datavorous/entrance-exam-dataset huggingface dataset:
If the roots of the equation $x^2-2 a x+a^2+a-3=0$ are real and less than 3 , then

Options:
A $a \lt 2$
B $2 \leq a \leq 3$
C $3 \lt a \leq 4$
D $a\gt4$

The correct answer is:
$a \lt 2$

Solution:
Given equation is $x^2-2 a x+a^2+a-3=0$
If roots are real, then $D \geq 0$
$\Rightarrow 4 a^2-4\left(a^2+a-3\right) \geq 0 \Rightarrow-a+3 \geq 0$
$\Rightarrow a-3 \leq 0 \Rightarrow a \leq 3$
As roots are less than 3 , hence $f(3)\gt0$
$9-6 a+a^2+a-3\gt0 \Rightarrow a^2-5 a+6\gt0$
$\Rightarrow(a-2)(a-3)\gt0 \Rightarrow$ either $a \lt 2$ or $a\gt3$
Hence $a \lt 2$ satisfy all.

Topic: Mathematics, Quadratic Equation, JEE Main
If the roots of the equation $x^{2}+a x+b=0$ are $c$ and $\mathrm{d}$, then one of the roots of the equation $x^{2}+(2 c+a) x+c^{2}+a c+b=0$ is

Options:
A $c$
B $d-c$
C $2 d$
D $2 c$

The correct answer is:
$d-c$

Solution:
$$
\begin{array}{l}
\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}+\mathrm{ax}+\mathrm{b}, \text { then } \\
\mathrm{f}(\mathrm{x}+\mathrm{c})=(\mathrm{x}+\mathrm{c})^{2}+\mathrm{a}(\mathrm{x}+\mathrm{c})+\mathrm{b} \\
=\mathrm{x}^{2}+(2 \mathrm{c}+\mathrm{a}) \mathrm{x}+\mathrm{c}^{2}+\mathrm{ac}+\mathrm{b}
\end{array}
$$
which shows that the roots of $\mathrm{f}(\mathrm{x})$ are transformed to $(x-c)$ i.e., roots of $\mathrm{f}(\mathrm{x}+\mathrm{c})=0$ are $\mathrm{c}-\mathrm{c}$ and $\mathrm{d}-\mathrm{c}$.
Hence, one of the roots of the equation $f(x+c)$ is $(d-c)$.

Topic: Mathematics, Quadratic Equation, JEE Main
If the roots of the equation x3+bx2+cx+d=0 are in arithmetic progression, then b, c and d satisfy the relation

Options:
A 2b2-27d=9bc
B 2b3-27d=9bc
C 2b2+27d=9bc
D 2b3+27d=9bc

The correct answer is:
2b3+27d=9bc

Solution:
Let Î±,Î²,Î³ be the required roots
Î±+Î²+Î³=-b, Î±Î²+Î²Î³+Î³Î±=c, Î±Î²Î³=-d
Also, Î±+Î³=2Î²â3Î²=-b
Î²(Î±+Î³)+Î³Î±=câ-b32-b3+Î³Î±=c
â2b29-dÎ²=câ2b29-d-b3=c
â2b29+3db=câ2b3+27d=9bc

Topic: Mathematics, Quadratic Equation, JEE Main
If the roots of the given equation $2 x^2+3(\lambda-2) x+\lambda+4=0$ be equal in magnitude but opposite in sign, then $\lambda=$

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

The correct answer is:
$2$

Solution:
Let roots are $\alpha$ and $-\alpha$, then sum of the roots
$\alpha+(-\alpha)=\frac{3(\lambda-2)}{2} \Rightarrow 0=\frac{3}{2}(\lambda-2)$
$\Rightarrow \lambda=2$

Topic: Mathematics, Quadratic Equation, JEE Main
If the roots of the quadratic equation \( (b-c) x^{2}+(c-a) x+(a-b)=0 \) are equal, then \( a, b, c \) are in

Options:
A A.P.
B G.P.
C H. P.
D None of these

The correct answer is:
A.P.

Solution:
Given equation: b-cx2+c-ax+a-b=0
Putting x=1, we get
b-c+c-a +a-b=0
So, x=1 is a root but both roots are equal so both roots are 1, 1.
Now sum of roots 1+1=-c-ab-c
⇒2=-c-ab-c⇒2b-2c=-c+a
⇒2b=a+c
⇒a, b, c are in A.P.

