datavorous/entrance-exam-dataset huggingface dataset:
If the lines $a x+y+1=0, x+b y+1=0$ and $x+y+c=0 \quad(a, b, c$ being distinct and different from 1) are concurrent, then $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=$

Options:
A 0
B 1
C $\frac{1}{a+b+c}$
D None of these

The correct answer is:
1

Solution:
If the given lines are concurrent, then
$\left|\begin{array}{lll}
a & 1 & 1 \\
1 & b & 1 \\
1 & 1 & c
\end{array}\right|=0 \Rightarrow\left|\begin{array}{ccc}
a & 1-a & 1-a \\
1 & b-1 & 0 \\
1 & 0 & c-1
\end{array}\right|=0$
$\left\{\right.$ Apply $C_2 \rightarrow C_2-C_1$ and $\left.C_3 \rightarrow C_3-C_1\right\}$
$\begin{aligned}
& \Rightarrow a(b-1)(c-1)-(b-1)(1-a)-(c-1)(1-a)=0 \\
& \Rightarrow \frac{a}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=0 \\
& \Rightarrow \frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=1
\end{aligned}$

Topic: Mathematics, Straight Lines, JEE Main
If the lines \( \frac{x-1}{1}=\frac{y-3}{1}=\frac{z-2}{\lambda} \) and \( \frac{x-1}{\lambda}=\frac{y-3}{2}=\frac{z-4}{1} \) intersect at a point, then the value of \( \lambda^{2}+4 \) is equal to

Options:
A \( 8 \)
B \( 10 \)
C \( 13 \)
D \( 5 \)

The correct answer is:
\( 8 \)

Solution:
Both the lines intersect at a point, that means, the vector joining their initial points and the vectors of the lines are coplanar.

⇒00211λλ21=0

⇒22-λ=0⇒λ=2.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the lines \( \frac{x-1}{-3}=\frac{y-2}{-2 k}=\frac{z-3}{2} \) and \( \frac{x-1}{k}=\frac{y-2}{1}=\frac{z-3}{5} \) are perpendicular, what would be the value of \( k \) and the equation of plane containing these lines?

Options:
A \( k=4 \) and \( -22 x+19 y=-96 \)
B \( k=2 \) and \( -56 y+5 z=-23 \)
C \( k=2 \) and \( -22 x+19 y+5 z=31 \)
D None of these

The correct answer is:
\( k=2 \) and \( -22 x+19 y+5 z=31 \)

Solution:
The given lines are

x−1−3=y−2−2k=z−32....(1)

x−1k=y−21=z−35.......(2)

For (1) and (2) to be ⊥, we must have −3k−2k +10=0

⇒−5k+10

⇒k=2.

The equation of lines becomes

x−1−3=y−2−4=z−32,

x−12=y−21=z−35

Now, the equation of plane containing these two lines is

x-x1y-y1z-z1l1m1n1l2m2n2=0

where x1=1, y1=2, z1=3,

l1=−3, m1=−4, n1=2

l2=2, m2=1, n2=5

∵x-1y-2z-3-3-42215=0

⇒x−1−20−2−y−2−15−4+z−3−3+8=0

⇒−22x+22+19y−38+5z−15=0

⇒22x−19y−5z+31=0.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the lines \( \frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}, \frac{x-1}{3 k}=\frac{y-5}{1}=\frac{z-6}{-5} \) are at right angles, then \( k= \)

Options:
A \( -10 \)
B \( \frac{10}{7} \)
C \( \frac{-10}{7} \)
D \( \frac{-7}{10} \)

The correct answer is:
\( \frac{-10}{7} \)

Solution:
Lines are perpendicular if a1a2+b1b2+c1c2=0
Hence, -3 3k+2k 1+2 -5=0⇒k=-107 .

