datavorous/entrance-exam-dataset huggingface dataset:
If the planes r→·2i^-λj^+3k^=0 and r→·λi^+5j^-k^=5 are perpendicular to each other, then the value of λ2+λ is

Options:
A 0
B 2
C 1
D 3

The correct answer is:
0

Solution:
We have,

r→·λi^+5j^-k^=5

r→·2i^-λj^+3k^=0

We know that two planes r→·n→1=d1 and r→·n→2=d2 are perpendicular if

n→1·n→2=0

Therefore,

λi^+5j^-k^·2i^-λj^+3k^=0

⇒2λ-5λ-3=0

⇒-3λ-3=0

⇒λ=-1

Therefore, λ2+λ=0

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the planes r→⋅i^+j^+k^=1,r→⋅i^+2aj^+k^=2 and r→⋅ai^+a2j^+k^=3 intersect in a line, then the possible number of real values of a is

Options:
A 1
B 2
C 4
D 0

The correct answer is:
0

Solution:
∵ all 3 planes intersect in a line
⇒11112a1aa21=0
⇒2a-a2-11-a+1a2-2a2=0
⇒2a-a2-1+a-a2=0
⇒-2a2+3a-1=0⇒a=1,12
for both values of a=1,12, two of the given 3 planes are parallel (rejected)

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the planes x-cy-bz=0, cx-y+az=0 and bx+ay-z=0 pass through a straight line, then the value of a2+b2+c2+2abc is

Options:

The correct answer is:
1

Solution:
Given planes are:
x-cy-bz=0 ...(i)
cx-y+az=0 ...(ii)
bx+ay-z=0 ...(iii)
Equation of the plane passing through the line of intersection of the plane (i) and (ii) may be taken as:
x-cy-bz+Î»cx-y+az=0
or x1+Î»c-yc+Î»+z-b+aÎ»=0 ...(iv)
If plane (iii) and (iv) are the same, then
â´ 1+cÎ»b=-(c+Î»)a=-b+aÎ»-1 or Î»=-a+bcac+b
and Î»=-(ab+c)1-a2
â´ -a+bcac+b=-ab+c1-a2
â a-a3+bc-a2bc=a2bc+ac2+ab2+bc
â 2a2bc+ab2+ac2+a3-a=0
â a2abc+c2+b2+a2-1=0
â a2+b2+c2+2abc=1

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the plates of a parallel plate capacitor connected to a battery are moved close to each other, then
A. the charge stored in it, increases.
B. the energy stored in it, decreases.
C. its capacitance increases.
D. the ratio of charge to its potential remains the same.
E. the product of charge and voltage increases.
Choose the most appropriate answer from the options given below:

Options:
A A, C and E only
B B, D and E only
C A, B and C only
D A, B and E only

The correct answer is:
A, C and E only

Solution:
Given $V=V=$ Constant
(i)
$\begin{aligned}
& C^{\prime}=\frac{\varepsilon_0 A}{d^{\prime}}, C=\frac{\varepsilon_0 A}{d} \\
& d^{\prime \prime} \lt d \\
& C^{\prime}\gtC
\end{aligned}$
Hence, final capacitance greater than initial capacitance,
(ii)
$\begin{aligned}
U^{\prime} & =\frac{1}{2} C^{\prime} V^2 \\
U & =\frac{1}{2} C V^2 \\
U^{\prime \prime} & \gtU
\end{aligned}$
Hence final energy is greater than initial energy
(iii)
$\begin{aligned}
& \frac{Q^{\prime}}{V^{\prime}}=C^{\prime} \text { and } \frac{Q}{V}=C \\
& \frac{Q^{\prime}}{V^{\prime}} \neq \frac{Q}{V}
\end{aligned}$
(iv) Product of charge and voltage
$\begin{aligned}
& X^{\prime}=Q^{\prime} V=C^{\prime} V^2 \\
& X=Q V=C V^2 \\
& X^{\prime}\gtX
\end{aligned}$

Topic: Physics, Capacitance, NEET, NEET 2024
If the point 1, 4 lies inside the circle x2+y2-6x+10y+p=0 and the circle does not touch or intersect the coordinate axes, then the set of all possible values of p is the interval

Options:
A 25, 39
B 25, 29
C 0, 25
D 9, 25

The correct answer is:
25, 29

Solution:
x2+y2-6x-10y+p=0

∴ Centre is 3, 5 and radius is 34-p.

