datavorous/entrance-exam-dataset huggingface dataset:
A light rod of length, l has two masses, m1 and m2 attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is

Options:
A m1m2m1+m2l2
B m1+m2m1m2l2
C m1+m2l2
D m1m2l2

The correct answer is:
m1m2m1+m2l2

Solution:
x1=m2lm1+m2 and x2=m1lm1+m2.

I=m1x12+m2x22=m1m2m1+m2l2.

Topic: Physics, Rotational Motion, JEE Main
A light rod of length L is hanging from the vertical smooth wall of a vehicle moving with acceleration 3galong the horizontal plane having a small mass attached at its one end and is free to rotate about an axis passing through the other end. The minimum velocity given to the mass at its equilibrium position with respect to the vehicle so that it can complete vertical circular motion is

Options:
A 5gL
B 4gL
C 8gL
D None of these

The correct answer is:
8gL

Solution:
The effective acceleration due to gravity in the reference frame of the vehicle is

geff=g2+3g2=2g

The angle made by geff with the verticle is

θ=tan-13gg=60°

At the lowest point (shown in the figure), let the speed is v, then

12mv2=m2g2L

⇒ v=8gL

Topic: Physics, Laws of Motion, JEE Main
A light rope is wound around a hollow cylinder of mass 5 kg and radius 70 cm. The rope is pulled with a force of 52.5 N. The angular acceleration of the cylinder will be _____ rad s-2.

Options:

The correct answer is:
15

Solution:
The formula to calculate the moment of inertia I of the cylinder about its radial axis is given by

I=Mr2 ...1

Also, the torque τ on the cylinder about the central axis because of the application of the external force is given by

τ= Fr ...2

Also, the torque can be expressed as

τ=Iα ...3

Substitute the expressions from equation (1) and (2) into equation (3) and simplify to obtain the angular acceleration.

Fr= Mr2α⇒α= FMr ...4

Substitute the values of the known parameters into equation (4) to calculate the required angular acceleration.

α= 52.5 N5 kg×0.70 m= 15 rad s-2

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2023 (13 Apr Shift 2)
A light spring balance hangs from the hook of the other light spring balance and a block of mass $M \mathrm{~kg}$ hangs from the former one. Then the true statement about the scale reading is

Options:
A Both the scales read $M / 2 \mathrm{~kg}$ each
B Both the scales read $M \mathrm{~kg}$ each
C The scale of the lower one reads $M \mathrm{~kg}$ and of the upper one zero
D The reading of the two scales can be anything but the sum of the reading will be $M \mathrm{~kg}$

The correct answer is:
Both the scales read $M \mathrm{~kg}$ each

Solution:
As the spring balance are massless therefore both the scales read $M \mathrm{~kg}$ each.

Topic: Physics, Laws of Motion, JEE Main
A light spring balance hangs from the hook of the other light spring balance and a block of mass M kg hangs from the former one. Then the true statement about the scale reading is

Options:
A
Both the scales read $\mathrm{M} \mathrm{kg}$ each

B
The scale of the lower one reads $\mathrm{M} \mathrm{kg}$ and of the upper one zero

C
The reading of the two scales can be anything but the sum of the reading will be $\mathrm{M} \mathrm{kg}$

D
Both the scales read $\mathrm{M} / 2 \mathrm{~kg}$ each

The correct answer is:
Both the scales read $\mathrm{M} \mathrm{kg}$ each

Solution:
No Solution Available

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2003
A light string is tied at one end to a fixed support and to a heavy string of equal length L at the other end A as shown in the figure ( Total length of both strings combined is 2L). A block of mass M is tied to the free end of heavy string. Mass per unit length of the strings are Î¼ and 16Î¼ and tension is T. Find lowest positive value of frequency such that junction point A is a node.