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the lines L1:x-13=y-λ1=z-32 and L2:x-31=y-12=z-23 are coplanar, then the equation of the plane passing through the point of intersection of L1 and L2 which is at a maximum distance from the origin is

Options:
A 3x+2y+z=5
B 2x+3y+z=6
C 4x+3y+5z=0
D 4x+3y+5z=50

The correct answer is:
4x+3y+5z=50

Solution:
âµL1 and L2 are coplanar, hence,
-2Î»-11312123=0âÎ»=2
Now point of intersection will be P4,3,5
â´ Equation of required plane is 4x+3y+5z=50

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the lines $\mathrm{p}_{1} \mathrm{x}+\mathrm{q}_{1} \mathrm{y}=1, \mathrm{p}_{2} \mathrm{x}+\mathrm{q}_{2} \mathrm{y}=1$ and $\mathrm{p}_{3} \mathrm{x}+ \mathrm{q}_{3} \mathrm{y}=1$ be concurrent, then the points $\left(\mathrm{p}_{1}, \mathrm{q}_{1}\right),$

$\left(p_{2}, q_{2}\right)$ and $\left(p_{3}, q_{3}\right)$

Options:
A are collinear
B form an equilateral triangle
C form a sealene triangle
D form a right angled triangle

The correct answer is:
are collinear

Solution:
The equations of the lines are

$$

\begin{array}{l}

\mathrm{p}_{1} \mathrm{x}+\mathrm{q}_{1} \mathrm{y}-1=0 \\

\mathrm{p}_{2} \mathrm{x}+\mathrm{q}_{2} \mathrm{y}-1=0

\end{array}

$$

and $\mathrm{p}_{3} \mathrm{x}+\mathrm{q}_{3} \mathrm{y}-1=0$

As they are concurrent,

$$

\left|\begin{array}{lll}

\mathrm{p}_{1} & \mathrm{q}_{1} & -1 \\

\mathrm{p}_{2} & \mathrm{q}_{2} & -1 \\

\mathrm{p}_{3} & \mathrm{q}_{3} & -1

\end{array}\right|=0 \Rightarrow\left|\begin{array}{lll}

\mathrm{p}_{1} & \mathrm{q}_{1} & 1 \\

\mathrm{p}_{2} & \mathrm{q}_{2} & 1 \\

\mathrm{p}_{3} & \mathrm{q}_{3} & 1

\end{array}\right|=0

$$

This is also the condition for the points $\left(\mathrm{p}_{1}, \mathrm{q}_{1}\right),\left(\mathrm{p}_{2}, \mathrm{q}_{2}\right)$ and $\left(\mathrm{p}_{3}, \mathrm{q}_{3}\right)$ to be collinear.

Topic: Mathematics, Straight Lines, JEE Main
If the lines r→=i^-j^+k^+λ3j^-k^ and r→=αi^-j^+μ2i^-3k^ are co-planar, the the distance of the plane containing these two lines from the point α,0,0 is

Options:
A 29
B 211
C 411
D 2

The correct answer is:
211

Solution:
Given,

r→=i^-j^+k^+λ3j^-k^ ...L1

r→=αi^-j^+μ2i^-3k^

L1 and L2 are coplanar then a2-a1b1b2=0

So, 03-120-31-α01=0

-32+31-α=0

3α=5

⇒α=53
Now, normal vector will be, n→=i^j^k^03-120-3=i^-9-j^2+k-6

=9,2,6

So, equation of plane is given by,

9x-1+2y+1+6z-1=0

9x+2y+6z-13=0

Now finding perpendicular distance from α,0,0

=9·53+0+0-1381+36+4=2121=211

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2022 (26 Jun Shift 2)
If the lines x-11=y-22=z+31 and x-a2=y+23=z-31 intersects at the point P, then the distance of the point P from the plane z=a is :

Options:
A 16
B 28
C 10
D 22

The correct answer is:
28

Solution:
Let

L1:x-11=y-22=z+31=λ

Point on L1≡λ+1,2λ+2,λ-3

And,

L2:x-a2=y+23=z-31=μ

Point on L2≡2μ+a,3μ-2,μ+3

For finding P, we must have

λ-3=μ+3⇒λ=μ+6 ...i

2λ+2=3μ-2⇒2λ=3μ-4 ...ii

Solving i & ii, we get

λ=22 and μ=16

Therefore,

P≡23,46,19

And,

2μ+a=λ+1

⇒32+a=22+1

⇒a=-9

Distance of point P23,46,19 from the plane z+9=0 is

=19+902+02+12=28 units

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (29 Jan Shift 2)
If the lines x-11=y-31=z-2Î» and x-1Î»=y-32=z-41 intersect at a point, then the value of Î»2+4 is equal to

Options:
A 8
B 10
C 13
D 5

The correct answer is:
8

Solution:
Both the lines intersect at a point
So, 00211λλ21=0⇒22-λ=0⇒λ=2

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the lines x-12=2-y-3=z-3α and x-45=y-12=zβ intersect, then the magnitude of the minimum value of 8αβ is _____.

