datavorous/entrance-exam-dataset huggingface dataset:
A plane electromagnetic wave of frequency 25GHz is propagating in vacuum along the z-direction. At a particular point in space and time, the magnetic filed is given by B→=5×10-8 j^ T. The corresponding electric field E→ is (speed of light =3×108 m s-1 )

Options:
A 1.66×10-16i^ Vm
B -1.66×10-16i^ Vm
C -15i^ Vm
D 15i^ Vm

The correct answer is:
15i^ Vm

Solution:
Given, the electromagnetic wave is propagating in z i.e., +k^ direction.

The relation between the magnitudes of electric field, magnetic field and the speed of light is given by

EB=C

⇒ E=B×C
⇒ E=5×10-8×3×108=15 V m-1

We know that the direction of propagation of electromagnetic wave is along E→×B→.

So, E→=15 i^ V m-1

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2020 (08 Jan Shift 2)
A plane electromagnetic wave of frequency 35 MHz travels in free space along the X-direction. At a particular point (in space and time) E→=9.6j^ V m-1. The value of magnetic field at this point is:

Options:
A 3.2×10-8k^ T
B 3.2×10-8i^
C 9.6j^ T
D 9.6×10-8k^ T

The correct answer is:
3.2×10-8k^ T

Solution:
At every instant, the ratio of the magnitude of the electric field to the magnitude of the magnetic field in an electromagnetic wave equals the speed of light.

Therefore, EB=c

⇒EB=3×108

⇒B=E3×108=9.63×108

⇒B=3.2×10-8 T

Now, direction of magnetic field can be calculated as

B^=v^×E^

=i^×j^=k^

So, B→=3.2×10-8k^T

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2024 (29 Jan Shift 2)
A plane electromagnetic wave of frequency 50 MHz travels in free space along the positive x - direction. At a particular point in space and time, E→=6.3 j^ V/m, The corresponding magnetic field B,→ at that point will be:

Options:
A 2.1×10-8k^ T
B 18.9×108k^ T
C 18.9×10-8k^ T
D 6.3×10-8k^ T

The correct answer is:
2.1×10-8k^ T

Solution:
As we know EB=C
E=6.3V/m
B=EC=6.33×108T=2.1×10-8T

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2019 (09 Jan Shift 1)
A plane electromagnetic wave of frequency 100 MHz is traveling in a vacuum along the x-direction. At a particular point in space and time, B→=2.0×10-8 k^T· (where, k^ is unit vector along z-direction) What is E→ at this point?

Options:
A 0.6j^ V m-1
B 6.0k^ V m-1
C 6.0j^ V m-1
D 0.6k^ V m-1

The correct answer is:
6.0j^ V m-1

Solution:
E=BC=6. (Direction of the wave) ‖(E→×B→)⇒i^=j^×k^
⇒E→=6j^ V m-1

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2021 (18 Mar Shift 1)
A plane electromagnetic wave of frequency 500 MHz is traveling in a vacuum along the y-direction. At a particular point in space and time, B→=8.0×10-8z^ T. The value of the electric field at this point is:(speed of light =3×108 ms-1) x^,y^,z^ are unit vectors along x,y and z direction.

Options:
A -24x^ V m-1
B 2.6x^ V m-1
C 24x^ V m-1
D -2.6x^ V m-1

The correct answer is:
-24x^ V m-1

Solution:
f=5×108 Hz

EM wave is travelling towards +j^

B→=8.0×10-8z^ T

E→=B→×C→=8×10-8z^×3×108y^

=-24x^ V m-1

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2021 (16 Mar Shift 1)
A plane electromagnetic wave of wavelength λ has an intensity I. It is propagating along the positive Y-direction. The allowed expressions for the electric and magnetic fields are given by:

Options:
A E→= 2 Iε0c cos2πλy-ctk^,
B→=+1cEi^
B E→= Iε0c cos⁡2πλy-ctk^,
B→=+1cEi^
C E→= 2 Iε0c cos2πλy+ctk^,
B→=+1cEi^
D E→= Iε0c cos2πλy-cti^,
B→=+1cEk^

The correct answer is:
E→= 2 Iε0c cos2πλy-ctk^,
B→=+1cEi^

Solution:
If E0 is the amplitude of the electric field then the intensity of the electromagnetic wave is given as I=12ε0E02×c. From the above relation, we get E0= 2 Ic ε0. The relation between electric and magnetic fields is B0= E0c. The direction of E → × B → will be along +j^.

