datavorous/entrance-exam-dataset huggingface dataset:
The shaded region in the given figure represents

Options:
A A∩ B∪C
B A∪B∩C
C A∩B-C
D A-B∪C

The correct answer is:
A-B∪C

Solution:
Shaded region contain elements of A not in B & not in C hence it is A-(B∪C)

Topic: Mathematics, Sets and Relations, JEE Main
The shape around the central form in ClF 4 + is

Options:
A square planar
B square pyramidal
C octahedral
D see-saw

The correct answer is:
see-saw

Solution:
Cl has 4 bonding pairs around it and 1 non-bonding pair is also present. If all 5 groups of electrons were involved in bonding, the shape would be trigonal bipyramidal. With one non-bonding pair, the shape becomes see-saw.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The shape of $2 p$ orbital is

Options:
A Spherical
B Ellipsoidal
C Dumb-bell
D Pyramidal

The correct answer is:
Dumb-bell

Solution:
The shape of $2 p$ orbital is dumb-bell.

Topic: Chemistry, Structure of Atom, JEE Main
The shape of $IF_7$ molecule is

Options:
A Octahedral
B Pentagonal bipyramidal
C Trigonal bipyramidal
D Tetrahedral

The correct answer is:
Pentagonal bipyramidal

Solution:
IF7 molecule show sp3d3 -hybridization72o.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The shape of $\mathrm{IF}_6^{-}$is :

Options:
A
Trigonally distorted octahedron

B
Pyramidal

C
Octahedral

D
Square antiprism

The correct answer is:
Trigonally distorted octahedron

Solution:
The structure of $\mathrm{IF}_6-$ is distorted octahedral This is due to presence of a "weak" lone pair.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2013 (23 Apr Online)
The shape of stress vs strain graph within elastic limit is:

Options:
A Parabolic
B Curve
C Straight line
D Ellipse

The correct answer is:
Straight line

Solution:
Within elastic limit it obeys Hooke's Law
i.e., Stress ∝Strain

Topic: Physics, Mechanical Properties of Solids, JEE Main
The shape of the molecule $\mathrm{CIF}_{3}$ is-

Options:
A trigonal planar
B pyramidal
C T-shaped
D Y-shaped

The correct answer is:
T-shaped

Solution:
Three bond pair \& two lone pair present in $\mathrm{CIF}_{3}$ molecule.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The shape of XeOF4 by VSEPR theory is:

Options:
A Trigonal bipyramidal
B Square pyramidal
C Pentagonal planar
D Octahedral

The correct answer is:
Square pyramidal

Solution:
sp3d2 hybridization
4 B.P + 1 B.P (Doubly bonded) + 1 L.P
Square pyramidal
Oxygen atom doubly bonded to Xe lone pair of electrons on axial position.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2015 (10 Apr Online)
The shape/structure ofXeF5- and XeO3F2, respectively are :

Options:
A pentagonal planar and trigonal bipyramidal
B octahedral and square pyramidal
C trigonal bipyramidal and pentagonal planar
D trigonal bipyramidal and trigonal bipyramidal

The correct answer is:
pentagonal planar and trigonal bipyramidal

Solution:
(i) XeF5-
St. No. =(5+2)=7 so hybridisation is=sp3d3 and structure is pentagonal planar.

(ii) XeO3F2 St. No. =5
so hybridisation is =sp3d and structure is trigonal bipyramidal.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2020 (02 Sep Shift 2)
The S.H.M. of a particle is given by the equation $y=3 \sin \omega t+4 \cos \omega t$. The amplitude is

Options:
A 7
B 1
C 5
D 12

The correct answer is:
5

Solution:
Resultant amplitude $=\sqrt{3^2+4^2}=5$

Topic: Physics, Oscillations, JEE Main
The shortest distance between line $y-x=1$ and curve $x=y^2$ is