Options:
A 1LTμ
B 52LTμ
C 32LTμ
D 12LTμ

The correct answer is:
12LTμ

Solution:
f1=n12LTμ, f2=n22LT16μ
f1=f2⇒n1=n24
n1=1, n2=4
fmin=12LTμ

Topic: Physics, Waves and Sound, JEE Main
A light string passing over a smooth light fixed pulley connects two blocks of masses m1 and m2. If the acceleration of the system is g8, then the ratio of masses is

Options:
A 97
B 81
C 43
D 53

The correct answer is:
97

Solution:
Acceleration of the system can be written as, a=m1-m2gm1+m2=g8

⇒8m1-8m2=m1+m2

⇒7m1=9m2

⇒m1m2=97

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2024 (31 Jan Shift 2)
A light string passing over a smooth light pulley connects two blocks of masses $m_1$ and $m_2$ (vertically). If the acceleration of the system is $\mathrm{g} / 8$, then the ratio of the masses is

Options:
A
8 : 1

B
9 : 7

C
4 : 3

D
5 : 3

The correct answer is:
9 : 7

Solution:
$a=\frac{m_1-m_2}{m_1+m_2} g ; \frac{1}{8}=\frac{m_1-m_2}{m_1+m_2} \Rightarrow m_1: m_2=9: 7$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2002
A light unstretchable string passing over a smooth light pulley connects two blocks of masses $m_1$ and $m_2$. If the acceleration of the system is $\frac{g}{8}$, then the ratio of the masses $\frac{m_2}{m_1}$ is :

Options:
A $8: 1$
B $5: 3$
C $4: 3$
D $9: 7$

The correct answer is:
$9: 7$

Solution:
$\begin{aligned} & a_{\text {sys }}=\left(\frac{m_2-m_1}{m_1+m_2}\right) g=\frac{g}{8} \\ & \Rightarrow \frac{m_2}{m_1}=\frac{9}{7}\end{aligned}$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2024 (09 Apr Shift 1)
A light wave has a frequency of $4 \times 10^{14} \mathrm{~Hz}$ and a wavelength of $5 \times 10^{-7}$ meters in a medium. The refractive index of the medium is

Options:
A 1.5
B 1.33
C 1.0
D 0.66

The correct answer is:
1.5

Solution:
$\mu=\frac{c}{v}=\frac{c}{v \lambda}=\frac{3 \times 10^8}{4 \times 10^{14} \times 5 \times 10^{-7}}=1.5$

Topic: Physics, Ray Optics, JEE Main
A light wave is incident normally on a glass slab of refractive index 1.5. If 4% of light gets reflected and the amplitude of the electric field of the incident light is 30 Vm, then the amplitude of the electric field for the wave propagating in the glass medium will be:

Options:
A 30Vm
B 6Vm
C 24Vm
D 10Vm

The correct answer is:
24Vm

Solution:
n= refractive index =εrμr
I=12ε0E02C
I′=12ε E2V
I′=0.96I
12εE2v=0.9612ε0E02C
E=0.96ε0ε0εrcvE0
(For most of the transparent medium μr≈1 )
⇒E=0.96 . 1εrμr. cvE0
=0.96.1n. n E0
E=0.96n E0E=0.961.5×30=24 Vm

Topic: Physics, Ray Optics, JEE Main, JEE Main 2019 (12 Jan Shift 1)
A light wave is incident normally over a slit of width $24 \times 10^{-5} \mathrm{~cm}$. The angular position of second dark fringe from the central maxima is $30^{\circ}$. What is the wavelength of light

Options:
A $6000 Å$
B $5000 Å$
C $3000 Å$
D $1500 Å$

The correct answer is:
$6000 Å$

Solution:
For second dark fringe $d \sin \theta=2 \lambda$
$\begin{aligned} & \Rightarrow \quad 24 \times 10^{-5} \times 10^{-2} \times \sin 30=2 \lambda \\ & \Rightarrow \quad \lambda=6 \times 10^{-7} \mathrm{~m}=6000 Å\end{aligned}$

Topic: Physics, Wave Optics, JEE Main
A light wave travelling linearly in a medium of dielectric constant 4, incidents on the horizontal interface separating medium with air. The angle of incidence for which the total intensity of incident wave will be reflected back into the same medium will be :
(Given : relative permeability of medium μr=1)

Options:
A 10°
B 20°
C 30°
D 60°

The correct answer is:
60°

Solution:
Relative refractive index is μ=μrεr=4×1=2

Now, for complete reflection, θ>θc, where, θc is critical angle.