∵ 1, 4 lies inside the circle.

So, 1+16-6-40+p<0

⇒p<29 ...1

Circle neither touches nor cut the coordinate axes.
So, radius of circle 34-p<5 (for not touching x- axis ) and 34-p<3 (for not touching y-axis)

⇒ 9<p and 25<p

⇒ p>25 ...2

From equations 1 & 2,

So, p∈25, 29

Topic: Mathematics, Circle, JEE Main, JEE Main 2014 (09 Apr Online)
If the point $(1, a)$ lies between the straight lines $x+y=1$ and $2(x+y)=3$ then a lies in interval

Options:
A
$\left(\frac{3}{2}, \infty\right)$

B
$\left(1, \frac{3}{2}\right)$

C
$(-\infty, 0)$

D
$\left(0, \frac{1}{2}\right)$

The correct answer is:
$\left(0, \frac{1}{2}\right)$

Solution:
Since, $(1, a)$ lies between $x+y=1$ and $2(x+y)=3$
$\therefore$ Put $x=1$ in $2(x+y)=3$.
We get the range of $y$. Thus,
$$
2(1+y)=3 \Rightarrow y=\frac{3}{2}-1=\frac{1}{2}
$$
Thus ' $a$ 'lies in $\left(0, \frac{1}{2}\right)$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2012 (12 May Online)
If the point $(2, \alpha, \beta)$ lies on the plane which passes through the points (3,4,2) and (7,0,6) and is perpendicular to the plane $2 x-5 y=15,$ then $2 \alpha-3 \beta$ is equal to :

Options:
A 12
B 7
C 5
D 17

The correct answer is:
7

Solution:
Let the normal to the required plane is $\vec{n}$, then

$\vec{n}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 4 & -4 & 4 \\ 2 & -5 & 0\end{array}\right|=20 \hat{i}+8 \hat{j}-12$

$\therefore \quad$ Equation of the plane

$(x-3) \times 20+(y-4) \times 8+(z-2) \times(-12)=0$

$5 x-15+2 y-8-3 z+6=0$

$5 x+2 y-3 z-17=0 \quad \ldots(1)$

Since, equation of plane (1) passes through $(2, \alpha, \beta)$, then $10+2 \alpha-3 \beta-17=0 \Rightarrow 2 \alpha-3 \beta=7$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2019 (11 Jan Shift 2)
If the point α, 733 lies on the curve traced by the mid-points of the line segments of the lines x cosθ+ysinθ=7, θ∈0, π2 between the co-ordinates axes, then α is equal to

Options:
A -7
B -73
C 73
D 7

The correct answer is:
7

Solution:
The straight line is shown below with the intercepts.

M72cosθ, 72sinθ≡h, k

∴ cosθ=72h sinθ=72k

∵ θ∈0, π2

⇒494h2+494k2=1 ∴ x and y are positive

∴ Locus =1x2+1y2=449

Now α, 733 lies on curve

⇒1α2+9147=449

1α2=149

∴ α=±7

∴ α=7

Hence this is the required option.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2023 (12 Apr Shift 1)
If the point Mh,k lie on the line 2x+3y=5 such that MA-MB is maximum where A1,2 and B2,3, then the value of h+k is

Options:

The correct answer is:
1.8

Solution:
As, MA-MB≤AB
∴MA-MBmaximum=AB, which is possible when M is the point of intersection of the line joining A1,2, B2,3 and 2x+3y=5.