Topic: Mathematics, Quadratic Equation, JEE Main
If the roots of the quadratic equation $x^2+p x+q=0$ are $\tan 30^{\circ}$ and $\tan 15^{\circ}$, respectively then the value of $2+q-p$ is

Options:
A
2

B
3

C
0

D
1

The correct answer is:
3

Solution:
$x^2+p x+q=0$
$\tan 30^{\circ}+\tan 15^{\circ}=-p$
$\tan 30^{\circ} \cdot \tan 15^{\circ}=q$
$\tan 45^{\circ}=\frac{\tan 30^{\circ}+\tan 15^{\circ}}{1-\tan 30^{\circ} \tan 15^{\circ}}=\frac{-p}{1-q}=1$
$\Rightarrow-p=1-q$
$\Rightarrow q-p=1 \quad \therefore 2+q-p=3$.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2006
If the roots of $x^{2}-2 m x+m^{2}-1=0$ lie between $-2$ and 4 , then which one of the following is correct?

Options:
A $-1 \leq \mathrm{m} \leq 3$
B $-3 \leq \mathrm{m} \leq 3$
C $-3 \leq \mathrm{m} \leq 5$
D $-1 \leq \mathrm{m} \leq 5$

The correct answer is:
$-3 \leq \mathrm{m} \leq 3$

Solution:
Since, the roots of $x^{2}-2 m x+m^{2}-1=0$ lie between
$-2$ and 4 i.e., $\mathrm{b} 2-4 \mathrm{ac} \geq 0$ and $-2 < \frac{-\mathrm{b}}{2 \mathrm{a}} < 4$
$(2 m)^{2}-4\left(m^{2}-1\right) \geq 0$...(1)
and $-2 < \frac{2 \mathrm{~m}}{2} < 4$
$\Rightarrow-2 < m < 4$
From (i) $4 m^{2}-4 m^{2}+4 \geq 0$
$\Rightarrow m \in R$
and $\mathrm{f}(-2)>0$, also $\mathrm{f}(4)>0$
$4+4 m+m^{2}-1>0,16-8 m+m^{2}-1>0$
$\Rightarrow \mathrm{m}^{2}+4 \mathrm{~m}+3>0, \mathrm{~m}^{2}-8 \mathrm{~m}+15>0$
$\Rightarrow(m+1)(m+3)>0,(m-3)(m-5)>0$
$\Rightarrow-3 < \mathrm{m} < -1$ and $3 < \mathrm{m} < 5$
Thus, the inteval in which it lies is $-1 \leq \mathrm{m} \leq 5$

Topic: Mathematics, Quadratic Equation, JEE Main
If the roots of $x^2+x+a=0$ exceed $a$, then

Options:
A $2 \lt a \lt 3$
B $a\gt3$
C $-3 \lt a \lt 3$
D $a \lt -2$

The correct answer is:
$a \lt -2$

Solution:
If the roots of the quadratic equation $a x^2+b x+c=0$ exceed a number $k$, then $a k^2+b k+c\gt0$ if $a\gt0, b^2-4 a c \geq 0$ and sum of the roots $\gt2 k$ Therefore, if the roots of $x^2+x+a=0$ exceed a number $a$, then $a^2+a+a\gt0,1-4 a \geq 0$ and $-1\gt2 a$
$\Rightarrow a(a+2)\gt0, a \leq \frac{1}{4}$ and $a \lt -\frac{1}{2}$
$\Rightarrow a\gt0$ or $a \lt -2, a \lt \frac{1}{4}$ and $a \lt -\frac{1}{2}$
Hence $a \lt -2$.