Options:
A
$\frac{3 \sqrt{2}}{8}$

B
$\frac{8}{3 \sqrt{2}}$

C
$\frac{4}{\sqrt{3}}$

D
$\frac{\sqrt{3}}{4}$

The correct answer is:
$\frac{3 \sqrt{2}}{8}$

Solution:
$$
P=\left(y^2, y\right)
$$
Perpendicular distance from $P$ to $x-y+1=0$ is $\frac{\left|y^2-y+1\right|}{\sqrt{2}}$
$$
\begin{aligned}
& \mathrm{y}^2-\mathrm{y}+1>0 \forall \mathrm{y} \in \mathrm{R} \\
& \therefore \text { Coefficient } \mathrm{y}^2>0 \\
& \therefore \text { Min value }=\frac{1}{\sqrt{2}}\left(\frac{4 \mathrm{ac}-\mathrm{b}^2}{4 \mathrm{a}}\right)=\frac{3}{4 \sqrt{2}}
\end{aligned}
$$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2011
The shortest distance between lines L1 and L2, where L1:x−12=y+1−3=z+42 and L2 is the line passing through the points A−4,4,3, B−1,6,3 and perpendicular to the line x−3−2=y3=z−11, is

Options:
A 121221
B 24117
C 141221
D 42117

The correct answer is:
141221

Solution:
Given: L2 passes through A-4,4,3 and B-1,6,3.

⇒L2:x+43=y−42=z−30

Also, L1:x-12=y+1-3=z+42

We know that, shortest distance between L1: x-x1a1=y-y1b1=z-z1c1 and L2: x-x2a2=y-y2b2=z-z2c2 is given by,

x2-x1y2-y1z2-z1a1b1c1a2b2c2a1b2-a2b12+b1c2-b2c12+c1a2-c2a12.

So, shortest distance, D=-4-11+44+32−323204+92+0-42+6-02

⇒D=-5572−32320169+16+36

⇒D=31×3+48221

⇒D=141221

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2024 (31 Jan Shift 2)
The shortest distance between the line passing through the point $\bar{i}+2 \bar{j}+3 \bar{k}$ and parallel to the vector $2 \bar{i}+3 \bar{j}+4 \bar{k}$ and the line passing through the point $2 \bar{i}+4 \bar{j}+5 \bar{k}$ and parallel to the vector $3 \bar{i}+4 \bar{j}+5 \bar{k}$, is

Options:
A 9
B $\frac{1}{\sqrt{6}}$
C 1
D $\sqrt{6}$

The correct answer is:
$\frac{1}{\sqrt{6}}$

Solution:
No solution. Refer to answer key.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The shortest distance between the line x-y=1 and the curve x2=2y is:

Options:
A 12
B 12
C 122
D 0

The correct answer is:
122

Solution:
Shortest distance between curves is always along common normal.

Let Px0, y0 be any point on the parabola x2=2y.

Clearly, slope of line x-y=1 is 1.

Hence, slope of perpendicular line is -1.

dydxP=Slope of tangent at P

Therefore, -1dydxP=Slope of normal=-1

⇒dydxP=1

⇒x0=1 ∴y0=12

⇒P1,12

∴ Shortest distance =1-12-112+12=122

option 3

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2021 (25 Feb Shift 2)
The shortest distance between the line x=y=z and the line of intersection of 2x+y+z-1=0 and 3x+y+2z-2=0 is

Options:
A 12 units
B 13 units
C 14 units
D 15 units

The correct answer is:
12 units

Solution:
P1=2x+y+z-1=0
P2=3x+y+2z-2=0
The line of intersection is parallel to
i^j^k^211312=i^-j^-k^
For the point on the line, putting z=0, we get,
2x+y=13x+y=2x=1, y=-1
So, the required point is 1,-1,0
Hence, the equation of the line of intersection is
x-11=y+1-1=z-0-1
Let, aâ=i^-j^
bâ=i^-j^-k^
câ=0â
dâ=i^-j^-k^
câ-aâ=-i^+j^
bâÃdâ=-2j^+2k^
Shortest distance =câ-aââbâÃdâbâÃdâ
=-222=12