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2018 (16 Apr Online)
A plane electromagnetic wave propagating along y-direction can have the following pair of electric field (E→) and magnetic field (B→) components.

Options:
A Ey,By or Ez,Bz
B Ey,Bx or Ex,By
C Ex,Bz or Ez,Bx
D Ex,By or Ey,Bx

The correct answer is:
Ex,Bz or Ez,Bx

Solution:

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2021 (18 Mar Shift 2)
A plane electromagnetic wave propagating in x-direction is described by
Ey=200 V m-1sin1.5×107t-0.05x; The intensity of the wave is :
(Use ϵ0=8.85×10-12 C2 N-1 m-2)

Options:
A 35.4 W m-2
B 53.1 W m-2
C 26.6 W m-2
D 106.2 W m-2

The correct answer is:
53.1 W m-2

Solution:
The magnitude of the electric field component of the electromagnetic wave is given by

E0=200 V m-1

The formula to calculate the intensity of the electromagnetic wave is given by

I=12ε0cE02 ...1

From equation (1), it follows that

I=12×8.85×10-12×3×108×2002 W m-2=53.1 W m-2

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2024 (27 Jan Shift 1)
A plane electromagnetic wave travels in free space along the x-direction. The electric field component of the wave at a particular point of space and time is E=6 V m-1 along y-direction. Its corresponding magnetic field component, B would be:

Options:
A 2×10-8 T along z-direction
B 2×10-8 T  along y-direction
C 6×10-8 T  along z-direction
D 6×10-8 T  along x-direction

The correct answer is:
2×10-8 T along z-direction

Solution:
B=EC=63×108=2×10-8 T
E→×B→ is along x -direction
j^×ai^+bj^+ck^=i^
c=1
B→ is along k^
B=2×10-8 T along z -direction

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2019 (08 Apr Shift 1)
A plane electromagnetic wave with a frequency of 30 MHz travels in free space. At a particular point in space and time, the electric field is 6 V m-1. The magnetic field at this point will be x×10-8 T. The value of x is,

Options:

The correct answer is:
2

Solution:
Recall the formula of the magnitude of the magnetic field in terms of the magnitude of the electric field, B=Ec=63×108=2×10-8 T, here c is the speed of light in a vacuum.

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2021 (27 Aug Shift 2)
A plane glass plate is placed over various coloured letters (violet, green, yellow, red). The letter which appears to raise more is

Options:
A Red
B Yellow
C Green
D Violet

The correct answer is:
Violet

Solution:
âµ Î¼=Real depthapparent depth
Where, Î¼ refractive index of medium
â Apparent depth=Real depthÎ¼
But, Î¼â1Î» âµÎ»v<Î»g<Î»y<Î»r
Thus, Î¼v is greater to all of them
â´ Apparent depth is minimum for violet.
It means that the violet letter appear raised more.

Topic: Physics, Ray Optics, JEE Main
A plane is in level flight at a constant speed such that the speed of the air below the wings is 144 km h-1 and above the wing is 180 km h-1. If each of its wings has an area of 25 m2, then the mass (in kg) of the plane is [Ïair=1 kg m-3, g=10 m s-2]

Options:

The correct answer is:
2250

Solution:
Given,

A=25 m2 [area of each wing]

v1=180×518 m s-1=50 m s-1 [speed of the air above the wing]

v2=144×518 m s-1=40 m s-1 [speed of the air below the wing]