Topic: Mathematics, Quadratic Equation, JEE Main
If the scalar projection of the vectors $x \mathbf{i}-\mathbf{j}+\mathbf{k}$ on the vector $2 \mathbf{i}-\mathbf{j}+5 \mathbf{k}$ is $\frac{1}{\sqrt{30}}$ then value of $x$ is equal to

Options:
A $\frac{-5}{2}$
B 6
C -6
D 3

The correct answer is:
$\frac{-5}{2}$

Solution:
Projection of $x \mathbf{i}-\mathbf{j}+\mathbf{k}$ on $2 \mathbf{i}-\mathbf{j}+5 \mathbf{k}$
$=\frac{(x \mathbf{i}-\mathbf{j}+\mathbf{k})(2 \mathbf{i}-\mathbf{j}+5 \mathbf{k})}{\sqrt{4+1+25}}=\frac{2 x+1+5}{\sqrt{30}}$
But, given $\frac{2 x+6}{\sqrt{30}}=\frac{1}{\sqrt{30}} \Rightarrow 2 x+6=1 \Rightarrow x=\frac{-5}{2}$.

Topic: Mathematics, Vector Algebra, JEE Main
If the screen is moved away from the plane of the slits in a Young's double slit experiment, then the:

Options:
A angular separation of the fringes increases
B angular separation of the fringes decreases
C linear separation of the fringes increases
D linear separation of the fringes decreases

The correct answer is:
linear separation of the fringes increases

Solution:
As per YDSE, $x=\frac{\lambda \mathrm{D}}{d}$,
(where, symbols have their usual meanings)
If the screen is moving away from the slits then D increases, and hence the fringe width $(x)$ increases.

Topic: Physics, Wave Optics, JEE Main
If the screw on a screw-gauge is given six rotations, it moves by 3mm on the main scale. If there are 50 divisions on the circular scale the least count of the screw gauge is:

Options:
A 0.001cm
B 0.02m
C 0.01cm
D 0.001mm

The correct answer is:
0.001cm

Solution:
Pitch =36=0.5 mm
L.C. =0.550 mm=1100 mm=0.1 mm=0.001 cm

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2020 (09 Jan Shift 1)
If the second term in the expansion $\left(\sqrt[13]{\mathrm{a}}+\frac{\mathrm{a}}{\sqrt{\mathrm{a}^{-1}}}\right)^{\mathrm{n}}$ is $14 \mathrm{a}^{5 / 2}$, then $\frac{{ }^{\mathrm{n}} \mathrm{C}_{3}}{{ }^{\mathrm{n}} \mathrm{C}_{2}}=$

Options:
A 4
B 3
C 12
D 6

The correct answer is:
4

Solution:
We have $\mathrm{T}_{2}=14 \mathrm{a}^{\frac{5}{2}}$
$$
\begin{array}{l}
\Rightarrow{ }^{\mathrm{n}} C_{1}\left(\mathrm{a}^{\frac{1}{13}}\right)^{\mathrm{n}-1}\left(\mathrm{a}^{\frac{3}{2}}\right)=14 \mathrm{a}^{\frac{5}{2}} \\
\Rightarrow \mathrm{na}^{\frac{\mathrm{n}-1}{13}+\frac{3}{2}}=14 \mathrm{a}^{\frac{5}{2}} \Rightarrow \mathrm{n}=14 \\
\Rightarrow \frac{{ }^{\mathrm{n}} \mathrm{C}_{3}}{{ }^{\mathrm{n}} \mathrm{C}_{2}}=\frac{{ }^{14} \mathrm{C}_{3}}{{ }^{14} \mathrm{C}_{2}}=\frac{12}{3}=4
\end{array}
$$

Topic: Mathematics, Binomial Theorem, JEE Main
If the second, third and fourth terms in the expansion of $(x+y)^n$ are 135,30 and $\frac{10}{3}$, respectively, then $6\left(n^3+x^2+y\right)$ is equal to _______

Options:

The correct answer is:
806

Solution:
${ }^{\mathrm{n}} \mathrm{C}_1 \mathrm{x}^{\mathrm{n}-1} \mathrm{y}=135$....(i)
${ }^n C_2 x^{n-2} y^2=30$....(ii)
${ }^{\mathrm{n}} \mathrm{C}_3 \mathrm{x}^{\mathrm{n}-3} \mathrm{y}^3=\frac{10}{3}$....(iii)
$\begin{aligned} & \text { By } \frac{(i)}{(\text { ii) }} \\ & \frac{{ }^n C_1}{{ }^n C_2} \frac{x}{y}=\frac{9}{2}...(iv) \\ & \text { By } \frac{(\text { ii) }}{(\text { iii) }} \\ & \frac{{ }^n C_2}{{ }^n C_3} \frac{x}{y}=9....(v)\end{aligned}$
$\begin{aligned} & \text { By } \frac{(\text { iv })}{(v)} \\ & \frac{{ }^n C_1{ }^n C_3}{{ }^n C_2{ }^n C_2}=\frac{1}{2} \\ & \frac{2 n^2(n-1)(n-2)}{6}=\frac{n(n-1)}{2} \frac{n(n-1)}{2} \\ & 4 n-8=3 n-3 \\ & \Rightarrow n=5 \\ & \text { put in (v) }\end{aligned}$
$\begin{aligned} & \frac{x}{y}=9 \\ & x=9 y \\ & \text { put in (i) } \\ & { }^5 C_1 x^4\left(\frac{x}{9}\right)=135 \\ & x^5=27 \times 9 \\ & \Rightarrow x=3, \quad y=\frac{1}{3} \\ & 6\left(n^3+x^2+y\right) \\ & =6\left(125+9+\frac{1}{3}\right) \\ & =806\end{aligned}$

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2024 (06 Apr Shift 1)
If the segment intercepted between the lines x+6y-13=0 and x–y+3=0 is bisected at 6,8, then the square of the length of segment is

Options:
A 268
B 244
C 212
D 252

The correct answer is:
244

Solution:
Let the segment is AB as shown in the figure
Let A=13-6y1,y1
B=y2-3,y2
∴ mid-point AB=6,8
∴13-6y1+y2-32=6.............i
And y1+y22=8.............ii
on solving i and ii we get

A=1,2 and B=11,14
⇒(AB)2=102+122=100+144=244

Topic: Mathematics, Straight Lines, JEE Main
If the series limit frequency of Balmer series is νB, then the series limit frequency of the Brackett series is

Options:
A 4νB25
B νB9
C νB4
D 9νB4

The correct answer is:
νB4

Solution:
As we know, wavelength of the light emitted during transition is given by,

1λ=R1n12-1n22.

Frequency can be written as,

ν=cR1n12-1n22

Now, for Balmer series2 to ∞,

νB=cR122-1∞2 ....Eq1

For Brackett series we can write4 to ∞,

ν'=cR142-1∞2 ....Eq2

From equation 2 and equation 1, we get

ν'νB=14⇒ν'=νB4

Topic: Physics, Atomic Physics, JEE Main
If the series limit frequency of the Lyman series is VL, then the series limit frequency of the Pfund series is:

Options:
A VL25
B 25 VL
C 16 VL
D VL16

The correct answer is:
VL25

Solution:
⇒V2=RCZ2 112-1∞2
V2=RCZ2
Series limit frequency of the Pfund series,
Vp=RCZ2152-1∞2
Vp =VL25

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2018 (08 Apr)
If the series limit wavelength of Lyman series for the hydrogen atom is $912 Å,$ then the series limit wavelength for Balmer series of hydrogen atoms is

Options:
A $912 Å$
B $912 \times 2 Å$
C $912 \times 4 Å$
D $\frac{912}{2} Å$

The correct answer is:
$912 \times 4 Å$

Solution:
$\frac{1}{\lambda}=R\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]$

For limiting wavelength of Lyman series

$$

n_{1}=1, n_{2}=\infty \quad \frac{1}{\lambda_{L}}=R

$$

For limiting wavelength of Balmer series $n_{1}=2, n_{2}=\infty$

$$

\begin{array}{l}

\frac{1}{\lambda_{B}}=R\left(\frac{1}{4}\right) \Rightarrow \lambda_{B}=\frac{4}{R} \\

\therefore \lambda_{B}=4 \lambda_{L}=4 \times 912 \AA

\end{array}

$$

Topic: Physics, Atomic Physics, JEE Main
If the series limit wavelength of the Lyman series of hydrogen atom is 912 Å, then the series limit wavelength of the Balmer series of the hydrogen atom is

Options:
A 912 Å
B 1824 Å
C 3648 Å
D 456 Å

The correct answer is:
3648 Å

Solution:
For series limit of Balmer series,
n2=2, n1=∞
1λ=R1n22-1n12
=122-1∞2=R4
∴λ=4R=410967800 m
=4×912×10-10 m
=3648 Å

