces are ∣x+1∣, ∣x−1∣, ∣y−1∣, ∣z−1∣
According to the given condition, we have
∣x+1∣2 + ∣x−1∣2 + ∣y+1∣2 + ∣y−1∣2 + ∣z+1∣2+ ∣z−1∣2=10
⇒ 2(x2+y2+z2)=10−6=4
⇒ x+ y+ z= 2

JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 15

A variable plane passes through a fixed point (a, b, c) and meets the coordinate axes in A, B, C. Locus of the point common to the planes through A, B, C and parallel to coordinate plane, is

Detailed Solution for JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 15

Let equation of plane is 
x/α + y/β + z/γ = 1.............(1)
Plane passes through point (a,b,c)
a/α + b/β + c/γ = 1................(2)
Again plane meets the coordinate points. 
Coordinates of points are (α,0,0)  
Coordinates of point B are (0, β, 0) and 
coordinates of point C are (0, 0, γ)  
∴ Equation of planes, parallel to coordinate axis and passing through  
Point A is x = α …..(3)  
Point B is y = β …..(4)  
Point C is z = γ …..(5)  
∴ Locus of the point of intersection is 
a/x + b/y + c/z = 1

JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 16

Two systems of rectangular axes have same origin. If a plane cuts them at distances a, b, c and a1, b1, cfrom the origin, then

JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 17

The angle between the plane 2x – y + z = 6 and a plane perpendicular to the planes x + y + 2z = 7 and x – y = 3 is

JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 18

The non zero value of ‘a’ for which the lines 2x – y + 3z + 4 = 0 = ax + y – z + 2 and x – 3y + z = 0 = x + 2y + z + 1 are co-planar is

Detailed Solution for JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 18

2x - y +3z + 4 = 0
x - 3y + z = 0
x + 2y + z + 1 = 0
x = 12/5
y = -1/5
z= -3
this point also satisfied by
ax - y + z + 2 = 0
a(12 / 5) - (-1/5) + (-3) = 0
⇒ 12a/5 + 1/5 -3 = 0
a= -2

JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 19

If the lines  and are concurrent then

Detailed Solution for JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 19

Let the point of intersection is (λ, 2λ, 3λ). 
Clearly, λ = 3μ + 1, 2λ = 2 – μ 
Solving, we get λ = 1, μ = 0 
Hence, the point of intersection is (1, 2, 3).
Therefore, (1+k)/3 = (2-1)/2 = (3-2)/h
= (1+k)/3 = 1/2 = 1/h
⇒ h = 2   k = 1/2

JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 20

The coplanar points A, B, C, D are (2 – x, 2, 2), (2, 2 – y, 2), (2, 2, 2 – z) and (1, 1, 1) respectively. Then

Detailed Solution for JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 20

We have four coplanar points.
The three vectors connecting two of them at a time are thus coplanar.
⇒{(−x, y, 0) (-x, 0, z) (1−x, 1, 1)} = 0
⇒−x (−z) −y (−x−z+xz)+0=0
⇒xz + xy + yz = xyz
⇒1/y + 1/z + 1/x = 1

JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 21

The direction ratios of a normal to the plane through (1, 0, 0), (0, 1, 0), which makes an angle of π/4 with the plane x + y = 3 are

JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 22

Let the points A(a, b, c) and B(a', b', c') be at distances r and r' from origin. The  line AB passes through origin when

JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 23

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

JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 24

If a line makes an angle of π/4 with the positive directions of each of x-axis and y-axis, then the angle that the line makes with the positive direction of the z-axis is

JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 25

If the angle θ between the line  and the plane  is such that sinθ = 1/3 The value of λ is

JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 26

A line makes the same angle θ with each of the x and z-axis. If the angle θ, which it makes with y-axis is such that sin2 β = 3 sin2 θ, then cos2θ equals

JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 27

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

Detailed Solution for JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 27

Let x1, y1, z1 be any point on the plane

JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 28

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

Detailed Solution for JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 28

The co-ordinates of any point on L1 in terms of parameter r1 are given by,x=y+a=z=r1
⇒x=r1,y=r1 - a,z=r1....(1)
Similarly the co-ordinates of any point on L2 in terms of parameter 2r2 are given by 
⇒x+a=2y=2z=2r2
⇒ x=2r2−a,y=r2, z=r2...(2)
Let 'A' be a point on L1 and 'B' be a point on L2
​Using (1) and (2), the direction ratios of AB are 
2r2−a−r1,r2−r1+a,r2−r1
​If the above line is same as the line whose direction cosines are proportional to (2,1,2) as given in the question, then  
(2r2−r1−a)/2= (r2−r1+a)/1 = (r2−r1)/2
Solving the first two of the above equation, we get r1=3a
Again solving the last two, we get r2=a
Using these values in (1) and (2), we get the coordinates of points as 
 
(3a,2a,3a) and (a,a,a).

JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 29

The lines  and  are coplanar if

Detailed Solution for JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 29

(x-x1)a1 = (y-y1)b1 = (z-z1)c1 &
(x-x2)a2 = (y-y2)b2 = (z-z2)c2 are coplanar if

k2 + 3k = 0
k = 0 or -3

JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 30

The equation of plane which meet the co-ordinate axes whose centroid is (a, b, c)

Detailed Solution for JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 - Question 30

A(a,0,0),B(0,b,0),C(0,0,c) are the points on coordinate axis and centriod (α,β,γ).
According to centroid formula,
α= a/3
​a=3α
b=3β
c=3γ
Equation of plane is,x/a+y/b+z/c =1
x/α+y/β+z/γ=3

204 videos|288 docs|139 tests
Information about JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 Page
In this test you can find the Exam questions for JEE Advanced Level Test: Three Dimensional 3D Geometry- 1 solved & explained in the simplest way possible. Besides giving Questions and answers for JEE Advanced Level Test: Three Dimensional 3D Geometry- 1, EduRev gives you an ample number of Online tests for practice

Up next

204 videos|288 docs|139 tests
Download as PDF

Up next

Download the FREE EduRev App
Track your progress, build streaks, highlight & save important lessons and more!