According to Bernoulli's theorem
P1+12ρairv12=P2+12ρv22

⇒P2-P1=12ρairv12-v22

⇒P1-P2=12×1×502-402

⇒∆P=450 Pa

The lifting force acting on the plane is

∆P×Atotal=450×50 N=22500 N

Mg=∆P×Atotal

⇒M=2250 kg

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A plane is in level flight at constant speed and each of its two wings has an area of 40 m2. If the speed of the air is 180 km h-1 over the lower wing surface and 252 km h-1 over the upper wing surface, the mass of the plane is ________kg. (Take air density to be 1 kg m–3 and g=10 m s–2)

Options:

The correct answer is:
9600

Solution:
Total wing span area, A=2×40=80 m2

Using Bernoulli's equation, we get uplift force as

AP2-P1=12ρV12-V22A

⇒mg=12×1702-502×80

⇒mg=2400×40

⇒m=9600 kg

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2024 (01 Feb Shift 1)
A plane is inclined at an angle \( \alpha=30^{\circ} \) with respect to the horizontal. A particle is projected with a speed \( \mathrm{u}=2 \mathrm{~m} \) s \( ^{-1} \), from the base of the plane, making an angle \( \theta=15^{\circ} \) with respect to the plane as shown in the figure. The distance from the base, at which the particle hits the plane is close to: (Take \( \mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{-2} \) )

Options:
A \( 20 \mathrm{~cm} \)
B \( 18 \mathrm{~cm} \)
C \( 14 \mathrm{~cm} \)
D \( 26 \mathrm{~cm} \)

The correct answer is:
\( 20 \mathrm{~cm} \)

Solution:
Range =2u2sinαcos⁡α+θgcos2⁡θ
α=angle of inclination=30°
θ=angle of projection from inclination=15°
u=2 m/s
R=222sin15°cos⁡15°+30°gcos2⁡30°=8sin⁡15°cos45°g cos230° sin15°=3-122
R=453-1 3≈0.2 m
R≈20 cm

Topic: Physics, Motion In Two Dimensions, JEE Main
A plane $\pi$ makes intercepts 3 and 4 respectively on $Z$-axis and $X$-axis. If $\pi$ is parallel to $Y$-axis, then its equation is :

Options:
A $3 x+4 z=12$
B $3 z+4 x=12$
C $3 y+4 z=12$
D $3 z+4 y=12$

The correct answer is:
$3 x+4 z=12$

Solution:
Given that $a=4, c=3$
Equation of the plane ' $\pi$ ' is
$\frac{x}{a}+\frac{y}{b}+\frac{z}{b}=1 \Rightarrow \frac{x}{4}+\frac{y}{b}+\frac{z}{3}=1$
Since, $\pi$ is parallel to $Y$-axis.
$\therefore \quad \frac{0 \cdot \frac{1}{a}+1 \cdot \frac{1}{b}+0 \cdot \frac{1}{c}}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}=0 \Rightarrow \frac{1}{b}=0$
Thus, the equation of plane $\pi$ is
$\frac{x}{4}+\frac{z}{3}=1 \Rightarrow 3 x+4 z-12=0$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
A plane mirror of length 30 cm is placed horizontally along with a vertical screen. A monochromatic point source of light is placed 20 cm to the left off the left edge of the mirror, at a height of 2 mm above the plane of the mirror. If the wavelength of light used is 6400 Aâ, then find the number of complete bright fringes formed.

Options:

The correct answer is:
37

Solution:
y=30cmtanÎ¸
= 3 0 cm 0.2 2 0 = 0.3 cm = 3 mm
=Î»Dd
= 6 4 0 0 Ã 1 0 - 1 0 Ã 0.5 4 Ã 1 0 - 3
= 8 0 0 Ã 1 0 - 7
= 0.08 mm
n=yw=30.08=3008=752
= 37.5
So no. of complete fringe will be 37.
(take a note first bright fringe will be at a distance of w2 above edge)

Topic: Physics, Wave Optics, JEE Main
A plane mirror produces a magnification of

Options:
A -1
B +1
C Zero
D Between 0 and $+\infty$

The correct answer is:
+1

Solution:
The magnification of a plane mirror is +1 . The plus sign indicates that the image formed is virtual and erect, meaning it is the same orientation as the object. So, the magnification of a plane mirror is +1 , not -1 .

