datavorous/entrance-exam-dataset huggingface dataset:
Let AB be a chord of length 12 of the circle
x-22+y+12=1694
If tangents drawn to the circle at points A and B intersect at the point P, then five times the distance of point P from chord AB is equal to _____.

Options:

The correct answer is:
72

Solution:
Plotting the diagram of given value we have,

Now given, radius =1694=132

Now by diagram M is the midpoint of AB so, AM=6

Now in ∆AMP, PM=6cotθ

And in ∆OAM

sin90°-θ=6132

⇒cosθ=1213

PM=6×125=725

⇒5PM=72

Topic: Mathematics, Circle, JEE Main, JEE Main 2022 (29 Jul Shift 2)
Let AB be a line segment with mid-point C and D be the mid-point of AC. Let C1 be the circle with diameter AB and C2 be the circle with diameter AC. Let E be a point on C1 such that EC is perpendicular to AB. Let F be a point on C2 such that DF is perpendicular to AB and E and F lie on opposite sides of AB. Then, the value of sin∠FEC is

Options:
A 110
B 210
C 113
D 213

The correct answer is:
110

Solution:
According to given informations in the question, if radius of circle C1 is 2a, centre c is at origin 0,0 and AB is along X-axis, then B2a,0, A-2a,0, E0,2 a and F-a,-a.

Let ∠FEC=θ

Then, slope of line EF=m=tanπ2-θ

⇒ cotθ=2a--a0--a=3 ∵m=y2-y1x2-x1

∴ sin∠FEC

=sinθ=11+cot2θ=11+9=110

Topic: Mathematics, Straight Lines, JEE Main
Let $a, b$ be the solutions of $x^2+p x+1=0$ and $c, d$ be the solution of $x^2+q x+1=0$. If $(a-c)(b-c)$ and $(a+d)(b+d)$ are the solution of $x^2+a x+\beta=0$, then $\beta$ is equal to

Options:
A
$p+q$

B
$p-q$

C
$p^2+q^2$

D
$q^2-p^2$

The correct answer is:

$q^2-p^2$

Solution:
Since, $a+b=-p, a b=1$ .....(i)
and $c+d=-q, c d=1$
Now $(a-c)(b-c)$ and $(a+d)(b+d)$ are the roots of $x^2+a x+\beta=0$
$$
\begin{aligned}
& (a-c)(b-c)(a+d)(b+d)=\beta \\
& \Rightarrow\left(a b-a c-b c+c^2\right)\left(a b+a d+b d+d^2\right)=\beta \\
& \Rightarrow\left\{1-c(a+b)+c^2\right\}\left\{1+d(a+b)+d^2\right\}=\beta \\
& \Rightarrow \quad\left(1+p c+c^2\right)\left(1-p d+d^2\right)=\beta \\
& \Rightarrow 1-p d+d^2+p c-p^2 c d+p c d^2+c^2-p c^2 d \\
& +c^2 d^2=\beta \\
& \Rightarrow 1-p d+d^2+p c-p^2+p d+c^2-p c+1=\beta \\
& {[\because c d=1]} \\
& \Rightarrow \quad 2+d^2+c^2-p^2=\beta \\
&
\end{aligned}
$$
$\begin{aligned} \Rightarrow \quad 2 c d+c^2+d^2-p^2 & =\beta \\ (c+d)^2-p^2 & =\beta \\ q^2-p^2 & =\beta \quad[\because(c+d)=-q]\end{aligned}$

Topic: Mathematics, Quadratic Equation, JEE Main
Let $\mathrm{a}, \mathrm{b}$ be the solutions of $\mathrm{x}^2+\mathrm{px}+1=0$ and $\mathrm{c}$, $\mathrm{d}$ be the solution of $\mathrm{x}^2+\mathrm{qx}+1=0$. If $(\mathrm{a}-\mathrm{c})(\mathrm{b}-\mathrm{c})$ and $(\mathrm{a}+\mathrm{d})(\mathrm{b}+\mathrm{d})$ are the solution of $\mathrm{x}^2+\mathrm{ax}+\beta=0$, then $\beta$ is equal to