Now, critical angle is sinθc=1μ=12

⇒θc=sin-112⇒θc=30°

So, the angle of incidence for which the total intensity of incident wave will be reflected back into the same medium is θ>30° i.e., 60°.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2022 (25 Jun Shift 1)
A light whose electric field vectors are completely removed by using a good polaroid, allowed to incident on the surface of the prism at Brewster's angle. Choose the most suitable option for the phenomenon related to the prism.

Options:
A Reflected and refracted rays will be perpendicular to each other.
B Wave will propagate along the surface of prism.
C No refraction, and there will be total reflection of light.
D No reflection. and there will be total transmission of light.

The correct answer is:
No reflection. and there will be total transmission of light.

Solution:
A light whose electric field vectors are completely removed by using a good Polaroid, allowed to incident on the surface of the prism at Brewster's angle. Then, no reflection of light but there will be total transmission of light.

Topic: Physics, Wave Optics, JEE Main, JEE Main 2022 (25 Jun Shift 2)
A light-emitting diode (LED) has a voltage drop of 2 V across it and passes a current of 10 mA. When it operates with 6 V battery through a limiting resistor R, what is the value of R (in ohm)?

Options:

The correct answer is:
400

Solution:
Current in the circuit = 10 mA = 10×10-3 A
and voltage in the circuit = 6-2= 4 V
from Ohm's law,
V=IR
∴ R=VI=410×10-3= 400 Ω

Topic: Physics, Semiconductors, JEE Main
A line 4x+y=1 passes thought the point A 2,-7 and meets the line BC at B whose equation is 3x-4y+1=0. If the equation of line AC such that AB=AC is xa+yb=1, then the value of a is equal to

Options:
A − 519 52
B 519 52
C -51989
D 51989

The correct answer is:
− 519 52

Solution:
tanâ¡Î¸=m1-m21+m1m2=-4-341-3=198
198=m-341+3m4
m=-4 or -5289
Equation AC is 52x+89y+519=0
âa=-51952

Topic: Mathematics, Straight Lines, JEE Main
A line $A B$ in three-dimensional space makes angles $45^{\circ}$ and $120^{\circ}$ with the positive $x$-axis and the positive $y$-axis respectively. If $A B$ makes an acute angle $\theta$ with the positive $z$-axis, then $\theta$ equals

Options:
A
$45^{\circ}$

B
$60^{\circ}$

C
$75^{\circ}$

D
$30^{\circ}$

The correct answer is:
$60^{\circ}$

Solution:
$$
\begin{aligned}
& \ell=\cos 45^{\circ}=\frac{1}{\sqrt{2}} \\
& m=\cos 120^{\circ}=-\frac{1}{2} \\
& n=\cos \theta
\end{aligned}
$$
where $\theta$ is the angle which line makes with positive z-axis.
Now $\ell^2+m^2+n^2=1$
$$
\begin{aligned}
& \Rightarrow \frac{1}{2}+\frac{1}{4}+\cos ^2 \theta=1 \\
& \cos ^2 \theta=\frac{1}{4} \\
& \Rightarrow \cos \theta=\frac{1}{2} ( $\theta$ Being acute)\\
& \Rightarrow \theta=\frac{\pi}{3}
\end{aligned}
$$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2010
A line drawn through the point P4,7 cuts the circle x2+y2=9 at the points A and B. Then PA·PB is equal to.