Options:

The correct answer is:
18

Solution:
Let,

L1=x-12=y-23=z-3α=λ

L2=x-45=y-12=z-0β=μ

For point of intersection

2λ+1=5μ+4 ... i

3λ+2=2μ+1 ... ii

αλ+3=βμ+0 ... iii

From i and ii, we get λ=μ=-1

Now, from iii, we get α-β=3

Let y=8αβ

⇒y=8αα-3

⇒y=8α2-3α+94-18

⇒y=8α-322-18

So, Minimum value of y=-18 at α=+32

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (06 Apr Shift 2)
If the lines x-12=y-1=z2 and x-y+z-2=0=Î»x+3z+5 are coplanar, then the value of 7Î» is equal to

Options:
A 31
B -52
C -39
D -31

The correct answer is:
-31

Solution:
A plane passing through the line of intersection of the given 2 planes is x-y+z-2+Î¼Î»x+3z+5=0
âx1+Î»Î¼-y+z1+3Î¼+5Î¼-2=0
Since the given line x-12=y-1=z2 must lie in this plane
â21+Î»Î¼+-1-1+1+3Î¼2=0
â2+2Î»Î¼+1+2+6Î¼=0
â2Î»Î¼+6Î¼+5=0â¦..i
The point 1,0,0 must satisfy the equation of the plane
â1+Î»Î¼-0+0+5Î¼-2=0
âÎ»Î¼+5Î¼-1=0â¦â¦ii
Solving i and ii we get,

Î¼=74, Î»=-317

â7Î»=-31

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the lines x1=y-1-1=z+22,x2=y-11=z+21,x-1=y-1-1=z+21 intersect the plane x+y-z=0 at P,Q,R respectively, then the area (in sq. units) of triangle PQR is

Options:
A 3
B 473
C 983
D 943

The correct answer is:
983

Solution:
Coordinates of P,Q,R are r1,-r1+1,2r1-2, 2r2,r2+1,r2-2, -r3,-r3+1,r3-2 respectively.
It must satisfy the plane, hence r1=32,r2=-32,r3=+1. So points P,Q,R are 32,-12,1,-3,-12,-72 –1,0,–1 respectively.
PQ→=-92i^-92k^
PR→=-52ı^+12j^-2k^
Area =12PQ→×PR→
=18ij^k^-90-9-51-4

=189i^+9j^-9k^=938 sq. units

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the lines x1=y2=z3,x-k3=y-3-1=z-4h and 2x+13=y-11=z-21 are concurrent, then the value of 2h-3k is equal to

Options:
A 3
B 2
C -4
D 4

The correct answer is:
4

Solution:
Any general point on the line x1=y2=z3 is λ,2λ,3λ
Any general point on the line x+123=y-12=z-22 is 3μ-12,2μ+1,2μ+2
If these two lines are intersecting, then
λ3μ-12=2λ2μ+1=3λ2μ+2⇒μ=12
So, the point of intersection is 1,2,3 and it must satisfy the third line.
1-k3=2-3-1=3-4h
1-k3=1=-1h⇒k=-2,h=-1

Topic: Mathematics, Straight Lines, JEE Main
If the lines x-21=y-31=z-4-k and x-1k=y-42=z-51 are coplanar, then k can have

Options:
A exactly two values.
B exactly three values.
C any value.
D exactly one value.

The correct answer is:
exactly two values.

Solution:
For x-x1a1=y-y1b1=z-z1c1
and x-x2a2=y-y2b2=z-z2c2 to be coplanar if
x2-x1y2-y1z2-z1a1b1c1a2b2c2=0.