Options:
A $p+q$
B $p-q$
C $p^2+q^2$
D $q^2-p^2$

The correct answer is:
$q^2-p^2$

Solution:
Since, $a+b=-p, a b=1$ $\qquad$ ...(i)
and $c+d=-q, c d=1$
Now, $(\mathrm{a}-\mathrm{c})(\mathrm{b}-\mathrm{c})$ and $(\mathrm{a}+\mathrm{d})(\mathrm{b}+\mathrm{d})$ are the roots of $\mathrm{x}^2+\mathrm{ax}+\beta=0$
$(\mathrm{a}-\mathrm{c})(\mathrm{b}-\mathrm{c})(\mathrm{a}+\mathrm{d})(\mathrm{b}+\mathrm{d})=\beta$
$\Rightarrow\left(a b-a c-b c+c^2\right)\left(a b+a d+b d+d^2\right)=\beta$
$\Rightarrow\left\{1-c(a+b)+c^2\right\}\left\{1+d(a+b)+d^2\right\}=\beta$
$\Rightarrow\left(1+\mathrm{pc}+\mathrm{c}^2\right)\left(1-\mathrm{pd}+\mathrm{d}^2\right)=\beta$
$\Rightarrow 1-p d+d^2+p c-p^2+p d+c^2-p c+1=\beta[\because c d=1]$
$\Rightarrow 1-p d+d^2+p c-p^2+p d+c^2-p c+1=\beta[\because c d=1]$
$\Rightarrow 2+d^2+c^2-p^2=\beta$
$\Rightarrow 2 c d+c^2+d^2-p^2=\beta$
$(\mathrm{c}+\mathrm{d})^2-\mathrm{p}^2=\beta$
$\mathrm{q}^2-\mathrm{p}^2=\beta[\because(\mathrm{c}+\mathrm{d})=-\mathrm{q}]$

Topic: Mathematics, Quadratic Equation, JEE Main
Let a^,b^ be unit vectors. If c→ be a vector such that the angle between a^ and c→ is π12, and b^=c→+2c→×a^, then 6c→2 is equal to:

Options:
A 63-3
B 63+3
C 3+3
D 63+1

The correct answer is:
63+3

Solution:
We have, b^=c→+2c→×a^,

b^·b^=c→+2c→×a^·c→+2c→×a^

⇒1=c→·c→+c→·2c→×a^+2c→×a^·c→+4c→×a^·c→×a^

⇒1=c→2+0+0+c→×a^2 ⇒1=c→2+4c→2a^2sin2π12

⇒1=c→2+4c→21-cosπ62

⇒1=c→21+21-32 ⇒1=c→23-3 ⇒c→2=13-3=3+36

∴36c→2=3+36×36=63+3

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2022 (24 Jun Shift 1)
Let aâ+bâ+câ=0â, where aâ, bâ and câ are three non-zero vectors such that the angle between aâ and bâ is Ï3. If the angle between bâ and câ is Ï18, then the angle between aâ and câ is

Options:
A 5π18
B 11π18
C π2
D π6

The correct answer is:
5π18

Solution:
From the figure, it is clear that a→, b→, c→ forms a triangle
∴ Angle between a→ & c→=π-2π3-π18=5π18

Topic: Mathematics, Vector Algebra, JEE Main
Let a,b,c>1,a3,b3 and c3 be in A.P. and logab, logca and logbc be in G.P. If the sum of first 20 terms of an A.P., whose first term is a+4b+c3 and the common difference is a-8b+c10 is -444, then abc is equal to

Options:
A 343
B 216
C 3438
D 1258

The correct answer is:
216

Solution:
Given,

a3,b3,c3 be in A.P.

So, a3+c3=2b3 ........1

Also given logab, logca & logbc are in G.P.

So, logbloga·logclogb=logalogc2

⇒(loga)3=(logc)3⇒a=c ..........(2)

Now from equation 1 & 2 we get, a=b=c

Now, T1=a+4b+c3=2a and d=a-8b+c10=-6a10=-35a

∴S20=2024a+19-35a

⇒S20=1020a-57a5

⇒S20=-74a

⇒-444=-74a⇒a=6

Hence, abc=63=216

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2023 (30 Jan Shift 2)
Let A, B, C and D are 4 points in space satisfying the equation ABââCDâ=kADâ2+BCâ2-ACâ2-BDâ2, then the value of k is equal to

Options:
A 12
B 1
C 32
D 2

The correct answer is:
12

Solution:
Let position vectors of A,B,C,D are a→,b→,c→,d→ respectively
AB→⋅CD→=b→-a→⋅d→-c→
=b→⋅d→-b→⋅c→-a→⋅d→+a→⋅c→ … 1
AB→2+BC→2-AC→2-BD→2
=d→2+a→2-2a→⋅d→+c→2+b→2-2b→⋅c→-a→2-c→2+2a→⋅c→-b→2-d→2+2b→⋅d→
=2b→⋅d→-b→⋅c→-a→⋅d→+a→⋅c→ … 2
From 1 & 2,
k=0.5

