datavorous/entrance-exam-dataset huggingface dataset:
If A=2294 and A2+aA+bÎ=O, then a+2b is equal to (where, Î is an identity matrix and O is a null matrix of order 2 respectively)

Options:
A 27
B -26
C 24
D 12

The correct answer is:
-26

Solution:
A=2294⇒A2=22942294=22125434
aA=2a2a9a4a, bΙ=b00b
A2+aA+bΙ=22+2a+b12+2a9a+5434+4a+b=O
⇒a=-6
⇒22+2a+b=0⇒b=-10
⇒a+2b=-26

Topic: Mathematics, Matrices, JEE Main
If A=2294 and I=1001, then 10 A-1, is equal to.

Options:
A A-4I
B 6I-A
C A-6I
D 4I-A

The correct answer is:
A-6I

Solution:
Characteristics equation of matrix A is A-xI=0.

2294-x1001=0
⇒2-x924-x=0 ⇒x2-6x-10=0

∴A2-6A-10I=0
10A-1=A-6I.

Topic: Mathematics, Matrices, JEE Main, JEE Main 2020 (08 Jan Shift 2)
If A=230-1, then the value of detA4+detA10-(Adj(2 A))10 is equal to ________.

Options:

The correct answer is:
16

Solution:
Given, A=230-1

⇒A2=230-1230-1=4301

⇒A4=43014301=161501

⇒A5=161501230-1=32330-1

⇒A10=32330-132330-1=1024102301 ....i

Now, 2A=460-2

⇒Adj2A=-20-64T

⇒Adj2A=-2-604

⇒Adj2A2=-2-604-2-604=4-12016

⇒Adj2A4=4-120164-12016=16-2400256

⇒Adj2A5=16-2400256-2-604=-32-105601024

⇒Adj2A10=-32-105601024-32-105601024=1024-104755201048576 .....ii

From equation i & ii

⇒A10-Adj2A10=1024102301-1024-104755201048576=010485750-1048575

⇒detA10-Adj2A10=010485750-1048575=0

Let E=detA4+detA10-(Adj(2 A))10

⇒E=detA4+0

⇒E=-2-04

⇒E=16

Topic: Mathematics, Matrices, JEE Main
If A=230-1, then the value of detA4+detA10-(Adj(2 A))10 is equal to ________.

Options:

The correct answer is:
16

Solution:
Given, A=230-1

⇒A2=230-1230-1=4301

⇒A4=43014301=161501

⇒A5=161501230-1=32330-1

⇒A10=32330-132330-1=1024102301 ....i

Now, 2A=460-2

⇒Adj2A=-20-64T

⇒Adj2A=-2-604

⇒Adj2A2=-2-604-2-604=4-12016

⇒Adj2A4=4-120164-12016=16-2400256

⇒Adj2A5=16-2400256-2-604=-32-105601024

⇒Adj2A10=-32-105601024-32-105601024=1024-104755201048576 .....ii

From equation i & ii

⇒A10-Adj2A10=1024102301-1024-104755201048576=010485750-1048575

⇒detA10-Adj2A10=010485750-1048575=0

Let E=detA4+detA10-(Adj(2 A))10

⇒E=detA4+0

⇒E=-2-04

⇒E=16

Topic: Mathematics, Matrices, JEE Main, JEE Main 2021 (17 Mar Shift 1)
If A=23-1-2 and B=∑r=110Ar, then the value of det B is equal to

Options:

The correct answer is:
0

Solution:
Note A2=ÎâAn=Î (if n is even) & An=A (if n is odd)
So, B=5A+Î=533-1-1âB=0

Topic: Mathematics, Determinants, JEE Main
If A=2-3-41 , then Adj3A2+12A is equal to:

Options:
A 72-84-6351
B 51638472
C 51846372
D 72-63-8451

The correct answer is:
51638472

Solution:
Given, A=2-3-41
⇒ A2=2-3-41 2-3-41
⇒ A2=4+12-6-3-8-412+1

⇒ A2=16-9-1213
⇒ 3A2=48-27-3639 and 12A=24-36-4812
∴ 3A2+12A=72-63-8451
Adj3A2+12A=51846372T=51638472

Topic: Mathematics, Matrices, JEE Main, JEE Main 2017 (02 Apr)
If $\mathrm{A}(2,3), \mathrm{B}(1,4), \mathrm{C}(0-2)$ and $\mathrm{D}(\mathrm{x}, \mathrm{y})$ are the vertices of a
parallelogram, then what is the value of $(\mathrm{x}, \mathrm{y})$ ?