Topic: Physics, Atomic Physics, JEE Main
If the set $R=\{(a, b): a+5 b=42, a, b \in \mathbb{N}\}$ has $m$ elements and $\sum_{n=1}^m\left(1-i^{n !}\right)=x+i y$, where $i=\sqrt{-1}$, then the value of $m+x+y$ is

Options:
A 12
B 4
C 8
D 5

The correct answer is:
12

Solution:
$\begin{aligned} & a+5 b=42, a, b \in N \\ & a=42-5 b, b=1, a=37 \\ & b=2, a=32 \\ & b=3, a=27 \\ & \vdots \\ & b=8, a=2\end{aligned}$
$\mathrm{R}$ has " 8 " elements $\Rightarrow \mathrm{m}=8$
$\begin{aligned} & \sum_{n=1}^8\left(1-i^{n !}\right)=x+i y \\ & \text { for } n \geq 4, i^{n !}=1 \\ & \Rightarrow(1-i)+\left(1-i^{2 !}\right)+\left(1-i^{3 !}\right) \\ & =1-I+2+1+1 \\ & =5-I=x+i y \\ & m+x+y=8+5-1=12\end{aligned}$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2024 (08 Apr Shift 1)
If the set Rez-z¯+zz¯2-3z+5z¯:z∈ℂ, Rez=3 is equal to the interval (α, β], then 24β-α is equal to

Options:
A 36
B 27
C 30
D 42

The correct answer is:
30

Solution:
Given that the setRez-z¯+zz¯2-3z+5z¯:z∈ℂ, Rez=3 is equal to the interval (α,β].

Let z=x+iy

=Rex+iy-x-iy+x2+y22-3x+iy+5x-iy

=Rex2+y2+i2y2+2x-8iy

=Rex2+y2+2yi21+x+8iy21+x2+8y2

=2x2+y21+x-16y241+x2+8y2

Now using, Rez=3⇒x=3 we get,

=89+y2-16y264+64y2

⇒fy=189-y21+y2

Let t=189-y21+y2

⇒8t+8ty2=9-y2

⇒y28t+1+8t-9=0

Now D≥0

⇒02-48t+18t-9≥0

⇒t--18t-98≤0

⇒t∈(-18,98] as t≠-18

Range of fy=(-0.125,1.125]

⇒α=- 0.125

⇒β=1.125

⇒β-α=1.25

⇒24β-α=30

Hence this is the required solution.

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2023 (15 Apr Shift 1)
If the shortest distance between the line joining the points 1,2,3 and 2,3,4, and the line x-12=y+1-1=z-20 is α, then 28α2 is equal to _____ .

Options:

The correct answer is:
18

Solution:
Equation of line joining 1,2,3 and 2,3,4 is

r→=i^+2j^+3k^+λi^+j^+k^

Also, vector equation is

r→=i^-j^+2k^+μ2i^-j^

So,

a→1=i^+2j^+3k^

a→2=i^-j^+2k^

Hence,

a→2-a→1=0i^-3j^-k^

And,

b→1=i^+j^+k^

b→2=2i^-j^+0k^

And,

b→1×b→2=i^j^k^1112-10=i^+2j^-3k^

⇒b→1×b→2=14

Required shortest distance is

α=a→2-a→1·b→1×b→2b→1×b→2

⇒α=-3j^-k^·i^+2j^-3k^14

⇒α=-6+314=314

Now, 28α2=228×914=18.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (25 Jan Shift 2)
If the shortest distance between the line x-1α=y+1-1=z1,α≠-1 , and x+y+z+1=0=2x-y+z+3 is 13,then value of α is :

Options:
A -1916
B 3219
C -1619
D 1932

The correct answer is:
3219

Solution:
Let us change the line into symmetric form.