Topic: Mathematics, Vector Algebra, JEE Main
Let A, B, C and D are any 4 points in space. If ABâÃCDâ+BCâÃADâ+CAâÃBDâ=Î» ÎABC, where ÎABC is the area of triangle ABC, then the value of Î» is equal to

Options:
A 2
B 12
C 4
D 14

The correct answer is:
4

Solution:
Let, the position vectors are Aaâ,Bbâ,Ccâ,Ddâ
So, ÎABC=12ABâÃACâ
=12bâ-aâÃcâ-aâ â¦ 1
Now, ABâÃCDâ=bâ-aâÃdâ-câ
=bâÃdâ-bâÃcâ-aâÃdâ+aâÃcâ
BCâÃADâ=câ-bâÃdâ-aâ
=câÃdâ-câÃaâ-bâÃdâ+bâÃaâ
CAâÃBDâ=aâ-câÃdâ-bâ=aâÃdâ-aâÃbâ-câÃdâ+câÃbâ
ABâÃCDâ+BCâÃADâ+CAâÃBDâ=2aâÃbâ+bâÃcâ+câÃaâ
=2bâ-aâÃcâ-aâ
=4ÎABC
âÎ»=4

Topic: Mathematics, Vector Algebra, JEE Main
Let a, b, c and d are in a geometric progression such that a<b<c<d, a+d=112 and b+c=48. If the geometric progression is continued with a as the first term, then the sum of the first six terms is

Options:
A 1156
B 1256
C 1356
D 1456

The correct answer is:
1456

Solution:
Let r be the common ratio
a+d=112⇒a+ar3=112
b+c=48⇒ar+ar2=48
Dividing the first equation by the second equation we get
a1+r3a1+rr=11248=1+r1-r+r2rr+1=73
⇒3r2-3r+3=7r
⇒3r2-10r+3=0
r=3 or 13
Given they are in ascending order ⇒r=3
r=3⇒a=1121+r3=11228=4
⇒a=4,b=12,c=36,d=108
⇒ Sum of first 6 terms =4(36-1)(3-1)=1456

Topic: Mathematics, Sequences and Series, JEE Main
Let a, b, c and d be non-zero numbers. If the point of intersection of the lines 4ax+2ay+c=0 & 5bx+2by+d=0 lies in the fourth quadrant and is equidistant from the two axes then

Options:
A 3bc-2ad=0
B 3bc+2ad=0
C 2bc-3ad=0
D 2bc+3ad=0

The correct answer is:
3bc-2ad=0

Solution:
Let (h,k) be the intersection of the two given lines its equidistant from both axes.

∴ h=k

Since it lies in the fourth quadrant, the point can be written as (h,-h), where h>0.

(h,-h) lies on both the lines.

∴ 4ah-2ah+c=0

∴ c=-2ah ∴ h=-c2a

5bh-2bh+d=0

∴ d=-3bh ∴ h=-d3b

∴ -c2a=-d3b

∴ 3bc-2ad=0

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2014 (06 Apr)
Let \(A, B, C\) are subsets of set \(X\). Then consider the validity of the following set theoretic statement :

Options:
A \(A \cup(B \backslash C)=(A \cup B) \backslash(A \cup C)\)
B \((A \backslash B) \backslash C=A \backslash(B \cup C)\)
C \((A \cup B) \backslash A=A \backslash B\)
D \(A \backslash C=B \backslash C\)

The correct answer is:
\((A \backslash B) \backslash C=A \backslash(B \cup C)\)

Solution:
Hint:
\(\begin{aligned}
:(A \backslash B) \backslash C & =\left(A \cap B^{\prime}\right) \cap C^{\prime} \\
& =A \cap\left(B^{\prime} \cap C^{\prime}\right) \\
& =A \cap(B \cup C)^{\prime} \\
& =A \backslash(B \cup C)
\end{aligned}\)

Topic: Mathematics, Sets and Relations, JEE Main
Let ABC be a triangle such that BC→=a→,CA→=b→, AB→=c→,a→=62,b→=23 and b→·c→=12 Consider the statements :