Options:
A 74
B 53
C 56
D 65

The correct answer is:
56

Solution:
As we know,
PA.PB=PT2=S12S1=16+49−9=56⇒S12=56⇒PA.PB=56

Topic: Mathematics, Circle, JEE Main, JEE Main 2017 (09 Apr Online)
A line in the 3-dimensional space makes an angle θ 0 < θ ≤ π 2 with both the X and Y-axes. Then, the set of all values of θ is in the interval :

Options:
A ( π 3 , π 2 ]
B ( 0 , π 4 ]
C π 4 , π 2
D π 6 , π 3

The correct answer is:
π 4 , π 2

Solution:
cos 2 α + cos 2 β + cos 2 γ = 1
α = β = θ⇒ cos 2 γ + 2 cos 2 θ = 1
cos 2 γ = 1 - 2 cos 2 θ
0 ≤ cos 2 γ ≤ 1
0 ≤ 1 - 2 cos 2 θ ≤ 1
- 1 ≤ - 2 cos 2 θ ≤ 0
0 ≤ cos 2 θ ≤ 1 2 (as θ > 0 , cos θ < 1 )
⇒ 0 ≤ cos θ ≤ 1 2
π 4 ≤ θ ≤ π 2

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2014 (09 Apr Online)
A line is a common tangent to the circle x-32+y2=9 and the parabola y2=4x. If the two points of contact a,b and c,d are distinct and lie in the first quadrant, then 2a+c is equal to ___ .

Options:

The correct answer is:
9

Solution:
Let coordinate of point At2,2t ∵a=1

equation of tangent at point A

yt=x+t2

x-ty+t2=0

centre of circle 3,0

Now Perpendicular distance = radius

3-0+t21+t2=3

3+t22=91+t2

9+t4+6t2=9+9t2

t=0,-3,3

So point A3,23

⇒a=3, b=23

(Since it lies in first quadrant)

For point B which is foot of perpendicular from centre 3,0 to the tangent x-3y+3=0

c-31=d-0-3=-3-0+34

⇒c=32 d=332

⇒23+32=9

Topic: Mathematics, Parabola, JEE Main, JEE Main 2021 (25 Feb Shift 2)
A line is drawn from the point P1,1,1 and perpendicular to a line whose direction ratios are 1,1,1 to intersect the plane x+2y+3z=4 at the point Q. The locus of Q is

Options:
A x1=y-5-2=z+21
B x-2=y-51=z+21
C x=y=z
D x2=y3=25

The correct answer is:
x1=y-5-2=z+21

Solution:
Let point Q is x1,y1,z1
D.R’s of PQ are x1-1,y1-1,z1-1
⇒1x1-1+1y1-1+1z1-1=0
⇒x1+y1+z1=3
Also x1+2y1+3z1=4
x1+y1+z1=3
⇒x1,y1,z1 is a point on the two planes
x+2y+3z=4
x+y+z=3
∴ the line on the intersection of these two planes will be the required locus

Now, the direction vector of the line is i^j^k^111123=i^-2j^+k^

Also, the point 0,5,-2 satisfy both the equation of planes

Hence, the equation of the required line is x1=y-5-2=z+21

Topic: Mathematics, Three Dimensional Geometry, JEE Main
A line is drawn through a fixed point $P(\alpha, \beta)$ to cut the circle $x^{2}+y^{2}=r^{2}$ at $A$ and $B$. Then
$P A \cdot P B$ is equal to

Options:
A $(\alpha+\beta)^{2}-r^{2}$
B $\alpha^{2}+\beta^{2}-r^{2}$
C $(\alpha-\beta)^{2}+r^{2}$
D None of the above