Equation of the line joining A & B is x-y+1=0

Hence, the point of intersection is h=25, k=75

Hence, h+k=95=1.8

Topic: Mathematics, Straight Lines, JEE Main
If the point of intersection of the plane 4x-5y+2z-6=0 with the line through the origin and perpendicular to the plane x-2y-4z=4 is P, then the distance of the point P from 1,2,3 is

Options:
A 63 units
B 8 units
C 65 units
D 72 units

The correct answer is:
65 units

Solution:
Equation of line through origin and perpendicular to the plane x-2y-4z=4 is
x-01=y-0-2=z-0-4=Î»
So, any point on this line is Î»,-2Î»,-4Î»
âµ the above point lies on the plane 4x-5y+2z-6=0
â 4Î»+10Î»-8Î»-6=0
â 6Î»=6âÎ»=1
Coordinates of the point P are 1,â2,â4
The distance of point P from
1,2,3 is 0+16+49=65

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the point on the curve y2=6x, nearest to the point 3,32 is (α,β), then 2(α+β) is equal to _________.

Options:

The correct answer is:
9

Solution:
We have,

y2=6x

Comparing with y2=4ax, we get a=32

Let the point on parabola be P≡at2,2at≡32t2,3t.

Normal at point P is

tx+y=3t+32t3

It passes through 3,32, hence

⇒3t+32=3t+32t3

⇒t3=1⇒t=1

Hence,

P≡32,3=(α,β)

Therefore,

2(α+β)=232+3=9

Topic: Mathematics, Parabola, JEE Main, JEE Main 2021 (20 Jul Shift 2)
If the point P3a2,1 lies between the two different lines x+y=a and x+y=2a, then the least integral value of a is equal to

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:
Point 32a, 1 lies between two different lines x+y=a and x+y=2a
∴32a+1-a32a+1-2a<0⇒a∈-∞,-2∪2,∞
⇒a∈2,∞

Hence, the least integral value of a is 3

Topic: Mathematics, Straight Lines, JEE Main
If the point P on the curve, 4x2+5y2=20 is farthest from the point Q0,-4, then PQ2 is equal to

Options:
A 36
B 48
C 21
D 29

The correct answer is:
36

Solution:
Give ellipse

x25+y24=1 ...1

Let P5cosθ, 2sinθ

Ellipse (1)

Given Q0, -4

PQ2=5cosθ2+2sinθ22

⇒PQ2=5cos2θ+4sin2θ+16cos2θ+16

⇒PQ2=4+cos2θ+16 sinθ+16

=1-sin2θ+16 sinθ

⇒PQ2=17-sin2θ+16 sinθ

⇒PQ2=17-sin2θ-16 sinθ

⇒PQ2=21sinθ-82-64

⇒PQ2=85-(sinθ-8)2=85-49

Max PQ2=36 (∵ maximum value of sinθ=1)

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2020 (05 Sep Shift 1)
If the points 1, 1, λ & (-3, 0, 1), are equidistant from the plane, 3x+4y-12z+13=0, then λ satisfies the equation:

Options:
A 3x2+10x+7=0
B 3x2+10x-13=0
C 3x2-10x+7=0
D 3x2-10x+21=0

The correct answer is:
3x2-10x+7=0

Solution:
1, 1, λ and -3, 0, 1 are equidistant from 3x+4y-12z+13=0 then
3+4-12λ+1332+42+122= -9+0-12+1332+42+122
20-12λ= -8
5-3λ = -2
25-30λ+9λ2=4
9λ2-30λ+21=0
3λ2-10λ+7=0

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2015 (10 Apr Online)
If the points \( (-2,0),\left(-1, \frac{1}{\sqrt{3}}\right) \) and \( (\cos \theta, \sin \theta) \) are collinear, then the number of values of \( \theta \in[0,2 \pi] \) is

Options:
A \( 0 \)
B \( 1 \)
C \( 2 \)
D infinite

The correct answer is:
\( 1 \)

Solution:
Let A≡-2,0, B=-1,13, C≡cosθ,sinθ

Equation to AB,

y-0=13-0-1+2x+2

⇒3y=x+2, Point C lies on this line.