S1: a→×b→+c→×b→-c→=622-1

S2: ∠ABC=cos-123. Then

Options:
A both S1 and S2are true
B only S1 is true
C only S2 is true
D both S1 and S2 are false

The correct answer is:
only S2 is true

Solution:
We know that by triangle law of vector addition,

a→+b→+c→=0

⇒b→+c→=-a→

Squaring both side we get,

b→2+c→2+2b→·c→=a→2

Now putting the given value a→=62, b→=23 & b→.c→=12

We get, c→2=36

So, c→=6

Now, S1: a→×b→+c→×b→-c→

⇒a→+c→×b→-c→

⇒0-6=-6

And S2: a→+b→+c→=0

a→+b→=-c→

Squaring both side we get,

a→2+b→2-2a→b→cos∠ACB=c→2

⇒72+12-246cos∠ACB=36

⇒cos∠ACB=48246=26=23

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2022 (25 Jul Shift 1)
Let ABC be a triangle whose circumcentre is at P. If the position vectors A, B, C and P are a→,b→,c→ and a→+b→+c→4 respectively, then the position vector of the orthocentre of this triangle, is :

Options:
A -a→ + b→ + c→2
B aâÂ +Â bâÂ +Â câ
C aâ + bâ + câ2
D 0â

The correct answer is:
aâ + bâ + câ2

Solution:
O is orthocentre and G is centroid and C is circumcentre. G divides OC in 2: 1 ratio.
Position vector of centriod G→=a→+b→+c→3
Position vector of circum center C→=a→+b→+c→4
Apply Section Formula,
G→=2C→+R→3

3G→=2C→+R→

R→=3G→-2C→=(a→+b→+c→)-2a→+b→+c→4

=a→+b→+c→2

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2016 (10 Apr Online)
Let A B C be an equilateral triangle with side length a. Let R and r denote the radii of the circumcircle and the incircle of triangle A B C respectively. Then, as a function of α the ratio Rr

Options:
A strictly increases
B strictly decreases

C remains constant

D strictly increases for a<1 and strictly decrease for a>1

The correct answer is:
remains constant

Solution:
For an equilateral triangle A B C having side length a. If R and r are radii of the circumcircle and the incircle of triangle A B C respectively, then
R=a2sec30°=a223=a3

and r=a2tan30°=a2×13=a23
∴Rr=a3a23=2 which is independent of a
and it is constant.

Topic: Mathematics, Properties of Triangles, KVPY, KVPY 2019 (SA)
Let $a, b, c$ be any real numbers. Suppose that there are real numbers $x, y, z$ not all zero such that $x=$ $c y+b z, y=a z+c x$ and $z=b x+a y$. Then $a^2+b^2+c^2+2 a b c$ is equal to

Options:
A
2

B
$-1$

C
0

D
1

The correct answer is:
1

Solution:
The system of equations $x-c y-b z=0, c x-y+a z=0$ and $b x+a y-z=0$ have non-trivial solution if $\left|\begin{array}{ccc}1 & -c & -b \\ c & -1 & a \\ b & a & -1\end{array}\right|=0 \Rightarrow 1\left(1-a^2\right)+c(-c-a b)-b(c a+b)=0$ $\Rightarrow a^2+b^2+c^2+2 a b c=1$.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2008
Let a,b,c be in arithmetic progression. Let the centroid of the triangle with vertices a,c,2,b and a,b be 103,73. If α,β are the roots of the equation ax2+bx+1=0, then the value of α2+β2-αβ is:

Options:
A -71256
B 69256
C 71256
D -69256

The correct answer is:
-71256

Solution:
a+2+a3=103

a=4

and c+b+b3=73

c+2b=7

also 2b=a+c

2b-a+2b=7

b=114

now 4x2+114x+1=0

α2+β2-αβ=α+β2-3αβ

=-11162-314

=121256-34=-71256

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2021 (24 Feb Shift 2)
Let $a, b, c$ be such that $b(a+c) \times 0$
If $\left|\begin{array}{ccc}a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1\end{array}\right|$
$+\left|\begin{array}{ccc}a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ (-1)^{n+2} a & (-1)^{n+1} b & (-1)^{n} c\end{array}\right|=0$
then the value of $n$ is