∴1-1-111-kk21=0⇒k=0, -3
Hence, there are exactly two values of k.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2013 (07 Apr)
If the lines x-32=y-52=z-4λ and x-2λ=y-64=z-52 intersect at a point α,β,γ, then the greatest value of λ is equal to

Options:
A 0
B 2
C -2
D 4

The correct answer is:
0

Solution:
Two given lines intersect ⇒ they must be coplanar
So, 3-25-64-522λλ42=0

⇒14-4λ+14-λ2-18-2λ=0
⇒4-4λ+4-λ2-8+2λ=0
⇒λ2+2λ=0⇒λ=0,-2

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the lines x=ay+b,z=cy+d and x=a′z+b′,  y=c′  z+d′ are perpendicular, then

Options:
A cc’+a+a’=0
B aa’+c+c’=0
C bb’+cc’+1=0
D ab’+bc’+1=0

The correct answer is:
aa’+c+c’=0

Solution:
Direction Ratios of the first line
i^j^k^1-a00c-1=a i^+j^+ck^
Hence, Direction Ratios of the first line are a, 1, c.
For Direction Ratios of the second line
i^j^k^10-a′01-c′=a′i^+c′j^+k^
Hence, Direction Ratios of the 2nd line =a′, c′, 1
Since, both the lines are perpendicular,
a.a′+c′+c=0

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2019 (09 Jan Shift 2)
If the lines x-k1=y-22=z-33 and x+13=y+22=z+31 are co-planar, then the value of k is _______.

Options:

The correct answer is:
1

Solution:
Given, lines x-k1=y-22=z-33 and x+13=y+22=z+31 are co-planar.

Therefore, k+146123321=0

⇒k+12-6-41-9+62-6=0

⇒k=1.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (25 Jul Shift 2)
If the lines x+y=a and x-y=b touch the curve y=x2-3x+2 at the points where the curve intersects the x-axis, then ab is equal to …

Options:

The correct answer is:
0.5

Solution:
y=x2-3x+2, x+y=a,x-y=b

⇒x2-3x+2=0

x1=2 x2=1

Therefore, points are(2,0) (1,0)

b=2 a=1

∴ab=12=0.5

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2020 (05 Sep Shift 2)
If the local minimum of the function fx=x3-3a2x+4 âa>0 occurs at x=Î» âÎ»>1, then a may take the value

Options:
A 32
B 23
C 45
D 0

The correct answer is:
23

Solution:
f ′ x =3 x 2 −3 a 2
=3x-ax+a
Hence, minimum occurs at x=a
∴a>1

Topic: Mathematics, Application of Derivatives, JEE Main
If the locus of the complex number z given by argz+i-argz-i=2π3 is an arc of a circle, then the length of the arc is

Options:
A 4π3
B 4π33
C 23π3
D 2π33

The correct answer is:
4π33

Solution:
Given, argz+iz-i=2π3

Let, z=x+iy

⇒x+iy+1x+iy-1=x2+y2-1+2ixx2+y-12

⇒tan-12xx2+y2-1=2π3

⇒x2+y2+23x-1=0

Hence, the given locus is a circle with centre -13,0 and radius 23 units
⇒ Length of the arc of the circle is 2π3×23=4π33 units

Topic: Mathematics, Complex Number, JEE Main
If the locus of the foot of the perpendicular drawn from centre upon any tangent to the ellipse x240+y210=1 is x2+y22=ax2+by2, then a–b is equal to

Options:
A 10
B 20
C 25
D 30

The correct answer is:
30

Solution:
Slope of OP =khâ slope of the tangent is -hk

Hence, the equation of tangent PQ is (y-k)x-h=-hk
â y=-hkx+h2+k2k
âh2+k2k2=40-hk2+10 condition of tangency isc2=a2m2+b2
âx2+y22=40x2+10y2
âaâb=30

Topic: Mathematics, Ellipse, JEE Main
If the locus of the image of the point λ2,2λ in the line mirror x-y+1=0 (where λ is a parameter) is x-a2=by-c where a,b,c∈I, then the value of a+bc+b is equal to

Options:

The correct answer is:
0.6

Solution:
Let, h,k be the image of the point λ2,2λ about the line x-y+1=0

h-λ21=k-2λ-1=-2λ2-2λ+12
h+1=2λ…..i
k=λ2+1…..ii
Put the value of λ from i in ii
k-1=h+122

⇒4y-1=x+12
Hence, a=-1, b=4 and c=1

Topic: Mathematics, Straight Lines, JEE Main
If the locus of the mid-point of the line segment from the point 3,2 to a point on the circle, x2+y2=1 is a circle of radius r, then r is equal to

Options:
A 14
B 1
C 13
D 12

The correct answer is:
12

Solution:
Given circle x2+y2=1

General coordinate on a given circle is cosθ, sinθ.