⇒3sinθ-cosθ=2

⇒32sinθ-12cosθ=1

⇒sinθ-π6=1, where θ∈0,2π

θ-π6=π2

⇒θ=π2+π6=2π3

So, only one value of θ is possible.

Topic: Mathematics, Straight Lines, JEE Main
If the points A:0,a, B:-2,0 and C:1,1 form an obtuse angle triangle (obtuse angled at angle A), then sum of all the possible integral values of a is

Options:
A 0
B 3
C 2
D 1

The correct answer is:
1

Solution:
Let, the point 0,a lie inside the circle whose end-points of the diameter are -2,0 and 1,1, but does not lie on this diameter
⇒0,a lie inside the circle x+2x-1+yy-1=0
⇒0+20-1+aa-1<0
⇒a2-a-2<0⇒a-2a+1<0
⇒a∈-1,2
But, a≠23 because at a=23, points 0,23,-2,0 and 1,1 are collinear
⇒a∈-1,2-23

Hence, the sum of all possible integral values of ‘a’ is 1

Topic: Mathematics, Circle, JEE Main
If the points A3-x,3,3,B3,3-y,3,C3,3,3-z and D2,2,2 are coplanar, then 1x+1y+1z is equal to

Options:
A -1
B 1
C 3
D 5

The correct answer is:
1

Solution:
DA→=<1-x,1,1>
DB→=<1,1-y,1>
DC→=<1,1,1-z>
∴A,B,C,D are coplanar ⇒DA→DB→DC→=0
1-x1111-y1111-z=0
⇒1-x1-y1-z-1-1-z-1+11-1-y=0
⇒1-x-y-z+yz+z+y=0
⇒-y-z+yz+xy+xz-xyz+z+y=0
⇒xy+yz+zx=xyz⇒1x+1y+1z=1

Topic: Mathematics, Vector Algebra, JEE Main
If the points having the position vectors $3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}}, 2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}},-\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}$ are coplanar, then $\lambda=$

Options:
A $\frac{46}{17}$
B 8
C -8
D $\frac{146}{17}$

The correct answer is:
$\frac{46}{17}$

Solution:
Let $A, B, C$ and $D$ are the position vectors of four points given as
$$
\begin{aligned}
& A(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-\hat{\mathrm{k}}), B(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}), C(-\hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \\
& \text { and } D(4 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+\lambda \hat{\mathrm{k}})
\end{aligned}
$$
Here,
$$
\begin{aligned}
& \mathrm{AB}=-\hat{\mathrm{i}}+5 \hat{\mathrm{j}}-3 \hat{\mathrm{k}} \\
& \mathrm{BC}=-3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}
\end{aligned}
$$
and
$$
C D=5 \hat{i}+4 \hat{j}+(\lambda-2) \hat{k}
$$
As given that, these points are coplanar, therefore $[\mathrm{AB} \quad \mathrm{BC} \mathrm{CD}]=0$
$$
\begin{aligned}
& \left|\begin{array}{ccc}
-1 & 5 & -3 \\
-3 & -2 & 6 \\
5 & 4 & \lambda-2
\end{array}\right|=0 \\
\Rightarrow \quad & -1(-2 \lambda+4-24)-5(-3(\lambda-2)-30) \\
\Rightarrow \quad & -3(-12+10)=0 \\
\Rightarrow \quad & 17 \lambda+20+20+15(\lambda-2)+150+36-30=0 \\
\Rightarrow \quad & 17 \lambda+146=0 \Rightarrow \lambda=-\frac{146}{17}
\end{aligned}
$$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the points of intersection of the ellipse x216+y2 b2=1 and the circle x2+y2=4 b, b>4 lie on the curve y2=3x2, then b is equal to :

Options:
A 12
B 5
C 6
D 10

The correct answer is:
12

Solution:
y2=3x2 and x2+y2=4b

Solving both the equations we get,

x2=b

As x216+y2b2=1

⇒x216+3x2b2=1

⇒b16+3b=1

⇒b2-16b+48=0

⇒b-12b-4=0

⇒b=12 ∵b>4

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2021 (16 Mar Shift 2)
If the points of intersection of two distinct conics x2+y2=4b and x216+y2b2=1 lie on the curve y2=3x2, then 33 times the area of the rectangle formed by the intersection points is _______.