The correct answer is:
$\alpha^{2}+\beta^{2}-r^{2}$

Solution:
The equation of any line through the point $P(\alpha, \beta)$ is
$$
\frac{x-\alpha}{\cos \theta}=\frac{y-\beta}{\sin \theta}=k \text { (say) }
$$
Any point on this line is
$$
(\alpha+k \cos \theta, \beta+k \sin \theta)
$$
This point lies on the given circle, if
$$
(\alpha+k \cos \theta)^{2}+(\beta+k \sin \theta)^{2}=r^{2}
$$
or $k^{2}+2 k(\alpha \cos \theta+\beta \sin \theta)$
$$
+\alpha^{2}+\beta^{2}-r^{2}=0
$$
Which being quadratic in $k$, gives two values of $k$. Let $P A=k_{1}, P B=k_{2}$, where $k_{1}, k_{2}$ are the roots of Eq. (i), then
$$
P A \cdot P B=k_{1} k_{2}=\alpha^{2}+\beta^{2}-r^{2}
$$

Topic: Mathematics, Circle, JEE Main
A line is drawn through the point $(1,2)$ to meet the coordinate axes at $P$ and $Q$ such that it forms a triangle $OPQ$, where $O$ is the origin. If the area of the triangle $OPQ$ is least, then the slope of the line $PQ$ is

Options:
A
$-\frac{1}{4}$

B
$-4$

C
$-2$

D
$-\frac{1}{2}$

The correct answer is:
$-2$

Solution:
Equation of line passing through $(1,2)$ with slope $m$ is $y-2=m(x-1)$
Area of $\Delta \mathrm{OPQ}=\frac{(m-2)^2}{2|m|}$
$\Delta=\frac{m^2+4-4 m}{2 m} \quad \Delta=\frac{m}{2}+\frac{2}{m}-2$
$\Delta$ is least if $\frac{m}{2}=\frac{2}{m} \quad \Rightarrow m^2=4 \quad \Rightarrow m=\pm 2 \quad \Rightarrow m=-2$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2012 (Offline)
A line L lies in the plane 2x-y-z=4 such that it is perpendicular to the line x-22=y-31=z-45. The line L passes through the point of intersection of the given line and given plane. Which of the following points does not satisfy line L?

Options:
A -5,-12,-272
B 3,1,6
C 0,292,-372
D -4,52,-292

The correct answer is:
3,1,6

Solution:
Any point on given line is 2Î»+2,Î»+3,5Î»+4
This must satisfy the equation of the plane
â4Î»+4-Î»-3-5Î»-4=4
â-2Î»=7âÎ»=-72
So, the coordinates of point are -5,-12,-272
A normal vector to given plane nâ=2i^-j^-k^
A vector parallel to given line bâ=2i^+j^+5k^
A vector parallel to required line
i^j^k^2-1-1215=i^-4-j^12+k^4
=-4i^+3j^-k^
Equation of required line is
x+51=y+123=z+272-1

Topic: Mathematics, Three Dimensional Geometry, JEE Main
A line L passing through 1,2,3 and perpendicular to the line L1:x-1-2=y+13=z-54 is also intersecting the line L1. If line L intersects the plane 2x+y+z+6=0 at point Î±,Î²,Î³, then the value of 2020Î±+Î²+2Î³ is equal to

Options:
A 2058
B 78
C 28
D -4012

The correct answer is:
28

Solution:
Given, L1 :x-1-2=y+13=z-54
Let, bâ=-2i^+3j^+4k^ (a vector parallel to the line)
From diagram coordinates of point are B-2Î»+1,3Î»-1,4Î»+5
Now, DR's of ABâ=<-2Î»,3Î»-3,4Î»+2>
ABââbâ=0â4Î»+9Î»-9+16Î»+8=0
â29Î»=1âÎ»=129
ABâ=-2i^29-84j^29+62k^29
So, the equation of line L is x-11=y-242=z-3-31
Any general point on this line is Î»+1,42Î»+2,-31Î»+3
This must satisfy the equation of the plane
â2Î»+2+42Î»+2-31Î»+3+6=0
â13Î»+13=0âÎ»=-1
So, Î±,Î²,Î³â¡0,-40,34
Hence, 2020Î±+Î²+2Î³=28