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The shortest distance between the line $\mathrm{y}-\mathrm{x}=1$ and the curve $\mathrm{x}=\mathrm{y}^2$ is

Options:
A
$\frac{3 \sqrt{2}}{8}$

B
$\frac{2 \sqrt{3}}{8}$

C
$\frac{3 \sqrt{2}}{5}$

D
$\frac{\sqrt{3}}{4}$

The correct answer is:
$\frac{3 \sqrt{2}}{8}$

Solution:
$$
\begin{aligned}
& x-y+1=0 \\
& x=y^2 \\
& 1=2 y \frac{d y}{d x} \Rightarrow \frac{d y}{d x}=\frac{1}{2 y}=\text { Slope of given line (1) } \\
& \frac{1}{2 y}=1 \Rightarrow y=\frac{1}{2} \Rightarrow y=\frac{1}{2} \Rightarrow x=\left(\frac{1}{2}\right)^2=\frac{1}{4} \Rightarrow(x, y)=\left(\frac{1}{4}, \frac{1}{2}\right)
\end{aligned}
$$
$\therefore$ The shortest distance is $\frac{\left|\frac{1}{4}-\frac{1}{2}+1\right|}{\sqrt{1+1}}=\frac{3}{4 \sqrt{2}}=\frac{3 \sqrt{2}}{8}$
Directions: Question number 86 to 90 are Assertion - Reason type questions. Each of these questions contains two statements
Statement-1 (Assertion) and Statement-2 (Reason).
Each of these questions also have four alternative choices, only one of which is the correct answer. You have to select the correct choice

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2009
The shortest distance between the line y=x and the curve y2=x–2 is

Options:
A 742
B 78
C 1142
D 2

The correct answer is:
742

Solution:
y2=x–2
Differentiating w.r.t. x we get,
2yy'= 1⇒y'=  12y
For the shortest distance, the tangent at point P will be parallel to the given line
⇒y'|p=12y1=1⇒y1=12
⇒x1=2+122=94 (∵y12=x1-2)
The shortest distance between the given curve & the line
= The perpendicular distance of point P from the line
=94-1212+12=742=742

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2019 (08 Apr Shift 1)
The shortest distance between the line \( y=x \) and the curve \( y^{2}=x-2 \) is

Options:
A \( \frac{7}{4 \sqrt{2}} \)
B \( \frac{7}{8} \)
C \( \frac{11}{4 \sqrt{2}} \)
D \( 2 \)

The correct answer is:
\( \frac{7}{4 \sqrt{2}} \)

Solution:
y2=x–2
Differentiating w.r.t. x we get,
2yy'= 1⇒y'=  12y
For the shortest distance, the tangent at point P will be parallel to the given line
⇒y'|p=12y1=1⇒y1=12
⇒x1=2+122=94 (∵y12=x1-2)
The shortest distance between the given curve & the line
= The perpendicular distance of point P from the line
=94-1212+12=742=742

Topic: Mathematics, Application of Derivatives, JEE Main
The shortest distance between the lines $\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{-1}$ and $\frac{x+3}{2}=\frac{y-6}{1}=\frac{z-5}{3}$ is :

Options:
A $\frac{18}{\sqrt{5}}$
B $\frac{22}{3 \sqrt{5}}$
C $\frac{46}{3 \sqrt{5}}$
D $6 \sqrt{3}$