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

Put z=1 , so we get x+y+2=0 and 2x-y+4=0

We will get x=-2, y=0

∴ The point -2,0,1 lies on the line and perpendicular vector will come from i^j^k^1112-11=2i^+j^-3k^

So the equation of line would be x+22=y1=z-1-3
And the other line x-1α=y+1-1=z1
Shortest distance would be D=a2→-a1→·b1→×b2→b1→×b2→

When a1→=-2i^+0j^+1k^

a2→=i^-j^+0k^

b1→=2i^+j^-3k^

b2→=αi^-j^+k^

D= 3-1-12 1-3α-1 1i^ j^ k^2 1-3α-1 1

= 31-3+12+3α+12+α-2i^-j^2+3α+k^-2-α

⇒ -6+2+3α+2+α4+(2+3α)2+(2+α)2=13

4α-24+4+12α+9α2+4+4α+α2=13

4α-210α2+16α+12=13

⇒ 2α19α-32=0

∴ α=3219

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2015 (10 Apr Online)
If the shortest distance between the lines $\frac{x+2}{2}=\frac{y+3}{3}=\frac{z-5}{4}$ and $\frac{x-3}{1}=\frac{y-2}{-3}=\frac{z+4}{2}$ is $\frac{38}{3 \sqrt{5}} \mathrm{k}$, and $\int_0^{\mathrm{k}}\left[x^2\right] \mathrm{d} x=\alpha-\sqrt{\alpha}$, where $[x]$ denotes the greatest integer function, then $6 \alpha^3$ is equal to________

Options:

The correct answer is:
48

Solution:
$\begin{aligned} & \frac{38}{3 \sqrt{5}} \hat{\mathrm{k}}=\frac{(5 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-9 \hat{\mathrm{k}})}{\sqrt{5}} \cdot\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 3 & 4 \\ 1 & -3 & 2\end{array}\right| \\ & \frac{38}{3 \sqrt{5}} \hat{\mathrm{k}}=\frac{19}{\sqrt{5}} \\ & \mathrm{k}=\frac{19}{\sqrt{5}} \\ & \mathrm{k}=\frac{3}{2} \\ & \int_0^{3 / 2}\left[\mathrm{x}^2\right]=\int_0^1 0+\int_1^{\sqrt{2}} 1+\int_{\sqrt{2}}^{3 / 2} 2 \\ & =\sqrt{2}-1+2\left(\frac{3}{2}-\sqrt{2}\right) \\ & =2-\sqrt{2} \\ & \alpha=2 \\ & \Rightarrow 6 \alpha^3=48\end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2024 (04 Apr Shift 1)
If the shortest distance between the lines $\frac{x-\lambda}{2}=\frac{y-4}{3}=\frac{z-3}{4}$ and $\frac{x-2}{4}=\frac{y-4}{6}=\frac{z-7}{8}$ is $\frac{13}{\sqrt{29}}$, then a value of $\lambda$ is :

Options:
A -1
B $-\frac{13}{25}$
C $\frac{13}{25}$
D 1

The correct answer is:
1

Solution:
$\left.\begin{array}{l}\overline{\mathrm{r}}_1=(\lambda \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\alpha(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \\ \overline{\mathrm{r}}_2=(2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+7 \hat{\mathrm{k}})+\beta(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})\end{array}\right\} \begin{gathered}\overline{\mathrm{b}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \\ \overline{\mathrm{a}}_2+\lambda \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\ \mathrm{a}_2=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}\end{gathered}$
Shortest dist. $=\frac{\left|\overline{\mathrm{b}} \times\left(\overline{\mathrm{a}}_2-\overline{\mathrm{a}}_1\right)\right|}{|\mathrm{b}|}=\frac{13}{\sqrt{29}}$
$\begin{aligned} & \left\lvert\, \frac{|(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times((2-\lambda) \hat{i}+4 \hat{k})|}{\sqrt{29}}=\frac{13}{\sqrt{29}}\right. \\ & |-8 \hat{\mathrm{j}}-3(2-\lambda) \hat{k}+12 \hat{i}+4(2-\lambda) \hat{j}|=13 \\ & |12 \hat{i}-4 \lambda \hat{j}+(3 \lambda-6) \hat{k}|=13\end{aligned}$
$\begin{aligned} & 144+16 \lambda^2+(3 \lambda-6)^2=169 \\ & 16 \lambda^2+(3 \lambda-6)^2=25=\lambda \Rightarrow=1\end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2024 (08 Apr Shift 2)
If the shortest distance between the lines $\frac{x-\lambda}{3}=\frac{y-2}{-1}=\frac{z-1}{1}$ and $\frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4}$ is $\frac{44}{\sqrt{30}}$, then the largest possible value of $|\lambda|$ is equal to _________