Topic: Mathematics, Three Dimensional Geometry, JEE Main
A line l passing through origin is perpendicular to the lines

l1:r→=3+ti^+-1+2tj^+4+2tk^

l2:r→=3+2si^+3+2sj^+2+sk^

If the co-ordinates of the point in the first octant on l2 at a distance of 17 from the point of intersection of l and l1 are a, b, c, then 18a+b+c is equal to ___ .

Options:

The correct answer is:
44

Solution:
l1:r→=3+ti^+-1+2tj^+4+2tk^

l2:r→=3+2si^+3+2sj^+2+sk^

DR of l1≡1,2,2

DR of l2≡2,2,1

l1×l2=ijk122221

=i2-4-j1-4+k2-4

=-2i+3j-2k

Hence, DR of l ( line ⊥ to l1 & l2)

=-2,3,-2

∴ l:r→=-2μi^+3μj^-2μk^

for intersection of l & l1

3+t=-2μ

-1+2t=3μ

4+2t=-2μ

⇒t=-1 & μ=-1

∴ Point of intersection P≡2, -3, 2

Let point on l2 be Q3+2s, 3+2s, 2+s

Given PQ=17 ⇒PQ2=17

⇒2s+12+6+2s2+s2=17

⇒9s2+28s+20=0

⇒s=-2,-109

s≠-2 as point lies on 1st octant.

∴a=3+2-109=79

b=3+2-109=79

c=2+-109=89

∴18a+b+c=18229=44

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (25 Feb Shift 2)
A line makes an angle Î¸ with the x-axis and the y-axis. If it makes an angle Î± with the z-axis such that sin2â¡Î±=3sin2â¡Î¸, then cosâ¡2Î¸ is equal to

Options:
A 32
B 34
C 15
D 12

The correct answer is:
15

Solution:
Direction cosines of the line are cosâ¡Î¸,cosâ¡Î¸,cosâ¡Î±
As we know that cos2â¡Î¸+cos2â¡Î¸+cos2â¡Î±=1
â2cos2â¡Î¸=sin2â¡Î±â¦â¦i
Also, sin2â¡Î±=3sin2â¡Î¸ (Given)
â2cos2â¡Î¸=3sin2â¡Î¸âtan2â¡Î¸=23
cosâ¡2Î¸=1-tan2â¡Î¸1+tan2â¡Î¸=1-231+23=15

Topic: Mathematics, Three Dimensional Geometry, JEE Main
A line makes angles α, β, γ with the coordinate axes. If α+β=π2, then cosα+cosβ+cosγ2 is equal to

Options:
A 1+sin2α
B 1-sin2α
C 1+cos2α
D None of these

The correct answer is:
1+sin2α

Solution:
We have, cos2α+cos2β+cos2γ=1

⇒cos2α+cos2π2-α+cos2γ=1 [given α+β=π2]

⇒cos2α+sin2α+cos2γ=1

⇒1+cos2γ=1

⇒cos2γ=0

⇒cosγ=0

∴ cosα+cosβ+cosγ2=cosα+sinα2

=1+2sinα·cosα=1+sin2α

Topic: Mathematics, Three Dimensional Geometry, JEE Main
A line makes the same angle $\theta$, with each of the $x$ and $z$ axis. If the angle $\beta$, which it makes with $y$-axis, is such that $\sin ^2 \beta=3 \sin ^2 \theta$, then $\cos ^2 \theta$ equals