Given, h=cosθ+32⇒cosθ=2h-32 ...i

k=sinθ+22⇒sinθ=2k-1 ...ii

Square i & ii then add we get,

h-322+k-12=122

⇒r=12

Topic: Mathematics, Circle, JEE Main, JEE Main 2021 (26 Feb Shift 2)
If the locus of the point, whose distances from the point $(2,1)$ and $(1,3)$ are in the ratio $5: 4$, is $a x^2+b y^2+c x y+d x+e y+170=0$, then the value of $a^2+2 b+3 c+4 d+e$ is equal to :

Options:
A 37
B 437
C -27
D 5

The correct answer is:
37

Solution:
let $\mathrm{P}(\mathrm{x}, \mathrm{y})$
$\begin{aligned}
& \frac{(x-2)^2+(y-1)^2}{(x-1)^2+(y-3)^2}=\frac{25}{16} \\
& 9 x^2+9 y^2+14 x-118 y+170=0 \\
& a^2+2 b+3 c+4 d+e \\
& =81+18+0+56-118 \\
& =155-118 \\
& =37
\end{aligned}$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2024 (06 Apr Shift 2)
If the $m^{\text {th }}$ term of a H.P. be $n$ and $n^{\text {th }}$ be $m$, then the $r^{\text {th }}$ term will be

Options:
A $\frac{r}{m n}$
B $\frac{m n}{r+1}$
C $\frac{m n}{r}$
D $\frac{m n}{r-1}$

The correct answer is:
$\frac{m n}{r}$

Solution:
Given $T_m=n, T_n=m$ for H.P. Therefore for the corresponding A.P. $m^{\text {th }}$ term $=\frac{1}{n}, n^{\text {th }}$ term $=\frac{1}{m}$
Let $a$ and $d$ be the first term and common difference of this A.P., then
$a+(m-1) d=\frac{1}{n} \ldots(i)$
$a+(n-1) d=\frac{1}{m} \ldots(ii)$
Solving these, we get
$a=\frac{1}{m n}, d=\frac{1}{m n}$
Now, $r^{\text {th }}$ term of corresponding A.P.
$=a+(r-1) d=\frac{1}{m n}+(r-1) \frac{1}{m n}$ $=\frac{1+r-1}{m n}=\frac{r}{m n}$
Therefore $r^{\text {th }}$ term of corresponding H.P. is
$\frac{m n}{r}$
Note : Students should remember this question as a fact.

Topic: Mathematics, Sequences and Series, JEE Main
If the magnetic dipole of an atom of diamagnetic material, paramagnetic material are denoted by $\mu_d, \mu_p$ and $\mu_f$ respectively, then:

Options:
A $\mu_d=0$ and $\mu_p \neq 0$
B $\mu_d \neq 0$ and $\mu_p=0$
C $\mu_p=0$ and $\mu_f \neq 0$
D $\mu_d \neq 0$ and $\mu_f \neq 0$

The correct answer is:
$\mu_d=0$ and $\mu_p \neq 0$

Solution:
$\mu_{\mathrm{p}} \neq \mu_f \neq 0$ and $\mu_a=0$

Topic: Physics, Magnetic Properties of Matter, JEE Main
If the magnetic field in a plane electromagnetic wave is given by B→=3×10-8sin⁡1.6×103x+48×1010tj^T, then what will be expression for electric field ?