Options:

The correct answer is:
43

Solution:
$\begin{aligned} & \overline{\mathrm{a}}_1=\lambda \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \\ & \overline{\mathrm{a}}_2=-2 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{p}}-=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{q}}-=-3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \\ & (\lambda+2) \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}=\overline{\mathrm{a}}_1-\overline{\mathrm{a}}_2 \\ & \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}-=-6 \hat{\mathrm{i}}-15 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\end{aligned}$
$\begin{aligned}
& \frac{44}{\sqrt{30}}=\frac{|-6 \lambda-12-105-9|}{\sqrt{(-6)^2+(-15)^2+3^2}} \\
& \frac{44}{\sqrt{30}}=\frac{|6 \lambda+126|}{3 \sqrt{30}} \\
& 132=|6 \lambda+126| \\
& \lambda=1, \lambda=-43 \\
& |\lambda|=43
\end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2024 (06 Apr Shift 2)
If the shortest distance between the lines
$\begin{array}{ll}
L_1: \vec{r}=(2+\lambda) \hat{i}+(1-3 \lambda) \hat{j}+(3+4 \lambda) \hat{k}, & \lambda \in \mathbb{R} \\
L_2: \vec{r}=2(1+\mu) \hat{i}+3(1+\mu) \hat{j}+(5+\mu) \hat{k}, \quad \mu \in \mathbb{R}
\end{array}$
is $\frac{m}{\sqrt{n}}$, where $\operatorname{gcd}(m, n)=1$, then the value of $m+n$ equals

Options:
A 390
B 384
C 377
D 387

The correct answer is:
387

Solution:
Shortes distance (CD) $=\left|\frac{\overline{\mathrm{AB}} \cdot \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}}{|\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}|}\right|$
$\begin{aligned} & =\left|\frac{(0 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \cdot(-15 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+9 \hat{\mathrm{k}})}{\sqrt{355}}\right| \\ & =\frac{0+14+18}{\sqrt{355}}=\frac{32}{\sqrt{355}} \\ & \therefore \mathrm{m}+\mathrm{n}=32+355=387\end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2024 (08 Apr Shift 1)
If the shortest distance between the lines r1→=αi^+2j^+2k^+λi^-2j^+2k^, λ∈R, α>0 and r2→=-4i^-k^+μ3i^-2j^-2k^, μ∈R is 9, then α is equal to_____.

Options:

The correct answer is:
6

Solution:
We have,

r1→=αi^+2j^+2k^+λi^-2j^+2k^

r2→=-4i^-k^+μ3i^-2j^-2k^

If r→=a→+λb→ and r→=c→+λd→, then shortest distance between two lines is

L=a→-c→·b→×d→b→×d→

Here,

b→×d→=i^j^k^1-223-2-2

⇒b→×d→=8i^+8j^+4k^

⇒b→×d→=42i^+2j^+k^

⇒b→×d→=4·3

And,

a→-c→=α+4i^+2j^+3k^

So, shortest distance is

α+4i^+2j^+3k^·42i^+2j^+k^4·3=9

⇒2α+4+4+3=27

⇒α+4=10

⇒α=6

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (20 Jul Shift 1)
If the shortest distance between the lines r→=-i^+3k^+λi^-aj^ and r→=-j^+2k^+μi^-j^+k^ is 23, then the integral value of a is equal to _____

Options:

The correct answer is:
2

Solution:
D.R. 's of a→1=-1,0,3

D.R. 's of a→2=0,-1,2

D.R. 's of b→1=1,-a,0

D.R. 's of b→2=1,-1,1

Now, a→2-a→1=i^-j^-k^

and b→1×b→2=i^j^k^1-a01-11

b→1×b2→=i^-a-j^+k^a-1

i.e. b1→×b2→=a2+1+a-12

So, a→2-a→1·b→1×b→2=2-2a

Shortest distance between the lines, 21-aa2+1+a-12=23

Squaring on both the sides, we get,

31-a2=a2-a+1

i.e. a=2,12

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2022 (24 Jun Shift 1)
If the shortest distance between the lines x-12=y-23=z-3λ and x-21=y-44=z-55 is 13, then the sum of all possible values of λ is:

Options:
A 16
B 6
C 12
D 15

The correct answer is:
16

Solution:
We know that, shortest distance between two line is given by a2-a1·b1×b2b1×b2

Here, a1=1,2,3 a2=2,4,5

b→2=2i^+3j^+λk^ and b→2=i^+4j^+5k^

So, S.D. =2-1i^+4-2j^+5-3k^·b→1×b→2b1×b2

Calculating b→1×b→2=i^j^k^23λ145

=i^15-4λ+j^λ-10+k^5

=15-4λi^+λ-10j^+5k^

So, b→1×b→2=15-4λ2+λ-102+25

Now again,

S.D. =i^+2j^+2k^·15-4λi^+λ-10j^+5k15-4λ2+λ-102+25

15-4λ+2λ-20+1015-4λ2+λ-102+25=13

squaring both side we get,

35-2λ2=225+16λ2-120λ+λ2+100-20λ+25

12λ2+75-60λ=17λ2-140λ+350

5λ2-80λ+275=0

λ2-16λ+55=0

λ-5λ-11=0

⇒λ=5,11

So sum of 5+11=16

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2022 (24 Jun Shift 2)
If the shortest distance between the lines x−λ−2=y−21=z−11 and x−31=y−1−2=z−21 is 1, then the sum of all possible values of λ is

Options:
A 0
B 23
C 33
D −23

The correct answer is:
23

Solution:
Given: Equations of line,

x-λ-2=y-21=z-11 and x-31=y-1-2=z-21

Now, we know that shortest distance between two lines is given by,

S.D=a→2-a→1·b1→×b2→b1→×b2→

⇒1=λ-31-1-2111-21i^j^k^-2111-21

⇒1=λ-33+3-33i^+3j^+3k^

⇒λ-3=3

⇒λ=0, 23

Hence, the sum is 23

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2024 (01 Feb Shift 1)
If the shortest distance between the lines x-41=y+12=z-3 and x-λ2=y+14=z-2-5 is 65, then the sum of all possible values of λ is :

Options:
A 5
B 8
C 7
D 10

The correct answer is:
8

Solution:
x-41=y+12=z-3

x-λ2=y+14=z-2-5

The shortest distance between the lines
=(a→-b→)·d1→×d2→d1→×d2→

a≡4,-1,0, b≡λ,-1,2

⇒a→-b→=4i^-j^-λi^-j^+2k^

⇒a→-b→=4-λi^-2k^

So, the shortest distance is given by,

D=λ-40212-324-5i^j^k^12-324-5

⇒D=(λ-4)(-10+12)-0+2(4-4)|2i^-1j^+0k^|

⇒65=2(λ-4)5

⇒3=|λ-4|

⇒λ-4=±3

⇒λ=7,1

So, sum of all possible values of λ is 8.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2024 (27 Jan Shift 1)
If the shortest distance between the lines x+62=y-63=z-64 and x-λ3=y-264=z+265 is 6, then sum of squares of all possible values(s) of λ is

Options:

The correct answer is:
384

Solution:
Given equations are

x-λ3=y-264=z+265 is passing through a point λ,26,36 and it's direction ratios are 3,4,5.

So,

a→1=λi^+26j^-26k^

b→1=3i^+4j^+5k^

And,

x+62=y-63=z-64 is passing through a point -6,6,6 and it's direction ratios are 1,2,3.

a→2=-6i^+6j^+6k^

b→2=2i^+3j^+4k^

Now,

a→2-a→1=-6i^+6j^+6k^-λi^+26j^-26k^=-6+λi^-6j^+36k^

b→1×b→2=i^j^k^345234

⇒b→1×b→2=i^-2j^+k^

So,

⇒b→1×b→2=6

We know that the shortest distance between two skew lines is a→2-a→1·b→1×b→2b→1×b→2=6

⇒-6-λi^-6j^+36k^·i^-2j^+k^6=6

⇒46-λ6=6

⇒λ=106, -26

So, required sum is

=106-262=384

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (24 Jan Shift 2)