Options:

The correct answer is:
432

Solution:
Given conics are x2+y2=4b and x216+y2b2=1.

The intersection point of these conics lie on y2=3x2.

So, putting y2=3x2 in both the conics.

⇒x2+3x2=4b, x216+3x2b2=1

⇒x2=b, b16+3bb2=1

⇒b16+3b=1

⇒b2+48=16b

⇒b2-16b+48=0

⇒b2-12b-4b+48=0

⇒b-4b-12

⇒b= 4, 12 ( b=4is rejected because curves coincide)

⇒b=12

⇒x=±12

⇒12+y2=48

⇒y=±6

Hence points of intersection are (±12,±6).

So, length and breadth of rectangle are 212 and 12.

So, the area of rectangle is given by, A=212×12=483

⇒33A=483×33

⇒33A=432

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2024 (29 Jan Shift 1)
If the points P and Q are respectively the circumcenter and the orthocentre of a ∆ABC, then PA→+PB→+PC→ is equal to _______

Options:
A 2QP→
B 2PQ→
C PQ→
D QP→

The correct answer is:
PQ→

Solution:
Let P be origin.

Now from the above triangle we can write PA→=a→, PA→=b→, PC→=c→.

Also, we know that centroid G divides orthocentre Q and Circumcentre P in the ratio 2:1.

Now PG→=a→+b→+c→3

⇒PQ→=3a→+b→+c→3

⇒PQ→=a→+b→+c→=PA→+PB→+PC→

Therefore, the required answer is PQ→.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (10 Apr Shift 2)
If the points whose position, vectors are $3 \mathbf{i}-2 \mathbf{j}-\mathbf{k}, \quad 2 \mathbf{i}+3 \mathbf{j}-4 \mathbf{k}, \quad-\mathbf{i}+\mathbf{j}+2 \mathbf{k}$ and $4 \mathbf{i}+5 \mathbf{j}+\lambda \mathbf{k}$ lie on a plane, then $\lambda=$

Options:
A $-\frac{146}{17}$
B $\frac{146}{17}$
C $-\frac{17}{146}$
D $\frac{17}{146}$

The correct answer is:
$-\frac{146}{17}$

Solution:
Let $\mathbf{a}=3 \mathbf{i}-2 \mathbf{j}-\mathbf{k}, \quad \mathbf{b}=2 \mathbf{i}+3 \mathbf{j}-4 \mathbf{k}, \quad \mathbf{c}=-\mathbf{i}+\mathbf{j}+2 \mathbf{k}$ and $\mathbf{d}=4 \mathbf{i}+5 \mathbf{j}+\lambda \mathbf{k}$.
Since the points are coplanar,
So,
$[\mathbf{d} \mathbf{b} \mathbf{c}]+[\mathbf{d} \mathbf{c} \mathbf{a}]+[\mathbf{d} \mathbf{a} \mathbf{b}]=[\mathbf{a} \mathbf{b} \mathbf{c}]$
$\begin{aligned}
& \Rightarrow\left|\begin{array}{ccc}
4 & 5 & \lambda \\
2 & 3 & -4 \\
-1 & 1 & 2
\end{array}\right|+\left|\begin{array}{ccc}
4 & 5 & \lambda \\
-1 & 1 & 2 \\
3 & -2 & -1
\end{array}\right|+\left|\begin{array}{ccc}
4 & 5 & \lambda \\
3 & -2 & -1 \\
2 & 3 & -4
\end{array}\right| \\
& =\left|\begin{array}{ccc}
3 & -2 & -1 \\
2 & 3 & -4 \\
-1 & 1 & 2
\end{array}\right|
\end{aligned}$
$\Rightarrow 40+5 \lambda+37-\lambda+94+13 \lambda=25 \Rightarrow \lambda=\frac{-146}{17} \text {. }$

Topic: Mathematics, Vector Algebra, JEE Main
If the points with position vectors αi^+10j^+13k^, 6i^+11j^+11k^,92i^+βj^-8k^ are collinear, then 19α-6β2 is equal to

Options:
A 36
B 25
C 49
D 16

The correct answer is:
36

Solution:
The given position vectors can be written as

A(α, 10, 13)

B(6, 11, 11)

C92, β, -8

Also given that these three points are collinear.