Options:
A any integer
B zero
C any even integer
D any odd integer

The correct answer is:
any odd integer

Solution:
We have,
$\left|\begin{array}{ccc}a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1\end{array}\right|$ + $\left|\begin{array}{ccc}a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ (-1)^{n+2} a & (-1)^{n+1} b & (-1)^{n} c\end{array}\right|=0$
$\Rightarrow\left|\begin{array}{ccc}a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1\end{array}\right|$ $+\left|\begin{array}{ccc}a+1 & a-1 & (-1)^{n+2} a \\ b+1 & b-1 & (-1)^{n+1} b \\ c-1 & c+1 & (-1)^{n} c\end{array}\right|=0$
$\Rightarrow\left|\begin{array}{ccc}a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1\end{array}\right|$$+\left|\begin{array}{ccc}(-1)^{n+2} a & a+1 & a-1 \\ (-1)^{n+1} b & b+1 & b-1 \\ (-1)^{n} c & c-1 & c+1\end{array}\right|=0$
$\Rightarrow\left|\begin{array}{ccc}a\left(1+(-1)^{n+2}\right) & a+1 & a-1 \\ b\left(-1+(-1)^{n+1}\right) & b+1 & b-1 \\ c\left(1+(-1)^{n}\right) & c-1 & c+1\end{array}\right|=0$
$\therefore n$ is any odd integer.

Topic: Mathematics, Determinants, JEE Main
Let $a, b, c$ be such that $b(a+c) \neq 0$. If $\left|\begin{array}{ccc}a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1\end{array}\right|+\left|\begin{array}{ccc}a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ (-1)^{n+2} a & (-1)^{n+1} b & (-1)^n c\end{array}\right|=0$, then the value of ' $n$ ' is

Options:
A
zero

B
any even integer

C
any odd integer

D
any integer

The correct answer is:
any odd integer

Solution:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
a & a+1 & a-1 \\
-b & b+1 & b-1 \\
c & c-1 & c+1
\end{array}\right|+(-1)^n\left|\begin{array}{ccc}
a+1 & b+1 & c-1 \\
a-1 & b-1 & c+1 \\
a & -b & c
\end{array}\right|=\left|\begin{array}{ccc}
a & a+1 & a-1 \\
-b & b+1 & b-1 \\
c & c-1 & c+1
\end{array}\right|+(-1)^n\left|\begin{array}{cccc}
a+1 & a-1 & a \\
b+1 & b-1 & -b \\
c-1 & c+1 & c
\end{array}\right| \\
& =\left|\begin{array}{ccc}
a & a+1 & a-1 \\
-b & b+1 & b-1 \\
c & c-1 & c+1
\end{array}\right|+(-1)^{n+1}\left|\begin{array}{ccc}
a+1 & a & a-1 \\
b+1 & -b & b-1 \\
c-1 & c & c+1
\end{array}\right|=\left|\begin{array}{ccc}
a & a+1 & a-1 \\
-b & b+1 & b-1 \\
c & c-1 & c+1
\end{array}\right|+(-1)^{n+2}\left|\begin{array}{ccc}
a & a+1 & a-1 \\
-b & b+1 & b-1 \\
c & c-1 & c+1
\end{array}\right|
\end{aligned}
$$
This is equal to zero only if $n+2$ is odd i.e. $n$ is odd integer.

Topic: Mathematics, Determinants, JEE Main
Let $a, b, c$ be such that $b(a+c) \neq 0$. If $\left|\begin{array}{ccc}a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1\end{array}\right|+\left|\begin{array}{ccc}a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ (-1)^{n+2} a & (-1)^{n+1} b & (-1)^n c\end{array}\right|=0$, then the value of ' $n$ ' is