Options:
A $(1,-3)$
B $(2,4)$
C (1,1)
D (0,0)

The correct answer is:
$(1,-3)$

Solution:
As given : $\mathrm{A}(2,3), \mathrm{B}(1,4), \mathrm{C}(0,-2)$ and $\mathrm{D}(\mathrm{x}, \mathrm{y})$ are the
vertices of a parallelogram. Diagonals of a parallelogram bisect each other So, mid-point are same for both diagonals $\mathrm{AC}$ and $\mathrm{BD}$.
$\frac{2+0}{2}=\frac{1+x}{2}$ and $\frac{3-2}{2}=\frac{4+y}{2}$
$\Rightarrow \quad x=1$ and $y=-3$
$\Rightarrow \quad \mathrm{D}(\mathrm{x}, \mathrm{y})=(1,-3)$

Topic: Mathematics, Straight Lines, JEE Main
If A2+3i and B3+4i are two vertices of a square ABCD (taken in anticlockwise order) in a complex plane, then the value of Z32-Z42 (where C is Z3 and D is Z4 ) is equal to

Options:
A 0
B 6
C 8
D 12

The correct answer is:
12

Solution:
Z4-2+3i3+4i-2+3i=eiπ/2=i
⇒Z4=2+3i+i-1=1+4i
and Z3-3+4i2+3i-3+4i=e-iπ/2=-i
⇒Z3=3+4i+i-1=2+5i
⇒Z32-Z42=29-17=12

Topic: Mathematics, Complex Number, JEE Main
If $A=\left[\begin{array}{ccc}2 & 4 & 5 \\ 4 & 8 & 10 \\ -6 & -12 & -15\end{array}\right]$. Then rank of $A$ is equal to

Options:
A $0$
B $1$
C $2$
D $3$

The correct answer is:
$1$

Solution:
$A=\left[\begin{array}{ccc}2 & 4 & 5 \\ 4 & 8 & 10 \\ -6 & -12 & -15\end{array}\right]_{3 \times 3}$
$|A|=0$, then rank cannot be 3 .
Considering a $2 \times 2$ minor, $\left[\begin{array}{ll}2 & 4 \\ 4 & 8\end{array}\right]$ its determinant is zero. Similarly considering
$\left[\begin{array}{cc}4 & 5 \\ 8 & 10\end{array}\right]$ $\left[\begin{array}{cc}4 & 8 \\ -6 & -12\end{array}\right]$ $\left[\begin{array}{cc}8 & 10 \\ -12 & 15\end{array}\right]$ $\left[\begin{array}{cc}2 & 5 \\ 4 & 10\end{array}\right]$ $\left[\begin{array}{cc}4 & 10 \\ -6 & -15\end{array}\right]$ their determinants is zero. Each rank can not be 2 . Thus rank $=1$.

Topic: Mathematics, Matrices, JEE Main
If $A=\{2,4\}, \quad B=\{3,4,5\}$, then $(\mathrm{A} \cap \mathrm{B}) \times(\mathrm{AUB})=$

Options:
A $\{(3,2),(3,4),(4,4),(5,4)\}$
B $\{(2,3),(2,4),(2,5)\}$
C $\{(4,2),(4,3),(4,4),(4,5)\}$
D $\{(4,3),(4,4),(4,5)\}$

The correct answer is:
$\{(4,2),(4,3),(4,4),(4,5)\}$

Solution:
$\mathrm{A} \cap \mathrm{B}=\{4\}$
$\mathrm{A} \cup \mathrm{B}=\{2,3,4,5\}$
Hence
$(\mathrm{A} \cap \mathrm{B}) \times(\mathrm{A} \cup \mathrm{B})$
$=\{(4,2),(4,3),(4,4),(4,5)\}$

Topic: Mathematics, Sets and Relations, JEE Main
If $A=\left[\begin{array}{cc}2 & -3 \\ 5 & -7\end{array}\right]$, then $A+A^{-1}=$

Options:
A $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
B $\left[\begin{array}{cc}-5 & 0 \\ 0 & -5\end{array}\right]$
C $\left[\begin{array}{ll}5 & 0 \\ 0 & 5\end{array}\right]$
D $\left[\begin{array}{cc}4 & 0 \\ 0 & -5\end{array}\right]$

The correct answer is:
$\left[\begin{array}{cc}-5 & 0 \\ 0 & -5\end{array}\right]$