Let us assume that the point B divides AB, BC in the ratio k:1.

Since, A, B, C are collinear

On applying section formula for z co-ordinate we get,

11=-8k+13k+1

⇒11k+11=-8k+13

⇒19k= 2

⇒k=219

∴Ratio=2 : 19

Now, α×19+92×22+19=6

⇒19α=117

⇒α=11719

Now similarly, 2β+19021=11

⇒β=412

∴19α-6β2=117-1232=36

Hence this is the correct option.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (08 Apr Shift 1)
If the position vector of \( A \& B \) are \( \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-7 \widehat{\mathrm{k}} \) and \( 5 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+4 \widehat{\mathrm{k}} \), then the direction cosine of ABalong \( Y \)-axis is

Options:
A \( \frac{4}{\sqrt{162}} \)
B \( -\frac{5}{\sqrt{162}} \)
C \( -5 \)
D 11

The correct answer is:
\( -\frac{5}{\sqrt{162}} \)

Solution:
Given:

OA→=i^+3j^-7k^ and OB→=5i^-2j^+4k^

AB→=OB→-OA→>

⇒AB→=5i^-2j^+4k^-i^+3j^-7k^

⇒AB→=4i^-5j^+11k^

Y-axis, cosβ=–516 + 25 + 121=–5162

Topic: Mathematics, Vector Algebra, JEE Main
If the position vector of one end of the line segment $A B$ be $2 \mathbf{i}+3 \mathbf{j}-\mathbf{k}$ and the position vector of its middle point be $3(\mathbf{i}+\mathbf{j}+\mathbf{k})$, then the position vector of the other end is

Options:
A $4 \mathbf{i}+3 \mathbf{j}+5 \mathbf{k}$
B $4 \mathbf{i}-3 \mathbf{j}+7 \mathbf{k}$
C $4 \mathbf{i}+3 \mathbf{j}+7 \mathbf{k}$
D $4 \mathbf{i}+3 \mathbf{j}-7 \mathbf{k}$

The correct answer is:
$4 \mathbf{i}+3 \mathbf{j}+7 \mathbf{k}$

Solution:
$\overrightarrow{O A}=2 \mathbf{i}+3 \mathbf{j}-\mathbf{k}, \quad \overrightarrow{O P}=3(\mathbf{i}+\mathbf{j}+\mathbf{k}), \quad \overrightarrow{O B}=?$
we have $\overrightarrow{O P}=\frac{\overrightarrow{O A}+\overrightarrow{O B}}{2}$
$\begin{aligned}
\Rightarrow \overrightarrow{O B} & =2 \overrightarrow{O P}-\overrightarrow{O A} \\
& =4 \mathbf{i}+3 \mathbf{j}+7 \mathbf{k}
\end{aligned}$
Trick : By inspection, middle point of $4 \mathbf{i}+3 \mathbf{j}+7 \mathbf{k}$ and $2 \mathbf{i}+3 \mathbf{j}-\mathbf{k}$ is $3(\mathbf{i}+\mathbf{j}+\mathbf{k})$.