Options:
A
zero

B
any even integer

C
any odd integer

D
any integer

The correct answer is:
any odd integer

Solution:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
a & a+1 & a-1 \\
-b & b+1 & b-1 \\
c & c-1 & c+1
\end{array}\right|+(-1)^n\left|\begin{array}{ccc}
a+1 & b+1 & c-1 \\
a-1 & b-1 & c+1 \\
a & -b & c
\end{array}\right|=\left|\begin{array}{ccc}
a & a+1 & a-1 \\
-b & b+1 & b-1 \\
c & c-1 & c+1
\end{array}\right|+(-1)^n\left|\begin{array}{cccc}
a+1 & a-1 & a \\
b+1 & b-1 & -b \\
c-1 & c+1 & c
\end{array}\right| \\
& =\left|\begin{array}{ccc}
a & a+1 & a-1 \\
-b & b+1 & b-1 \\
c & c-1 & c+1
\end{array}\right|+(-1)^{n+1}\left|\begin{array}{ccc}
a+1 & a & a-1 \\
b+1 & -b & b-1 \\
c-1 & c & c+1
\end{array}\right|=\left|\begin{array}{ccc}
a & a+1 & a-1 \\
-b & b+1 & b-1 \\
c & c-1 & c+1
\end{array}\right|+(-1)^{n+2}\left|\begin{array}{ccc}
a & a+1 & a-1 \\
-b & b+1 & b-1 \\
c & c-1 & c+1
\end{array}\right|
\end{aligned}
$$
This is equal to zero only if $n+2$ is odd i.e. $n$ is odd integer.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2009
Let a, b, c be the three distinct positive real numbers such that 2alogea=bclogeb and bloge2=alogec Then 6a+5bc is equal to ______.

Options:

The correct answer is:
8

Solution:
Given equation: 2alna=bclnb

⇒lna·ln2a=lnb·lnbc

⇒lna·ln2+lna=lnb·lnb+lnc

let lna=x, lnb=y, lnc=z, x≠y≠z

⇒xln2+x=yy+z

⇒xln2=y2-x2+yz ........(i)

Similarly, from second equation

ln2·lnb=lnc·lna

⇒ln2=xzy ......(ii)

Substitute eqii in eqi,

We get,

⇒y3-yx2+y2z=x2z

⇒y2y+z-x2y+z=0

⇒y2-x2y+z=0

⇒x-yx+yy+z=0

∵ x≠y⇒x+yy+z=0

Now x=-y, y=-z and x=z

But ln2=xzy

⇒ln2=xx-x

⇒ln2=-lna

=-ln 2⇒a=12 and

⇒y=-z

⇒lnb=-lnc

⇒bc=1

Now 6a+5bc=612+51=8

Hence this is the required answer.

Topic: Mathematics, Basic of Mathematics, JEE Main, JEE Main 2023 (10 Apr Shift 1)
Let a, b, c be three distinct real numbers, none equal to one. If the vectors ai^+j^+k^, i^+bj^+k^ and i^+j^+ck^ are coplanar, then 11-a+11-b+11-c is equal to

Options:
A 2
B -1
C -2
D 1

The correct answer is:
1

Solution:
Given that the vectors are coplanar.

That means the determinant is equal to zero.

⇒a111b111c=0

⇒R1→R1–R2 and R2→R2–R3

⇒a-11-b00b-11-c11c=0

Expand the determinant along C1.

⇒(a-1) [c(b-1)-(1-c)]+1[(1-b) (1-c)]=0

⇒c(a-1) (b-1)-(a-1) (1-c)+(1-b) (1-c)=0

⇒c(1-a) (1-b)+(1-a) (1-c)+(1-b) (1-c)+1-a1-b1-c=1-a1-b1-c

⇒(1-a) (1-b)+(1-a) (1-c)+(1-b) (1-c)=1-a1-b1-c

⇒11-a+11-b+11-c=1

Hence this is the required option.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (12 Apr Shift 1)
Let A,B,C be three events and A-,B-,C- be their corresponding complementary event. If the probabilities of events B, A∩B∩C- and A-∩B∩C- are 56,12 and 14 respectively, then the probability of the event B∩C is

Options:
A 112
B 14
C 16
D 13

The correct answer is:
112

Solution:
From the venn diagram, note that PB=PB∩C+PA∩B∩C-+PA-∩B∩C-
Hence, PB∩C=56-12-14=112

Topic: Mathematics, Probability, JEE Main
Let a→,b→,c→ be three mutually perpendicular vectors of the same magnitude and equally inclined at an angle θ, with the vector a→+b→+c→. Then 36cos22θ is equal to

Options:

The correct answer is:
4

Solution:
We have,

a→=b→=c→

a→·b→=b→·c→=c→·a→=0

Since, vectors are mutually perpendicular and of equally magnitude, hence let

a→=i^, b→=j^, c→=k^

Now,

a→+b→+c→·a→=i^+j^+k^·i^

⇒i^+j^+k^·i^=3·1·cosθ

⇒1+0+0=3·1·cosθ

⇒3cosθ=1

⇒cosθ=13

⇒cos2θ=2cos2θ-1=23-1

⇒cos2θ=-13

⇒cos22θ=19

⇒36cos22θ=4

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2021 (20 Jul Shift 1)
Let a→,b→,c→ be three non-zero, non-coplanar vectors and p→,q→,r→ be three vectors given by p→=a→+2b→-2c→, q→=3a→+b→-3c→ and r→=a→-4b→+4c→. If the volume of parallelepiped determined by a→, b→ and c→ is v1 cubic units and volume of tetrahedron determined by p→, q→ and r→ is v2 cubic units, then v1v2 is equal to