Topic: Physics, Ray Optics, JEE Main
A plane mirror reflecting a ray of incident light is rotated through an angle $\theta$ about an axis through the point of incidence in the plane of the mirror perpendicular to the plane of incidence, then

Options:
A The reflected ray does not rotate
B The reflected ray rotates through an angle $\theta$
C The reflected ray rotates through an angle 20
D The incident ray is not fixed

The correct answer is:
The reflected ray rotates through an angle 20

Solution:
The reflected ray rotates through an angle $2 \theta$ D
The incident ray is fixed
By keeping the incident ray is fixed, if plane mirror rotates through an angle $\theta$ reflected ray rotates through an angle $2 \theta$.

Topic: Physics, Ray Optics, JEE Main
A plane P=0 is the perpendicular bisector of the line joining the points 2,3,4 and 6,7,8. The perpendicular distance of P=0 from the origin is

Options:
A 43 units
B 53 units
C 63 units
D 83 units

The correct answer is:
53 units

Solution:
From diagram mid-point of AB must lie on the plane
M2+62,3+72,4+82≡4,5,6
AB is perpendicular to the required plane
Direction ratios of AB→=<4,4,4>=<1,1,1>
Equation of the required plane is
1x-4+1y-5+1z-6=0
⇒x+y+z=15
The perpendicular distance from the origin to the plane =0+0+0-153=53 units

Topic: Mathematics, Three Dimensional Geometry, JEE Main
A plane P=0 passes through the line of intersection of the planes x+y+z+3=0 and x-y+z-2=0. If the plane P divides the line joining 3,0,2 and 0,3,-1 in the ratio 2:1 internally and the equation of the plane is ax-2y+bz=c where a,b,câN, then the value of 3a+4b-5c is equal to

Options:
A 22
B 32
C 42
D 10

The correct answer is:
32

Solution:
The equation of the plane is x+y+z+3+λx-y+z-2=0
⇒x1+λ+y1-λ+z1+λ+3-2λ=0
A point which divides 3,0,2 and 0,3,-1 in 2:1 internally is

2×0+1×33,2×3+0×13,2×-1+1×23
≡1,2,0
This point must satisfy the equation of the plane

⇒1+λ+2-2λ+0+3-2λ=0
⇒6-3λ=0⇒λ=2
Equation of the required plane is 3x-y+3z-1=0
⇒6x-2y+6z=2
⇒a=6,b=6,c=2
Hence, 3a+4b-5c=18+24-10=32

Topic: Mathematics, Three Dimensional Geometry, JEE Main
A plane P=0 passing through the point 1,1,1 is perpendicular to the planes 2x-y+2z=5 and 3x+6y-2z=7. If the distance of the point 1,2,3 from the plane P=0 is k units, then the value of 34k2 is equal to

Options:
A 817
B 16
C 64
D 128

The correct answer is:
128

Solution:
Normal vector to the plane P=0 is i^ j^ k^2-1 23 6-2
=i^-10-j^-10+k^15
=-10i^+10j^+15k^
Equation of the plane through 1,1,1 is
-2x-1+2y-1+3z-1=0
⇒-2x+2y+3z-3=0
So, the distance from the point 1,2,3 is -2+4+9-317=817 units

Topic: Mathematics, Three Dimensional Geometry, JEE Main
A plane P=0, which is perpendicular to line x-22=y+22=z-11 is passing through the point at which the above line meets the plane x+y+z=21, then the distance of plane P=0 from origin is

Options:
A 73
B 5
C 323
D 373

The correct answer is:
373

Solution:
Any general point on the line is 2λ+2, 2λ-2, λ+1
This point most satisfy x+y+z=21
Hence 2λ+2+2λ-2+λ+1=21⇒λ=4
So, the point of intersection of line and plane is (10, 6, 5)
So, equation of plane P = 0 is 2x-10+2(y-6)+1(z-5)=0
⇒2x+2y+z-37=0
Hence, distance of P=0 from origin is
=373=373

Topic: Mathematics, Three Dimensional Geometry, JEE Main
A plane P1 has the equation 4x-2y+2z=3 and P2 has the equation -x+ky-2z=7. If the angle between P1 and P2 is 2π3, then the value of k can be