Topic: Mathematics, Vector Algebra, JEE Main
If the position vectors of the vertices $A, B$ and $C$ are 6i, $6 \mathbf{j}$ and $\mathbf{k}$ respectively w.r.t. origin $O$, then the volume of the tetrahedron $O A B C$ is

Options:
A 6
B 3
C $\frac{1}{6}$
D $\frac{1}{3}$

The correct answer is:
6

Solution:
Given that, the position vectors of the vertices $A, B$ and $C$ are
$$
\begin{array}{l}
\mathbf{O A}=6 \mathbf{i}=6 \mathbf{i}+0 \mathbf{j}+0 \mathbf{k} \\
\mathbf{O B}=6 \mathbf{j}=0 \mathbf{i}+6 \mathbf{j}+0 \mathbf{k} \\
\mathbf{O C}=\mathbf{k}=0 \mathbf{i}+0 \mathbf{j}+\mathbf{k}
\end{array}
$$
Now, volume of the tetrahedron
$$
\begin{array}{l}
=\frac{1}{6}[\mathbf{O A} \mathbf{O B} \mathbf{O C}] \\
=\frac{1}{6}\left|\begin{array}{lll}
6 & 0 & 0 \\
0 & 6 & 0 \\
0 & 0 & 1
\end{array}\right| \\
=\frac{1}{6}(6 \times 6 \times 1)=6
\end{array}
$$

Topic: Mathematics, Vector Algebra, JEE Main
If the position vectors of the vertices $A, B$ and $C$ of a $\triangle \mathrm{ABC}$ are respectively $4 \hat{i}+7 \hat{j}+8 \hat{k}, 2 \hat{i}+3 \hat{j}+4 \hat{k}$ and $2 \hat{i}+5 \hat{j}+7 \hat{k}$, then the position vector of the point, where the bisector of $\angle A$ meets $B C$ is

Options:
A
$\frac{1}{2}(4 \hat{i}+8 \hat{j}+11 \hat{k})$

B
$\frac{1}{3}(6 \hat{i}+13 \hat{j}+18 \hat{k})$

C
$\frac{1}{4}(8 \hat{i}+14 \hat{j}+9 \hat{k})$

D
$\frac{1}{3}(6 \hat{i}+11 \hat{j}+15 \hat{k})$

The correct answer is:
$\frac{1}{3}(6 \hat{i}+13 \hat{j}+18 \hat{k})$

Solution:
Suppose angular bisector of $A$ meets $B C$ at $\mathrm{D}(x, y, z)$
Using angular bisector theorem,
$$
\begin{aligned}
\frac{A B}{A C} &=\frac{B D}{D C} \\
\frac{B D}{D C} &=\frac{\sqrt{(4-2)^2+(7-3)^2+(8-4)^2}}{\sqrt{(4-2)^2+(7-5)^2+(8-7)^2}} \\
&=\frac{\sqrt{2^2+4^2+4^2}}{\sqrt{2^2+2^2+1^2}}=\frac{6}{3}=2
\end{aligned}
$$
$$
D(x, y, z)=\left(\frac{6}{3}, \frac{13}{3}, \frac{18}{3}\right)
$$
Therefore, position vector of point
$$
P=\frac{1}{3}(6 i+13 j+18 k)
$$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
If the position vectors of the vertices of a \(\triangle A B C\) are \(\mathbf{O A}=3 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}, \mathbf{O B}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\) and \(\mathbf{O C}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}}\), then the length of the altitude of \(\triangle A B C\) drawn from \(A\) is

Options:
A \(\sqrt{\frac{3}{2}}\)
B \(\frac{3}{\sqrt{2}}\)
C \(\frac{\sqrt{3}}{2}\)
D \(\frac{3}{2}\)

The correct answer is:
\(\frac{3}{\sqrt{2}}\)