Options:
A 13
B 34
C 14
D 12

The correct answer is:
12

Solution:
v1=a→b→c→
v2=16p→q→r→=1612-231-31-44a→b→c→
v2=1614-12-212+3-2-12-1a→b→c→
v2=16-8-30+26a→b→c→
v2=2a→b→c→
v1v2=12=0.5

Topic: Mathematics, Vector Algebra, JEE Main
Let a→, b→, c→ be three vectors mutually perpendicular to each other and have same magnitude. If a vector r→ satisfies a→×{r→-b→×a→}+b→×{r→-c→×b→}+c→×{r→-a→×c→}=0→, then r→ is equal to:

Options:
A 13(a→+b→+c→)
B 13(2a→+b→-c→)
C 12(a→+b→+c→)
D 12(a→+b→+2c→)

The correct answer is:
12(a→+b→+c→)

Solution:
Given:a→×{r→-b→×a→}+b→×{r→-c→×b→}+c→×{r→-a→×c→}=0→

⇒(a→.a→)(r→-b→)-(a→·(r→-b→))a→+(b→·b→)(r→-c→)-(b→·(r→-c→))b→+(c→.c→)(r→-a→)-(c→.(r→-a→))c→=0→

⇒|a→|2(r→-b→)-(r→·a→)a→+|b→|2(r→-c→)-(r→·b→)b→+|c→|2(r→-a→)-(r→·c→)c→=0→

⇒|a→|2[3r→-(a→+b→+c→)]-[(r→·a→)a→)+(r→·b→)b→+(r→·c→)c→]=0→ ∵|a→|2=|b→|2=|c→|2

⇒|a→|2[3r→-(a→+b→+c→)-(r→)]=0→∵(r→·a→)a→+(r→·b→)b→+(r→·c→)c→=a→2r→ , |a→|2=|b→|2=|c→|2

⇒ 3r→-(a→+b→+c→)-r→=0→

∴r→=a→+b→+c→2

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2021 (31 Aug Shift 2)
Let a→, b→, c→ be three vectors of magnitude 3, 4, 5 respectively, satisfying a→b→c→=60. If a→+2b→+3c→·a→×c→×b→+b→=λ, then λ is equal to

Options:
A 16
B 32
C 20
D 40

The correct answer is:
32

Solution:
Given, aâbâcâ=3Â·4Â·5=60
Also, aâbâcâ=60âaâbâcâ=aâbâcâ
âaâ,bâ,câ are mutually perpendicular vectors
Also, aâÃcâÃbâ+bâ=aâÂ·bâcâ-câÂ·bâaâ+bâ
=bâ
Hence, aâ+2bâ+3câÂ·bâ=2bâ2=32

Topic: Mathematics, Vector Algebra, JEE Main
Let a→,b→,c→ be three vectors such that a→=31,4b→=c→=2 and 2a→×b→=3c→×a→. If the angle between b→ and c→ is 2π3, then a→×c→a→·b→2 is equal to _____ .

Options:

The correct answer is:
3

Solution:
Given,

a→=31, 4b→=c→=2, 2a→×b→=3c→×a→ and angle between b→ & c→ is given as 2π3

Now solving, 3c→×a→+2b→×a→=0

⇒3c→×2b→×a→=0

Means 3c→×2b→ & a→ are parallel vector,

So, let 3c→×2b→=λa→

Now squaring both sides we get,
9c→2+4b→2+12b→·c→=λ2a→2

⇒36+1+12×12×2cos2π3=λ231

⇒λ2=1

⇒λ=±1

Now putting the value of λ in 3c→×2b→=λa→ we get,

3c→+2b→=±a→ …1

Now taking dot product with b→ in above equation we get,

3b→·c→+2b→·b→=±a→·b→

⇒a→·b→=±-32+12=±-1

⇒a→·b→2=1

Again taking 3c→×a→=2a→×b→ and sqauring both side,

⇒c→×a→2=49a→×b→2

⇒c→×a→2=49a→2b→2-a→·b→2

⇒c→×a→2=49314-1

⇒c→×a→2=49×274=3

Hence, the value of a→×c→a→·b→2=31=3.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (31 Jan Shift 2)
Let \( A B C D \) be a parallelogram such that \( \overrightarrow{A B}=\vec{q}, \overrightarrow{A D}=\vec{p} \) and \( \angle B A D \) be an acute angle. If \( \vec{r} \) is the vector that coincides with the altitude directed from the vertex \( B \) to the side \( A D \), then \( \vec{r} \) is given by