Options:
A Eâ=60sinâ¡1.6Ã103x+48Ã1010tk^Vm
B E→=9sin⁡1.6×103x+48×1010tk^Vm
C Eâ=3Ã10-8sinâ¡1.6Ã103x+48Ã1010tj^Vm
D Eâ=3Ã10-8sinâ¡1.6Ã103x+48Ã1010ti^Vm

The correct answer is:
E→=9sin⁡1.6×103x+48×1010tk^Vm

Solution:
E0B0=C (speed of light in vacuum)
E0=B0C=3×10-8×3×108
=9NC
So,

E=9sin⁡ωt+kx+ϕ

E→=9sin⁡1.6×103x+48×1010tk ^Vm

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2020 (07 Jan Shift 1)
If the magnetic field of a plane electromagnetic wave is given by (The speed of light =3×108 m/s)
B=100×10-6sin⁡2π×2×1015t-xc then the maximum electric field associated with it is:

Options:
A 3×104 N C-1
B 4×104 N C-1
C 4.5×104 N C-1
D 6×104 N C-1

The correct answer is:
3×104 N C-1

Solution:
We know that relation between electric field E and magnetic field B is given by:

EB=c⇒E=Bc=100×10-6×3×108 V m-1=3×104 V m-1

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2019 (10 Jan Shift 1)
If the magnetic moment and the pole strength of a bar magnet are 3.6×10-3 A m2 and 120×10-3 A m, then its magnetic length is

Options:
A 3 cm
B 0.3 cm
C 3.3 cm
D 33 cm

The correct answer is:
3 cm

Solution:
For a bar magnet, the magnetic moment is

M = pole strength×magnetic length

So the magnetic length will be
lmag=Mm
⇒lmag=3.6×10-3120×10-3=3×10-2 m=3 cm

Topic: Physics, Magnetic Properties of Matter, JEE Main
If the magnetic monopole exists, then which of the Maxwell's equation to be modified?

Options:
A \( \vec{d} \cdot \vec{A}=\frac{Q_{\text {enclosed }}}{\varepsilon_{0}} \)
B \( \vec{B} \cdot \overrightarrow{d A}=0 \)
C \( \vec{B} \cdot \overrightarrow{d l}=\mu_{0} I_{\text {enclosed }}+\mu_{0} \varepsilon_{0} \frac{d}{d t} \int_{s} \vec{E} \cdot \overrightarrow{d A} \)
D \( \vec{E} \cdot \overrightarrow{d l}=-\frac{d}{d t} \phi_{B} \)

The correct answer is:
\( \vec{B} \cdot \overrightarrow{d A}=0 \)

Solution:
Gauss’s law for magnetism states that no magnetic monopoles exists and that the total flux through a closed surface must be zero.

According to the gauss law, magnetic flux coming out of the closed surface is directly proportional to the magnetic charge enclosed inside it. If magnetic monopole exist, then the charge enclosed may not be zero, and we need to modify gauss law for magnetism.

Topic: Physics, Electromagnetic Waves, JEE Main
If the magnitude of the projection of the vector i^-j^+2k^ on the vector perpendicular to the plane containing the vectors 2i^-j^+3k^ and i^+j^-2k^ is k, then the value of 1k2 is equal to

Options:

The correct answer is:
14.75

Solution:
A vector perpendicular to the plane is i^j^k^2-1311-2
=i^-1-j^-7+k^3
=-i^+7j^+3k^=b→ let
Let, a→=i^-j^+2k^
Projection of a→ on b→=a→⋅b→b→=-1-7+659=259=k

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the magnitude of the vector product of the vector \(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}\) with a unit vector along the sum of the vectors \(2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}\) and \(\lambda \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\) is equal to \(\sqrt{2}\), then the value of ' \(\lambda\) ' is