Options:
A
$\frac{2}{3}$

B
$\frac{1}{5}$

C
$\frac{3}{5}$

D
$\frac{2}{5}$

The correct answer is:
$\frac{3}{5}$

Solution:
$I=\cos \theta, m=\cos \theta, n=\cos \beta$
$\cos ^2 \theta+\cos ^2 \theta+\cos ^2 \beta=1 \Rightarrow 2 \cos ^2 \theta=\sin ^2 \beta=3 \sin ^2 \theta \quad$ (given)
$\cos ^2 \theta=3 / 5$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2004
A line meets $x$-axis and $y$-axis at the points $A$ and $B$ respectively. If the middle point of $A B$ be $\left(x_1, y_1\right)$, then the equation of the line is

Options:
A $y_1 x+x_1 y=2 x_1 y_1$
B $x_1 x+y_1 y=2 x_1 y_1$
C $y_1 x+x_1 y=x_1 y_1$
D $x_1 x+y_1 y=x_1 y_1$

The correct answer is:
$y_1 x+x_1 y=2 x_1 y_1$

Solution:
Obviously, $x_1=\frac{a}{2}$ and $y_1=\frac{b}{2}$
Therefore the equation of line $A B$ is $\frac{x}{a}+\frac{y}{b}=1$
$\Rightarrow \quad \frac{x}{2 x_1}+\frac{y}{2 y_1}=1 \Rightarrow x y_1+y x_1=2 x_1 y_1$

Topic: Mathematics, Straight Lines, JEE Main
A line parallel to 6i^-3j^+2k^ intersects the line x1=y2=z2 at A and the line x+11=y+22=z+32 at B, then the length of AB is equal to

Options:
A 37 units
B 5439 units
C 785 units
D 9425 units

The correct answer is:
785 units

Solution:
Both the given lines are parallel to bâ=i^+2j^+2k^
Let the angle between the line and 6i^-3j^+2k^ is Î¸
â´cosÎ¸=i^+2j^+2k^â6i^â3j^+2k^3Ã7=421
Let the 1st line is passing through aâ=0â and the 2nd line is passing through câ=-i^-2j^-3k^
AC= distance between the parallel lines =aâ-câÃbâbâ
=i^+2j^+3k^Ãi^+2j^+2k^3=53
â´AB=ACsecâ¡90-Î¸=ACcosecÎ¸
=53Ã21425=785

Topic: Mathematics, Three Dimensional Geometry, JEE Main
A line parallel to the straight line 2x-y=0 is tangent to the hyperbola x24−y22=1 at the point x1, y1. Then x12+5y12 is equal to

Options:
A 6
B 8
C 10
D 5

The correct answer is:
6

Solution:
Tangent at x1,y1

xx1-2yy1-4=0

This is parallel to 2x-y=0

⇒ x12y1=2

⇒ x1=4y1 ...1
Point x1,y1 lie on hyperbola x124-y122-1=0 ...2

Using equation 1 and 2, 4y12-y122=1⇒y12=27 & x12=327

We get x12+5y12=6

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2020 (02 Sep Shift 1)
A line passes through $(2,2)$ and is perpendicular to the line $3 x+y=3$. What is its $y$-intercept?

Options:
A $1 / 3$
B $2 / 3$
C 1
D $4 / 3$

The correct answer is:
$4 / 3$

Solution:
Equation of line perpendicular to $3 x+y-3=0$ is $x-3 y+c=0$
Since, it passes through $(2,2)$
$$
\therefore \mathrm{c}=4
$$
$\therefore$ Equation of line perpendicular to $3 x+y-3=0$ is $x-3 y+4=0$
For $y$ - intercept, put $x=0$
$$
\therefore y=\frac{4}{3}
$$

Topic: Mathematics, Straight Lines, JEE Main
A line passes through A4,−6,−2 and B16,−2,4. The point Pa,b,c where a,b,c are non-negative integers, on the line AB lies at a distance of 21 units, from the point A. The distance between the points Pa,b,c and Q4,−12,3 is equal to ______.