Options:
A \( \vec{r}=\vec{q}-\left(\frac{\vec{p} \cdot \vec{q}}{\vec{p} \cdot \vec{p}}\right) \vec{p} \)
B \( \vec{r}=3 \vec{q}+\frac{3(\vec{p} \cdot \vec{q})}{(\vec{p} \cdot \vec{p})} \vec{p} \)
C \( \vec{r}=3 \vec{q}-\frac{3(\vec{p} \cdot \vec{q})}{(\vec{p} \cdot \vec{p})} \vec{p} \)
D \( \vec{r}=-\vec{q}+\left(\frac{\vec{p} \cdot \vec{q}}{\vec{p} \cdot \vec{p}}\right) \vec{p} \)

The correct answer is:
\( \vec{r}=-\vec{q}+\left(\frac{\vec{p} \cdot \vec{q}}{\vec{p} \cdot \vec{p}}\right) \vec{p} \)

Solution:
By Triangle law of Vector addition

r→=BA→+AQ→
r→=-q→ + λp→

∵ r→ is Perpendicular to the vector p→

⇒ r→·p→=0 ⇒ -q→+λp→·p→=-q·→p→+λp·→p→=0

⇒ λ=q·→p→p·→p→

∴ r→=-q→+q·→p→p→·p→p→

Topic: Mathematics, Vector Algebra, JEE Main
Let ABCD be a quadrilateral. If E and F are the mid points of the diagonals AC and BD respectively and AB→- BC→+AD→-DC→=k FE→, then k is equal to

Options:
A 4
B -2
C 2
D -4

The correct answer is:
-4

Solution:
Given,

ABCD be a quadrilateral. If E and F are the mid points of the diagonals AC and BD respectively and AB→- BC→+AD→-DC→=k FE→,

Now plotting the diagram we get,

Now let position vector of A, B, C and D are a,→ b,→ c→ and d→ respectively

Now by midpoint formula,

Position vector of E=OC→+OA→2=c→+a→2

And Position vector of F=b→+d→2

Now, AB→-BC→+AD→-DC→

⇒b→-a→-c→-b→+d→-a→-c→-d→

⇒2b→-2a→-2c→+2d→

⇒2b→+d→-2a→+c→

⇒4b→+d→2-a→+c→2=4OF→-OE→

⇒4EF→=-4FE→

∴ k=-4

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (15 Apr Shift 1)
Let ABCD be a quadrilateral in which AB is parallel to CD, AB is perpendicular to AD and AB=3CD. The area of the quadrilateral ABCD is 4 sq. units. The diameter (in units) of a circle touching all the sides of the quadrilateral is

Options:
A 1
B 5
C 2
D 3

The correct answer is:
3

Solution:
From the diagram,

Area of the trapezium ABCD=12a+3a2r=4 sq. units
âar=1
Equation of BC is y=-r2x-3r
y+r2x-3r=0
âµ BC is a tangent to the circle
âr+r3-3r1+r4=r
âr4+4-4r2=1+r4
âr=32 units

Topic: Mathematics, Circle, JEE Main
Let $a, b, c, d$ be numbers in the set $\{1,2,3,4,5,6\}$ such that the curves $y=2 x^{3}+a x+b$ and $y=2 x^{3}+c x+d$ have no point in common. The maximum possible value of $(a-c)^{2}+b-d$ is $-$

Options:
A 0
B 5
C 30
D 36

The correct answer is:
5

Solution:
$y=2 x^{3}+a x+b$
$y=2 x^{3}+c x+d$
No Solution
$\begin{array}{l}
2 x^{3}+a x+b \neq 2 x^{3}+c x+d \\
a x+b \neq c x+d \quad \text { for no real } x \\
(a-c) x \neq d-b \\
x \neq \frac{d-b}{a-c} \quad a=c \\
(a-c)^{2}+(b-d)=0+6-1=5
\end{array}$

Topic: Mathematics, Quadratic Equation, JEE Main