The correct answer is:
$\frac{18}{\sqrt{5}}$

Solution:
Give lines are $\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{-1}$ and $\frac{x+3}{2}=\frac{y-6}{1}=\frac{z-5}{3}$
Lines passes through the points
$\vec{a}_1=(3,2,1) \text { and } \vec{a}_2=(-3,6,5) \text {, }$
$\vec{b}_1=2 \hat{i}+3 \hat{j}-\hat{k}$
$\vec{b}_1=2 \hat{i}+\hat{j}-3 \hat{k}, \vec{a}_2-\vec{a}_1=6 \hat{i}-4 j-4 \hat{k}$
Shortest distance $=\frac{\left|\left(\vec{a}_2-\vec{a}_1\right)\left(\vec{b}_1 \times \vec{b}_2\right)\right|}{\left|\left(\vec{b}_1 \times \vec{b}_2\right)\right|}$
$\vec{b}_1 \times \vec{b}_2=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 2 & 1 & 3\end{array}\right|=10 \hat{i}-8 \hat{j}-4 \hat{k}$
$\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)=60+32+16=108$
$\left|\vec{b}_1 \times \vec{b}_2\right|=\sqrt{100+64+16}=\sqrt{180}$
$S . D=\frac{108}{\sqrt{180}}=\frac{108}{6 \sqrt{5}}=\frac{18}{\sqrt{5}}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The shortest distance between the lines $\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$ is

Options:
A $\sqrt{30}$
B $2 \sqrt{30}$
C $5 \sqrt{30}$
D $3 \sqrt{30}$

The correct answer is:
$3 \sqrt{30}$

Solution:
$\begin{aligned} & \text { S.D. }=\frac{\left|\begin{array}{ccc}6 & 15 & -3 \\ 3 & -1 & 1 \\ -3 & 2 & 4\end{array}\right|}{\sqrt{(-4-2)^2+(12+3)^2+(6-3)^2}} \\ & =\frac{270}{\sqrt{270}}=\sqrt{270}=3 \sqrt{30} \\ & \end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The shortest distance between the lines $\frac{x-3}{4}=\frac{y+7}{-11}=\frac{z-1}{5}$ and $\frac{x-5}{3}=\frac{y-9}{-6}=\frac{z+2}{1}$ is:

Options:
A $\frac{178}{\sqrt{563}}$
B $\frac{187}{\sqrt{563}}$
C $\frac{185}{\sqrt{563}}$
D $\frac{179}{\sqrt{563}}$

The correct answer is:
$\frac{187}{\sqrt{563}}$

Solution:
$\begin{aligned} & \overrightarrow{\mathrm{n}}=\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}} \\ & \overrightarrow{\mathrm{n}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 4 & -11 & 5 \\ 3 & -6 & 1\end{array}\right|=19 \hat{\mathrm{i}}+11 \hat{\mathrm{j}}+9 \hat{\mathrm{k}} \\ & \text { S.d. }=\text { projection of } \overrightarrow{\mathrm{AB}} \text { on } \overrightarrow{\mathrm{n}} \\ & =\left|\frac{\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{n}}}{|\overrightarrow{\mathrm{n}}|}\right|=\left|\frac{(2 \hat{\mathrm{i}}+16 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}) \cdot(19 \hat{\mathrm{i}}+11 \hat{\mathrm{j}}+9 \hat{\mathrm{k}})}{\sqrt{361+121+81}}\right| \\ & =\frac{38+176-27}{\sqrt{563}} \\ & \text { S.d. }=\frac{187}{\sqrt{563}}\end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2024 (09 Apr Shift 1)
The shortest distance between the lines $\mathbf{r}=3 \mathbf{i}+5 \mathbf{j}+7 \mathbf{k}+\lambda(\mathbf{i}+2 \mathbf{j}+\mathbf{k}) \quad$ and $\mathbf{r}=-\mathbf{i}-\mathbf{j}-\mathbf{k}+\mu(7 \mathbf{i}-6 \mathbf{j}+\mathbf{k})$ is

Options:
A $\frac{16}{5 \sqrt{5}}$
B $\frac{26}{5 \sqrt{5}}$
C $\frac{36}{5 \sqrt{5}}$
D $\frac{46}{5 \sqrt{5}}$