Options:
A -1
B 1
C 0
D 2

The correct answer is:
1

Solution:
\(\begin{aligned}
\text {Let } \mathbf{a} & =\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \\
\mathbf{b} & =2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-5 \hat{\mathbf{k}} \\
\mathbf{c} & =\lambda \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \\
\mathbf{b}+\mathbf{c} & =(2+\lambda) \hat{\mathbf{i}}+6 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}
\end{aligned}\)
\(\hat{\mathbf{d}}=\) unit vector along \((\mathbf{b}+\mathbf{c})=\frac{\mathbf{b}+\mathbf{c}}{|\mathbf{b}+\mathbf{c}|}\)
\(\begin{aligned}= & \frac{(2+\lambda) \hat{\mathbf{i}}+6 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}}{\sqrt{(2+\lambda)^2+36+4}} \\ \mathbf{d} & =\frac{(2+\lambda) \hat{\mathbf{i}}+6 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}}{\sqrt{\lambda^2+4 \lambda+44}}\end{aligned}\)
\(\begin{aligned} & \hat{\mathbf{a}} \times \hat{\mathbf{d}} \\ & =\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 1 & 1 \\ \frac{(2+\lambda)}{\sqrt{\lambda^2+4 \lambda+44}} & \frac{6}{\sqrt{\lambda^2+4 \lambda+44}} & \frac{-2}{\sqrt{\lambda^2+4 \lambda+44}}\end{array}\right|\end{aligned}\)
\(\begin{aligned}
& =\frac{1}{\sqrt{\lambda^2+4 \lambda+44}}\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
1 & 1 & 1 \\
2+\lambda & 6 & -2
\end{array}\right| \\
& =\frac{1}{\sqrt{\lambda^2+4 \lambda+44}}|\hat{\mathbf{i}}(-8)-\hat{\mathbf{j}}(-2-2-\lambda)+\hat{\mathbf{k}}(6-2-\lambda)| \\
& \hat{\mathbf{a}} \times \hat{\mathbf{d}}=\frac{1}{\sqrt{\lambda^2+4 \lambda+44}}|-8 \hat{\mathbf{i}}+(4+\lambda) \mathbf{j}+(4-\lambda) \hat{\mathbf{k}}| \\
& |\hat{\mathbf{a}} \times \hat{\mathbf{d}}|=\frac{1}{\sqrt{\lambda^2+4 \lambda+44}} \sqrt{64+(4+\lambda)^2+(4-\lambda)^2} \\
& \sqrt{2}=\frac{\sqrt{64+16+\lambda^2+16+\lambda^2}}{\sqrt{\lambda^2+4 \lambda+44}}
\end{aligned}\)
Squaring on both sides,
\(2=\frac{2 \lambda^2+96}{\lambda^2+4 \lambda+44}\)
\(\begin{aligned}
2 \lambda^2+8 \lambda+88 & =2 \lambda^2+96 \\
8 \lambda & =8 \\
\lambda & =1
\end{aligned}\)
Hence, option (b) is correct.

Topic: Mathematics, Vector Algebra, JEE Main
If the mass of a body is \( M \) on the surface of the earth, the mass of the same body on the
surface of the moon is

Options:
A \( \mathrm{M} / 6 \)
B \( \mathrm{M} \)
C \( 6 \mathrm{M} \)
D Zero

The correct answer is:
\( \mathrm{M} \)

Solution:
Mass is fundamental property and it does not depend on external conditions. So mass of an object always remains
same.
Therefore, if \( M \) is mass of body on surface of Earth, then mass of same body on the surface of moon is \( M \).

Topic: Physics, Motion In Two Dimensions, KCET, KCET 2015
If the mass of neutron =1.7Ã10-27 kg, then the de-Broglie wavelength of neutron of energy 3 eV is (h=6.6Ã10-34 J-s)

Options:
A 1.6×10-16 m
B 1.6×10-11 m
C 1.4×10-10 m
D 1.4×10-11 m

The correct answer is:
1.6×10-11 m

Solution:
Given E=3 eV=3Ã1.6Ã10-19 J
We know that Î»=h2mE
=6.6Ã10-342Ã1.7Ã10-27Ã3Ã1.6Ã10-19=1.65Ã10-11 m

Topic: Physics, Dual Nature of Matter, JEE Main
If the mass of the Sun were ten times smaller and the universal gravitational constant were ten times larger in magnitude, which of the following is not correct?

Options:
A Time period of a simple pendulum on the Earth would decrease
B Walking on the ground would become more difficult
C Raindrops will fall faster
D g on the Earth will not change

The correct answer is:
g on the Earth will not change

Solution:
msâ²=ms10and Gâ²=10G
gE=Gâ²MER2=10GMER2=10g
â´ Raindrops will fall faster, Walking on the ground would become more difficult, Time period of a simple pendulum on the earth would decrease.