Options:

The correct answer is:
22

Solution:
Given,

A line passes through A4,−6,−2 and B16,−2,4,

So, the equation of line will be x−412=x+64=z+26

⇒x−46=x+62=z+23

So, the direction cosines will be 662+22+32,262+22+32,362+22+32

≡67,27,37

Now, given point Pa,b,c lies on the line AB and at a distance of 21 from point A

So, by using the equation of line AB we get,

⇒a−467=b+627=c+237=21

⇒21×67+4,27×21−6,37×21−2≡a,b,c

⇒22,0,7≡a,b,c

Hence, the distance between P22,0,7 & Q4,-12,3 is given by, 324+144+16=22

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2024 (31 Jan Shift 2)
A line passes through the point $(3,4)$ and cuts off intercepts from the coordinates axes such that their sum is 14. The equation of the line is

Options:
A $4 x-3 y=24$
B $4 x+3 y=24$
C $3 x-4 y=24$
D $3 x+4 y=24$

The correct answer is:
$4 x+3 y=24$

Solution:
Given $a+b=14 \Rightarrow a=14-b$
Hence the equation of straight line is $\frac{x}{14-b}+\frac{y}{b}=1$.
Also, it passes through $(3,4)$
$\therefore \frac{3}{14-b}+\frac{4}{b}=1 \Rightarrow b=8$
or 7
Therefore equations are $4 x+3 y=24$ and $x+y=7$.
Trick : This question can be checked with the options as the line $4 x+3 y=24$ passes through $(3,4)$ and also cuts the intercepts from the axes whose sum is 14 .

Topic: Mathematics, Straight Lines, JEE Main
A line passes through the point A2,3,5 and is parallel to the vector 2i^-j^+k^. If P is a point on this line such that AP=26, then the coordinates of point P can be

Options:
A 4,2,6
B 6,1,7
C –2,5,–3
D 2,3,5

The correct answer is:
6,1,7

Solution:
Equation of the line through A2,3,5 and parallel to 2i^-j^+k^ is
x-22=y-3-1=z-51
Coordinates of point P on this line are 2Î»+2,-Î»+3,Î»+5
AP=4Î»2+Î»2+Î»2=26
â6Î»2=24âÎ»=Â±2
Coordinates of point P can be 6,1,7 or â2,5,3

Topic: Mathematics, Three Dimensional Geometry, JEE Main
A line passes through the point of intersection of $2 x+y=5$ and $x+3 y+8=0$ and parallel to the line $3 x+4 y=7$ is

Options:
A $3 x+4 y+3=0$
B $3 x+4 y=0$
C $4 x-3 y+3=0$
D $4 x-3 y=3$

The correct answer is:
$3 x+4 y+3=0$

Solution:
Point of intersection
$\begin{aligned}
& \quad y=-\frac{21}{5} \text { and } x=\frac{23}{5} \\
& \therefore \quad 3 x+4 y=\frac{3(23)+4(-21)}{5}=\frac{69-84}{5}=-3
\end{aligned}$
Hence required line is $3 x+4 y+3=0$

Topic: Mathematics, Straight Lines, JEE Main
A line passing through the point 2,2 encloses an area of 4 sq. units with coordinate axes. The sum of intercepts made by the line on the x and y axis is equal to

Options:
A –2
B 4
C –4
D 2

The correct answer is:
–4

Solution:
Let equation of line is
xa+yb=1⇒2a+2b=1
⇒ 2b+a=ab ... 1
Case-1: If both intercepts are positive
Then, 12ab=4⇒ab=8⇒b+a=4
⇒ a4-a8⇒a2-4a+8=0 ⇒a4-a=8
⇒ No value of a is possible (Discriminant is negative)
Case-2: If intercepts are of opposite sign, then -12ab=4⇒ab=-8
⇒b+a=-82=-4
⇒a-4-a=-8⇒a2+4a-8=0
⇒ a is possible hence b is possible
⇒ the sum of intercepts is equal to -4

Topic: Mathematics, Straight Lines, JEE Main