The correct answer is:
$\frac{46}{5 \sqrt{5}}$

Solution:
The given lines are $\mathbf{r}=\mathbf{a}_1+\lambda \mathbf{b}_1, \mathbf{r}=\mathbf{a}_2+\mu \mathbf{b}_2$ where,
$$
\begin{aligned}
& \mathbf{a}_1=3 \mathbf{i}+5 \mathbf{j}+7 \mathbf{k}, \mathbf{b}_1=\mathbf{i}+2 \mathbf{j}+\mathbf{k} \\
& \mathbf{a}_2=-\mathbf{i}-\mathbf{j}-\mathbf{k}, \mathbf{b}_2=7 \mathbf{i}-6 \mathbf{j}+\mathbf{k} \\
& \quad\left|\mathbf{b}_1 \times \mathbf{b}_2\right|=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 2 & 1 \\
7 & -6 & 1
\end{array}\right| \\
\Rightarrow \quad & |\mathbf{i}(2+6)-\mathbf{j}(1-7)+\mathbf{k}(-6-14)| \\
\Rightarrow & |8 \mathbf{i}+6 \mathbf{j}-20 \mathbf{k}| \\
\Rightarrow & \sqrt{64+36+400}=\sqrt{500}=10 \sqrt{5}
\end{aligned}
$$
Now,
$$
\begin{aligned}
& {\left[\left(\mathbf{a}_2-\mathbf{a}_1\right) \mathbf{b}_1 \mathbf{b}_2\right]=\left(\mathbf{a}_2-\mathbf{a}_1\right) \cdot\left(\mathbf{b}_1 \times \mathbf{b}_2\right)} \\
& =(-4 \mathbf{i}-6 \mathbf{j}-8 \mathbf{k}) \cdot(8 \mathbf{i}+6 \mathbf{j}-20 \mathbf{k}) \\
& =-32-36+160 \\
& =160-68=92
\end{aligned}
$$
Shortest distance
$$
\begin{aligned}
& =\frac{\left[\left(\mathbf{a}_2-\mathbf{a}_1\right) \cdot\left(\mathbf{b}_1 \times \mathbf{b}_2\right)\right]}{\left|\mathbf{b}_1 \times \mathbf{b}_2\right|} \\
& =\frac{92}{10 \sqrt{5}}=\frac{46}{5 \sqrt{5}}
\end{aligned}
$$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The shortest distance between the lines x-10=y+1-1=z1and x+y+z+1=0,2 x-y+z+3=0 is

Options:
A 1
B 13
C 12
D 12

The correct answer is:
13

Solution:
We know that shortest distance between two skew lines exists along the line which is perpendicular to both the lines.

Now finding the equation of a plane P, whose normal is perpendicular to the given first line and the line obtained by planes x+y+z+1=0, 2x-y+z+3=0.

So, P is:

x+y+z+1+λ(2x-y+z+3)=0

It should be parallel to given line. ⇒ 0(1+2λ)-1(1-λ)+1(1+λ)=0
⇒λ=0
Thus, Plane, P is x+y+z+1=0

So, the shortest distance between the lines will be same as distance between the plane P and given line x-10=y+1-1=z1

Shortest distance = Perpendicular distance of (1,-1,0) from this plane =|1-1+0+1|12+12+12=13

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2020 (06 Sep Shift 1)
The shortest distance between the lines x+1=2 y=-12z and x=y+2=6z-6 is

Options:
A 2
B 3
C 52
D 32

The correct answer is:
2

Solution:
Given,

Equation of the lines

x+1=2 y=-12z and x=y+2=6z-6

⇒x+11=y12=z-112 and x1=y+21=z-116

We know that, shortest distance between the lines is given by S.D=b→-a→·p→×q→p→×q→

Here p→=i^+j^2+-k^12 & q→=i^+j^+k^6 and a→=i^ & b→=2j^-k^

⇒S.D=-i^+2j^-k^·p→×q→p→×q→

Now solving p→×q→≡i^j^k^112-1121116

⇒p→×q→≡16i^-14j^+12k^

⇒p→×q→≡2i^-3j^+6k^

Hence, S.D=-i^+2j^-k^·2i^-3j^+6k^22+32+62=-147=2

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (25 Jan Shift 2)
The shortest distance between the lines x+21=y-2=z-52 and x-41=y-12=z+30 is