Topic: Physics, Gravitation, JEE Main
If the matrix $\left[\begin{array}{rr}2 & 3 \\ 5 & -1\end{array}\right]=A+B$, where $A$ is symmetric and $B$ is skew-symmetric, then $B$ is equal to

Options:
A $\left[\begin{array}{cc}2 & 4 \\ 4 & -1\end{array}\right]$
B $\left[\begin{array}{rr}0 & -2 \\ 2 & 0\end{array}\right]$
C $\left[\begin{array}{rr}0 & 1 \\ -1 & 0\end{array}\right]$
D $\left[\begin{array}{rr}0 & -1 \\ 1 & 0\end{array}\right]$

The correct answer is:
$\left[\begin{array}{rr}0 & -1 \\ 1 & 0\end{array}\right]$

Solution:
Given,
$\left[\begin{array}{rr}
2 & 3 \\
5 & -1
\end{array}\right]=A+B$
We know that, every square matrix can be expressed uniquely as the sum of symmetric and skew-symmetric matrix.
Let
$C=\left[\begin{array}{rr}
2 & 3 \\
5 & -1
\end{array}\right]$
Since, $A$ is symmetric and $B$ is skew-symmetric matrix.
$\begin{aligned}
\therefore \quad C &=\frac{1}{2}\left(C+C^{\prime}\right)+\frac{1}{2}\left(C-C^{\prime}\right)=A+B \\
\Rightarrow B &=\frac{1}{2}\left(C-C^{\prime}\right)=\frac{1}{2}\left\{\left(\begin{array}{rr}
2 & 3 \\
5 & -1
\end{array}\right)-\left(\begin{array}{rr}
2 & 5 \\
3 & -1
\end{array}\right)\right\} \\
&=\frac{1}{2}\left(\begin{array}{rr}
0 & -2 \\
2 & 0
\end{array}\right)=\left(\begin{array}{rr}
0 & -1 \\
1 & 0
\end{array}\right)
\end{aligned}$

Topic: Mathematics, Matrices, JEE Main
If the matrix A=02K-1 satisfies AA3+3I=2I, then the value of K is

Options:
A 12
B 1
C -1
D -12

The correct answer is:
12

Solution:
AA3+3I=2I

A2=02K-102K-1

=2K-2-K2K+1

A3=2K-2-K2K+10-1K2

=-2K4K+22K2+K-2K-2K-1

A4=2K-2-K2K+12K-2-K2K+1

=4K2+2K -4K-4K-2-2K2-2K2-K 2K+4K2+1+4K

A4+3AI=2I

2I=4K2+2K-8K-2-4K2-K4K2+6K+1+302K-11001

2I=4K2+2K-8K-2-4K2-K4K2+6K+1+302K-1

2I=4K2+2K6-8K-23K-K-4K24K2+6K+1-3

4-8K=0

K=12

Topic: Mathematics, Matrices, JEE Main, JEE Main 2021 (27 Aug Shift 1)
If the matrix A=α02α ∀α∈R, α>0 and 2A2-2A=144, then the value of A+trA is equal to (where trA represent the trace of the matrix A i.e. the sum of all the principal diagonal elements of the matrix A and A is the determinant value of A)

Options:
A 8
B 15
C 20
D 24

The correct answer is:
15

Solution:
2A2-2A=144
22AA-I=144
⇒α2α-12=36
αα-1=6 or -6
α2-α-6=0 or α2-α+6=0
α-3α+2=0 or α=1±1-242×1
α=3 or -2 (rejected) or 12±i232 (rejected)
∴trA=2α=6
A=α2=9
A+trA=15

Topic: Mathematics, Determinants, JEE Main
If the matrix A=10002030-1 satisfies the equation A20+αA19+βA=100040001 for some real numbers α and β, then β-α is equal to ______.

Options:

The correct answer is:
4

Solution:
A=10002030-1

A2=100040001,A3=10008030-1

A4=1000160001

Hence,

A20=10002200001,A19=1000219030-1

So A20+αA19+βA=1+α+β000220+α.219+2β03α+3β01-α-β=100040001

Therefore α+β=0 and 220+219α-2α=4

⇒α=41-2182218-1=-2

Hence β=2

So, β-α=4

Topic: Mathematics, Matrices, JEE Main, JEE Main 2021 (26 Feb Shift 2)