Options:
A 1
B 2
C -17
D 17

The correct answer is:
17

Solution:
A normal vector to plane P1 is <2,-1,1>
A normal vector to plane P2 is <-1,k,-2>
Now, cos⁡2π3=-2-k-26k2+5=-12
⇒2k+42=6k2+30
⇒4k2+32k+64=6k2+30
⇒2k2-32k-34=0
⇒k2-16k-17=0
⇒k-17k+1=0
⇒k=17,-1

Topic: Mathematics, Three Dimensional Geometry, JEE Main
A plane P contains the line of intersection of the plane r→·i^+j^+k^=6 and r→·2i^+3j^+4k^=-5. If P passes through the point (0, 2, -2),then the square of distance of the point (12, 12, 18) from the plane P is

Options:
A 620
B 155
C 310
D 1240

The correct answer is:
620

Solution:
Given,

A plane P contains the line of intersection of the plane r→·i^+j^+k^=6 and r→·2i^+3j^+4k^=-5,

Now let, P1 : x+y+z-6=0

And P2 : 2x+3y+4z+5=0

So, by family of planes concept we get,

P : x+y+z-6+λ2x+3y+4z+5=0

Also given plane P passes through (0, 2, -2)

So, 0+2-2-6+λ4+6-8+5=0

⇒-6+λ3=0

⇒λ=2

Hence, equation of plane will be,

P : 5x+7y+9z+4=0

Now distance of point 12,12,18 from plane will be, 512+712+918+425+49+81

And square of distance will be,

=512+712+918+425+49+812

=310×310155=620

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (06 Apr Shift 2)
A plane P contains the line x+2y+3 z+1=0=x-y-z-6, and is perpendicular to the plane -2x+y+z+8=0. Then which of the following points lies on P?

Options:
A (2,-1,1)
B (0,1,1)
C (-1,1,2)
D (1,0,1)

The correct answer is:
(0,1,1)

Solution:
A plane which containing the line

x+2y+3z+1=0=x-y-z-6 is

x+2y+3z+1+λx-y-z-6=0

⇒(1+λ)x+(2-λ)y+(3-λ)z+1-6λ=0

This plane is perpendicular to the plane -2x+y+z+8=0 then

-2(1+λ)+1(2-λ)+1(3-λ)=0

⇒λ=34

Equation of plane is

74x+54y+94z=144

⇒7x+5y+9z=14

Clearly, (0,1,1) lies on the plane.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (26 Aug Shift 1)
A plane P is parallel to two lines whose direction ratios are -2,1,-3, and -1,2,-2 and it contains the point 2,2,-2. Let P intersect the co-ordinate axes at the points A,B,C making the intercepts α,β,γ. If V is the volume of the tetrahedron OABC, where O is the origin and p=α+β+γ, then the ordered pair V,p is equal to

Options:
A 48,-13
B 24,-13
C 48,11
D 24,-5

The correct answer is:
24,-13

Solution:
Given a plane P is parallel to two lines whose direction ratios are -2,1,-3, and -1,2,-2,

So normal of plane P is given by, i^j^k^-21-3-12-2=4i^-j^-3k^

Now equation of plane P which passes through 2,2,-2 will be,

4x-2-y-2-3z+2=0

⇒4x-y-3z-12=0

Now, coordinates of plane making intercepts at X,Y & Z axis will be A3,0,0,B0,-12,0,C0,0,-4

⇒α=3,β=-12,γ=-4

⇒p=α+β+γ=-13

Now finding volume of tetrahedron OABC

V=16OA→·OB→×OC→=163i^·-12j^×-4k^

⇒V=163i^·48i^=8×3=24

So, V,p=24,-13

Topic: Mathematics, Three Dimensional Geometry, JEE Main
A plane P is parallel to two lines whose direction ratios are -2,1,-3, and -1,2,-2 and it contains the point 2,2,-2. Let P intersect the co-ordinate axes at the points A,B,C making the intercepts α,β,γ. If V is the volume of the tetrahedron OABC, where O is the origin and p=α+β+γ, then the ordered pair V,p is equal to