Options:
A 8
B 6
C 7
D 9

The correct answer is:
9

Solution:
Given,

x+21=y-2=z-52

x-41=y-12=z+30

Now we know that,

Shortest distance between two skew lines is given by,

SD=a2→-a1→n1→×n2→n1→×n2→

Now finding, n1→×n2→=i^j^k^1-22120

⇒n1→×n2→=-4i^+2j^+4k^

And a2→-a1→=4i^+j^-3k^--2i^+0j^+5k^=6i^+j^-8k^

Now putting the value in formula we get,

SD=6i^+j^-8k^-4i^+2j^+4k^16+16+4

⇒SD=-24+2-326=546=9

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (10 Apr Shift 1)
The shortest distance between the lines x-22=y-32=z-01 and x+4-1=y-78=z-54 lies in the interval

Options:
A 0,1
B 1,2
C 2,3
D 3,4

The correct answer is:
2,3

Solution:
Let, a→=2i^+3j^
b→=-4i^+7j^+5k^ ⇒b→-a→=-6i^+4j^+5k^
c→=2i^+2j^+k^
d→=-i^+8j^+4k^

⇒c→×d→=-9j^+18k^
Hence, shortest distance =b→-a→⋅c→×d→c→×d→
=-36+9095=65∈2,3

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The shortest distance between the lines x-23=y+12=z-62 and x-63=1-y2=z+80 is equal to ______

Options:

The correct answer is:
14

Solution:
Given equations are

x-23=y+12=z-62 is passing through a point 2,-1,6 and it's direction ratios are 3,2,2.

So,

a→1=2i^-j^+6k^

b→1=3i^+2j^+2k^

And,

x-63=y-1-2=z+80 is passing through a point 6,1,-8 and it's direction ratios are 3,-2,0.

a→2=6i^+j^-8k^

b→2=3i^-2j^+0k^

Now,

a→2-a→1=4i^+2j^-14k^

b→1×b→2=i^j^k^3223-20

⇒b→1×b→2=4i^+6j^-12k^

So,

⇒b→1×b→2=16+36+144=14

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

=4i^+2j^-14k^·4i^+6j^-12k^14

=16+12+16814=14 units

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (24 Jan Shift 1)
The shortest distance between the lines x2=y2=z1 and x+2-1=y-48=z-54, lies in the interval:

Options:
A 3,4
B 2,3
C 1,2
D 0,1

The correct answer is:
2,3

Solution:
Let the direction vectors of the two lines be given by-a1i^+b1j^+c1k^=2i^+2j^+k^ and a2i^+b2j^+c2k^=-i^+8j^+4k^

The shortest distance between the two lines is given by-

x2-x1i^+y2-y1j^+z2-z1k^.a1i^+b1j^+c1k^×a2i^+b2j^+c2k^a1i^+b1j^+c1k^×a2i^+b2j^+c2k^

=x2−x1y2- y1z2−z1a1b1c1a2b2c2ijka1b1c1a2b2c2

=−245221−184ijk221−184=5492+182=65

Now 69<65<64⇒2<65<3

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2016 (09 Apr Online)
The shortest distance between the lines x-32=y-23=z-1-1 and x+32=y-61=z-53 is

Options:
A 185
B 2235
C 4635
D 63

The correct answer is:
185

Solution:
Given equation of linesx-32=y-23=z-1-1 &

x+32=y-61=z-53

Now let the point be A=3,2,1 on first line

and B=-3,6,5 on second line, so BA→=6i^-4j^-4k^

And direction ratio's be n1→=2i^+3j^-k^ &

n2→=2i^+i^-3k^

n1→×n2→=i^j^k^23-1213=10i^-8j^-4k^

We know that shortest Distance =BA→n1→n2→n1→×n2→

BA→n1→n2→=6i^-4j^-4k^.10i^-8j^-4k^=60+32+16=108

n1→×n2→=100+64+16=180

S.D=108180=10865=185

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2022 (27 Jun Shift 2)
The shortest distance between the lines x-33=y-8-1=z-31 and x+3-3=y+72=z-64 is