Solution:
Since, length of altitude of \(\triangle A B C\) drawn from \(A\) is
\(\begin{aligned}
& h=\frac{(\text { Area of } \triangle A B C)}{\frac{1}{2}|\mathbf{B C}|}=\frac{\frac{1}{2}|\mathbf{A B} \times \mathbf{A C}|}{\frac{1}{2}|\mathbf{B C}|} \\
& \because \quad \mathbf{A B}=-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \\
& \mathbf{A C}=-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}} \\
& \text { and } \quad \mathbf{B C}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}
\end{aligned}\)
\(\begin{aligned}
& \text {So, } \mathbf{A B} \times \mathbf{A C}=\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
-2 & 1 & 1 \\
-1 & 2 & -1
\end{array}\right| \\
& =\hat{\mathbf{i}}(-\mathbf{l}-2-\mathbf{j}(2+1)+\hat{\mathbf{k}}(-4+1)=-3 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} \\
& \therefore|\mathbf{A B} \times \mathbf{A C}|=3 \sqrt{3} \text { and }|\mathbf{B C}|=\sqrt{6} \\
& \therefore h=\frac{3 \sqrt{3}}{\sqrt{2} \sqrt{3}}=\frac{3}{\sqrt{2}}
\end{aligned}\)
Hence, option (2) is correct.

Topic: Mathematics, Vector Algebra, JEE Main
If the position vectors of the vertices of a triangle are $2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \quad \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}$ and $3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}$, then the triangle is

Options:
A equilateral
B isosceles
C right angled isosceles
D right angled

The correct answer is:
right angled

Solution:
Let $\mathbf{A}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{B}=\hat{\mathbf{i}}-3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}$ $\mathbf{C}=3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}$
$A B=-\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}, \mathrm{BC}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$
and
$\begin{aligned}
& \mathbf{C A}=-\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}} \\
& a=|\mathbf{B C}|=\sqrt{4+1+1}=\sqrt{6} \\
& b=|\mathbf{C A}|=\sqrt{1+9+25}=\sqrt{35} \\
& c=|\mathbf{A B}|=\sqrt{1+4+36}=\sqrt{41}
\end{aligned}$
Now, $c^2=a^2+b^2$
$\begin{aligned}
\Rightarrow \quad 41 & =6+35 \\
& =41=41
\end{aligned}$
$\therefore$ It is right angled triangle

Topic: Mathematics, Vector Algebra, JEE Main
If the potential barrier across a p-n junction is 0.6 V. Then the electric field intensity, in the depletion region having the width of 6×10-6 m, will be _____ ×105 N C-1

Options:

The correct answer is:
1

Solution:
Electric field intensity is E=Vd=Potential barrier across junctionwidth of depletion region

=0.6 V6×10-6 m=1×105 V m-1

=1×105 N C-1

Topic: Physics, Semiconductors, JEE Main, JEE Main 2022 (29 Jul Shift 1)
If the potential difference between B and D is zero, the value of x is 1n Ω. The value of n is ______.

Options:

The correct answer is:
2

Solution:
As potential difference between B & D is zero therefore it is a balanced wheatstone bridge. So, we can write

⇒R1R2=R3R4⇒6×36+31+2=x×1x+1x

⇒23=1x+1

⇒x=0.5=12

Hence, required n=2.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (30 Jan Shift 2)
If the potential difference used to accelerate electrons is doubled, by what factor does the de-Broglie wavelength associated with the electrons change?

Options:
A Wavelength is decreased to $\frac{1}{3}$ times.
B Wavelength is increased to $\frac{1}{2}$ times.
C Wavelength is increased to $\frac{1}{\sqrt{2}}$ times.
D Wavelength is decreased to $\frac{1}{\sqrt{2}}$ times.

The correct answer is:
Wavelength is decreased to $\frac{1}{\sqrt{2}}$ times.

Solution:
The de Broglie wavelength is given by
$\lambda(\text { in } \mathrm{nm})=\frac{1.23}{\sqrt{V}}$
Hence it becomes $\frac{1}{\sqrt{2}}$ times

Topic: Physics, Dual Nature of Matter, JEE Main
If the potential energy between two molecules is given by U=Ar6+Br12, then at equilibrium, separation between molecules, and the potential energy are:

Options:
A B2A1/6,-A22B
B BA1/6,0
C 2BA1/6,A24B
D B2A16,A22B

The correct answer is:
2BA1/6,A24B

Solution:
U=Ar6+Br12

f=-dUdr

For equilibrium

f=0

so r=2BA16

Then, U=A24B

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2020 (06 Sep Shift 1)