Options:
A 48,-13
B 24,-13
C 48,11
D 24,-5

The correct answer is:
24,-13

Solution:
Given a plane P is parallel to two lines whose direction ratios are -2,1,-3, and -1,2,-2,

So normal of plane P is given by, i^j^k^-21-3-12-2=4i^-j^-3k^

Now equation of plane P which passes through 2,2,-2 will be,

4x-2-y-2-3z+2=0

⇒4x-y-3z-12=0

Now, coordinates of plane making intercepts at X,Y & Z axis will be A3,0,0,B0,-12,0,C0,0,-4

⇒α=3,β=-12,γ=-4

⇒p=α+β+γ=-13

Now finding volume of tetrahedron OABC

V=16OA→·OB→×OC→=163i^·-12j^×-4k^

⇒V=163i^·48i^=8×3=24

So, V,p=24,-13

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2022 (28 Jul Shift 2)
A plane P meets the coordinate axes at A,B and C respectively. The centroid of ∆ABC is given to be 1,1,2. Then the equation of the line through this centroid and perpendicular to the plane P is :

Options:
A x-12=y-11=z-21
B x-11=y-11=z-22
C x-12=y-12=z-21
D x-11=y-12=z-22

The correct answer is:
x-12=y-12=z-21

Solution:
LetA(α,0,0),B(0,β,0),C(0,0,γ)

then

Gα3,β3,γ3≡(1,1,2)

α=3,β=3,γ=6

∴

eequation of plane is

xα+γβ+zγ=1

⇒x3+y3+z6=1

⇒2x+2y+z=6

∴

required line

x-12=y-12=z-21

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2020 (06 Sep Shift 2)
A plane P passes through the point 1,1,1 and is parallel to the vectors aâ=-i^+j^ and bâ=i^-k^. The distance of the point 332,33,3 from the plane is equal to

Options:
A 3 units
B 92 units
C 33 units
D 3 units

The correct answer is:
92 units

Solution:
Equation of the plane is x-1y-1z-1-11010-1=0
⇒x-1-1-y-11+z-1-1=0
⇒x+y+z-3=0
The distance of point 332,33,3 from x+y+z-3=0 is
332+33+3-33=92 units

Topic: Mathematics, Three Dimensional Geometry, JEE Main
A plane passes through 1,-2,1 and is perpendicular to two planes 2x-2y+z=0 and x-y+2z=4. The distance of the plane from the point 0,2,2 is

Options:
A 32 units
B 42 units
C 32 units
D 22 units

The correct answer is:
32 units

Solution:
Normal vector perpendicular to the required plane is
i^j^k^2-211-12
=i^-3-j^3+k^0
=-3i^-3j^
Equation of the plane through 1,-2,1 is
1x-1+1y+2+0z-1=0
x+y+1=0
Distance from 0,2,2 is 2+12=32

Topic: Mathematics, Three Dimensional Geometry, JEE Main
A plane passes through the point -2,-2,2 and contains the line joining the points 1,-1,2 and 1,1,1. Then the image of -7,2,3 in the plane is

Options:
A 1,-1,5
B -5,-4,-2
C -6,-1,-3
D 1323,723,623

The correct answer is:
-6,-1,-3

Solution:
The plane passes through the points -2,-2,2, 1,-1,2, 1,1,1
Its equation is x+2y+2z-231033-1=0
âx-3y-6z+8=0
Image of -7,2,3 in the plane is
x+71=y-2-3=z-3-6=-2-7-6-18+81+9+36
x+71=y-2-3=z-3-6=2323=1
x=-6,y=-1,z=-3

Topic: Mathematics, Three Dimensional Geometry, JEE Main
A plane passes through the points A1,2,3,B2,3,1 and C2,4,2. If O is the origin and P is 2,-1,1, then the projection of OP→ on this plane is of length:

Options:
A 25
B 27
C 23
D 211

The correct answer is:
211

Solution:
Normal to plane n→=i^j^k^11-2011

=3i^-j^+k^

OP→=2i^-j^+k^

cosθ=6+1+1611=866⇒sinθ=266

∴ Projection of OP→ on plane =|OP→|sinθ

=211

option 4

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (25 Feb Shift 2)
A plane passing though the points (0, -1, 0) and (0, 0, 1) and making an angle π4 with the plane y–z+5=0, also passes through the point

Options:
A (2, -1, 4)
B (2, 1, 4)
C (-2, -1, -4)
D (-2, 1, -4)

The correct answer is:
(2, 1, 4)

Solution:
The equation of a plane passing through a point x1, y1, z1 and having direction ratios of the normal to the plane as <a, b, c> is ax-x1+by-y1+cz-z1=0.

Thus, the equation of the plane passing through 0, -1, 0 is
ax-0+by+1+cz-0=0

⇒ax+by+cz+b=0

∵ It passes through 0, 0, 1

⇒c=-b

∴ Equation of plane is ax+by-bz+b=0 …i

∵ It makes angle π4 with plane y-z+5=0

The angle θ between the planes a1x+b1y+c1z+d1=0 and a2x+b2y+c2z+d2=0 is defined as cosθ=a1a2+b1b2+c1c2a12+b12+c12×a22+b22+c22

⇒cosπ4=0+b+ba2+b2+b2×2

⇒12=2ba2+2b2×2

Squaring both sides, we get

⇒a2+2b2=4b2

⇒a2=2b2

⇒a=±2b

∴ Equation of plane is ± 2bx+by-bz+b=0

⇒±2x+y-z+1=0

∴ Option 2, 1, 4 satisfies the given equation.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2019 (09 Apr Shift 1)
A plane passing through $(-1,2,3)$ and whose normal makes equal angles with the coordinate axes is

Options:
A $x+y+z+4=0$
B $x-y+z+4=0$
C $x+y+z-4=0$
D $x+y+z=0$

The correct answer is:
$x+y+z-4=0$

Solution:
A plane passing through the point $(-1,2,3)$, then its equation is
$$
a(x+1)+b(y-2)+c(z-3)=0
$$
where $\langle a, b, c\rangle$ are direction ratios of normal to the plane $A B C$.
So, the normal makes equal angles with coordinate axes i.e.,
$$
(a, b, c)=\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)
$$
Now, from Eq. (i),
$$
\begin{array}{cc}
\frac{1}{\sqrt{3}}(x+1)+\frac{1}{\sqrt{3}}(y-2)+\frac{1}{\sqrt{3}}(z-3)=0 \\
\Rightarrow \quad x+y+z-4 & =0
\end{array}
$$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
A plane passing through the point 3,1,1 contains two lines whose direction ratios are 1, – 2, 2 and 2, 3,–1 respectively. If, this plane also passes through the point α,–3,5, then α is equal to

Options:
A 5
B -10
C 10
D -5

The correct answer is:
5

Solution:
Plane is a(x-3)+b(y-1)+c(z-1)=0

Direction ratios of lines are1, -2, 2, 2, 3, -1.

Since both lines lie in to the plane implies dot product of direction ratios of line with direction ratio of normal of plane will be zero.

⇒a-2b+2c=0

⇒2a+3b-c=0

a-4=b--5=c7⇒a4=b-5=c-7

∴ Direction ratios of normal of plane 4,-5,-7

∴ Equation of plane is 4x-5y-7z=0

α,-3,5 lies on plane

∴α=5

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2020 (02 Sep Shift 2)
A plane polarized light is incident on a polariser with its pass axis making angle $\theta$ with $\mathrm{x}$-axis, as shown in the figure. At four different values of $\theta, \theta=8^{\circ}, 38^{\circ}, 188^{\circ}$ and $218^{\circ}$, the observed intensities are same. What is the angle between the direction of polarization and $\mathrm{x}$-axis

Options:
A
$203^{\circ}$

B
$45^{\circ}$

C
$98^{\circ}$

D
$128^{\circ}$

The correct answer is:
$203^{\circ}$

Solution:
No Solution Available

Topic: Physics, Wave Optics, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)