Options:
A 230
B 7230
C 330
D 3

The correct answer is:
330

Solution:
w→=i^+j^–3k^
p→=3i^-j^+k^
q→=-3i^+2j^+4k^
p→×q→=ijk3-11-324=6i^-15j^+3k^
S.D.=|AB.p×q|p×q=36+225+936+225+9=330

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2020 (08 Jan Shift 1)
The shortest distance between the lines x-44=y+25=z+33 and x-13=y-34=z-42 is

Options:
A 63
B 26
C 62
D 36

The correct answer is:
36

Solution:
Given:

x-44=y+25=z+33

x-13=y-34=z-42

So,

a→1=4i^-2j^-3k^a→2=i^+3j^+4k^

So,

a→2-a→1=-3i^+5j^+7k^

And,

b→1×b→2=i^j^k^453342=-2i^+j^+k^

⇒b→1×b→2=-22+12+12=6

Shortest distance between the lines

=a→2-a→1·b→1×b→2b→1×b→2

=-3i^+5j^+7k^·-2i^+j^+k^6

=6+5+76=6×36 units

=36 units

Hence this is the correct option.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (08 Apr Shift 1)
The shortest distance between the lines x-51=y-22=z-4-3 and x+31=y+54=z-1-5 is

Options:
A 73
B 53
C 63
D 43

The correct answer is:
63

Solution:
Shortest distance between two lines

x-x1a1=y-y1b1=z-z1c1 and x-x2a2=y-y2b2=z-z2c2 is given as

x1-x2y1-y2z1-z2a1b1c1a2b2c2a1b2-a2b12+b1c2-b2c12+c1a2-c2a12

=5--3 2--5 4-112-314-54-22+-10+122+-3+52

=87312-314-522+22+22

=8-10+12-7-5+3+34-24+4+4

=16+14+612=3623

=183=63

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (01 Feb Shift 1)
The shortest distance between the lines x+7-6=y-67=z and 7-x2=y-2=z-6 is

Options:
A 229
B 1
C 3729
D 292

The correct answer is:
229

Solution:
Given line L1:x+7-6=y-67=z-01

Let any point on it is given by a→1-7,6,0 and L1 is parallel to b→1-6,7,1

Also given line L2:x-7-2=y-21=z-61

Let any point on it be a→27,2,6 and L2 is parallel to b→2-2,1,1

Now finding b→1×b→2=i^j^k^-671-211=3i^+2j^+4k^

Now shortest distance between L1 and L2 will be

=a→2-a→1·b→1×b→2b→1×b→2=-14i^+4j^-6k^·3i^+2j^+4k^9+4+16

=229

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2022 (25 Jul Shift 2)
The shortest distance between the point 32,0 and the curve y=x,x>0 , is

Options:
A 32
B 54
C 32
D 52

The correct answer is:
52

Solution:
Let P be the nearest to 32, 0, then normal at P will pass through 32, 0.

Any point on the parabola y2=4ax can be taken as at2, 2at

Let, co-ordinates of P on the curve y=x, ⇒y2=x can be taken as t24, t2
The equation of the normal to the parabola y2=4ax at a point at2, 2at is y+tx=2at+at3.

Hence, the equation of normal to y=x or y2=x at t24, t2 is y+tx=t2+t34
This line passes through 32, 0, hence
3t2=t2+t34

⇒t3-4t=0

⇒tt2-4=0

⇒t=0, 2, -2 but -2 is rejected as we have to take point in the first quadrant.

Thus, the points are 0, 0 or 1, 1 and the distances of these points from 32, 0 are respectively

32-02+0-02=32 and 32-12+1-02=52 units.
Hence, nearest point is 1,1 and the minimum distance is 52 units.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2019 (10 Jan Shift 1)
The shortest distance between the skew lines \( \hat{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+t(\hat{i}+3 \hat{j}+2 \hat{k}) \) and
\( \hat{r}=(4 \hat{i}+5 \hat{j}+6 \hat{k})+t(2 \hat{i}+3 \hat{j}+\hat{k}) \) is

Options:
A \( \sqrt{6} \)
B \( 3 \)
C \( 2 \sqrt{3} \)
D \( \sqrt{3} \)

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

Solution:
Given skew lines are
r→=i^+2j^+3k^+ti^+3j^+2k^
and r→=4i^+5j^+6k^+t2i^+3j^+k^
Here, a→1= i^+2j^+3j^,
b→1= i^+3j^+2k^
a→2=4i^+5j^+6k^
and b→2=2i^+3j^+ k^
Now, 0 a→2- a→1=4i^+ 5j^+6k^-i^+2j^+3k^
=3i^+3j^+3k^
and b1×b2=i^j^k^132231
= i^ 3-6- j^ 1-4+ k^ 3-6
= -3i^+3j^-3k^
∴ The shortest distance between skew lines
=a→2- a→1-b→2×b→2b→1×b→2
=3i^+3j^+3k^ . -3i^+3j^-3k^-32+32+-32
=-9+9-99+9+9=933= 3

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The shortest distance between the z - axis and the line x+y+2z-3=0=2x+3y+4z-4, is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
We have,

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

Change it into symmetrical form

Put z=1, then

x+y-1=0

2x+3y=0

On solving, we get

x=3, y=-2, z=1

3, -2 , 1

Direction ratio of the line is

=i^j^k^112234=-2i^+k^

Hence, equation of line in symmetrical form is

x-3-2= y+20=z-11

Equation of z axis is, x0= y0= z1

Shortest Distance =3-21-201001i^j^k^-201001

=-42j^=2

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2015 (11 Apr Online)
The shortest distance between two Na+ ions in rock-salt arrangement having edge length equal to a 2 picometers is-

Options:
A 2a
B 2a2
C a2
D a

The correct answer is:
a

Solution:
Distance between two Na + is 12 times of edge length

Topic: Chemistry, Solid State, JEE Main
The shortest distance from the plane $12 x+4 y+3 z=327$ to the sphere $x^2+y^2+z^2+4 x-2 y-6 z=155$ is

Options:
A
39

B
26

C
$11 \frac{4}{13}$

D
13

The correct answer is:
13

Solution:
Shortest distance $=$ perpendicular distance $=\left|\frac{-2 \times 12+4 \times 1+3 \times 3-327}{\sqrt{144+9+16}}\right|=26$
$\therefore$ Shortest distance
$$=26-\sqrt{4+1+15+9}=26-13=13 \quad[\because 26-r]$$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2003
The shortest distance of 0,3 from the parabola y=x2 is

Options:
A 3 units
B 112 units
C 102 units
D 52 units

The correct answer is:
112 units

Solution:
Any point on the parabola is x,x2
Distance between (x,x2) and (0,3) is
D=x2+x2-32
D2=x2-522+3-14 is minimum if
x2-52=0
Hence, Dmin=114=112 units

Topic: Mathematics, Parabola, JEE Main
The shortest wavelength in Lyman series is 91.2 nm . The longest wavelength of the series is

Options:
A 121.6 nm
B 182.4 nm
C 234.4 nm
D 364.8 nm

The correct answer is:
121.6 nm

Solution:
The wavelength (Î») of lines is given by
1Î» =R112-1n2
For Lyman series, the shortest wavelength is for n=â and longest is for n= 2.
â´ 1Î»s= R112 â¦(i)
1Î»L= R11-122=34R ...(ii)
Dividing Eq.(ii) by Eq. (i) , we get
Î»LÎ»s=43
Given, Î»s= 91.2 nm
â Î»L = 91.2 Ã43 = 121.6 nm

Topic: Physics, Atomic Physics, JEE Main