Solution:
$\quad A=\left[\begin{array}{cc}2 & -3 \\ 5 & -7\end{array}\right]$
$\therefore \quad|A|=-14+15=1 \neq 0$
So, $A^{-1}$ exists.
$\begin{aligned}
&\therefore \quad \operatorname{adj} A=\left[\begin{array}{ll}
-7 & 3 \\
-5 & 2
\end{array}\right] \\
&\Rightarrow \quad A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)=\left[\begin{array}{cc}
-7 & 3 \\
-5 & 2
\end{array}\right]
\end{aligned}$
$\begin{aligned}
\Rightarrow A+A^{-1} &=\left[\begin{array}{cc}
2 & -3 \\
5 & -7
\end{array}\right]+\left[\begin{array}{ll}
-7 & 3 \\
-5 & 2
\end{array}\right] \\
&=\left[\begin{array}{cc}
-5 & 0 \\
0 & -5
\end{array}\right]
\end{aligned}$

Topic: Mathematics, Matrices, JEE Main
If $$
\begin{aligned}
&\left.\mid \begin{array}{ccc}
a^2 & b^2 & c^2 \\
(a+\lambda)^2 & (b+\lambda)^2 & \left(c+\lambda^2\right) \\
(a-\lambda)^2 & \left(b-\lambda^2\right) & \left(-\lambda^2\right.
\end{array}\right) \\
&=k \lambda\left|\begin{array}{ccc}
a^2 & b^2 & c^2 \\
a & b & c \\
1 & 1 & 1
\end{array}\right|, \lambda \neq 0
\end{aligned}
$$
then $\mathrm{k}$ is equal to:

Options:
A
$4 \lambda \mathrm{abc}$

B
$-4 \lambda \mathrm{abc}$

C
$4 \lambda^2$

D
$-4 \lambda^2$

The correct answer is:
$4 \lambda^2$

Solution:
Let $\Delta=\left|\begin{array}{ccc}a^2 & b^2 & c^2 \\ (a+\lambda)^2 & (b+\lambda)^2 & (c+\lambda)^2 \\ (a-\lambda)^2 & (b-\lambda)^2 & (c-\lambda)^2\end{array}\right|$
$$
\begin{aligned}
&\text { Apply } \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_3 \\
&\Delta=\left|\begin{array}{ccc}
a^2 & b^2 & c^2 \\
(a+\lambda)^2-(a-\lambda)^2 & (b+\lambda)^2-(b-\lambda)^2 & (c+\lambda)^2-(c-\lambda)^2 \\
(a-\lambda)^2 & (b-\lambda)^2 & (c-\lambda)^2
\end{array}\right| \\
&=\left|\begin{array}{ccc}
a^2 & b^2 & c^2 \\
4 a \lambda & 4 b \lambda & 4 c \lambda \\
(a-\lambda)^2 & (b-\lambda)^2 & (c-\lambda)^2
\end{array}\right| \\
&\left(\because(x+y)^2-(x-y)^2=4 x y\right) \\
&
\end{aligned}
$$
Taking out 4 common from $\mathrm{R}_2$
$=4\left|\begin{array}{ccc}a^2 & b^2 & c^2 \\ a \lambda & b \lambda & c \lambda \\ a^2+\lambda^2-2 a \lambda & b^2+\lambda^2-2 b \lambda & c^2+\lambda^2-2 c \lambda\end{array}\right|$
$$
\begin{aligned}
&\text { Apply } \mathrm{R}_3 \rightarrow\left[\mathrm{R}_3-\left(\mathrm{R}_1-2 \mathrm{R}_2\right)\right] \\
&=4\left|\begin{array}{lll}
a^2 & b^2 & c^2 \\
a \lambda & b \lambda & c \lambda \\
\lambda^2 & \lambda^2 & \lambda^2
\end{array}\right|
\end{aligned}
$$
Taking out $\lambda$ common from $\mathrm{R}_2$ and $\lambda^2$ from $\mathrm{R}_3$.
$$
=4 \lambda\left(\lambda^2\right)\left|\begin{array}{ccc}
a^2 & b^2 & c^2 \\
a & b & c \\
1 & 1 & 1
\end{array}\right|
$$
$$
\begin{aligned}
&=k \lambda\left|\begin{array}{ccc}
a^2 & b^2 & c^2 \\
a & b & c \\
1 & 1 & 1
\end{array}\right| \\
&\Rightarrow k=4 \lambda^2
\end{aligned}
$$

Topic: Mathematics, Determinants, JEE Main, JEE Main 2014 (12 Apr Online)
If $A^2-A+I=0$, then $A^{-1}=$

Options:
A $A^{-2}$
B $A+1$
C $I-A$
D $A-I$

The correct answer is:
$I-A$

Solution:
$A^2-A+I=0$
$\Rightarrow \quad I=A-A^2 \Rightarrow I=A(I-A)$
$\Rightarrow A^{-1} I=A^{-1}(A(I-A)) \Rightarrow A^{-1}=I-A$.

Topic: Mathematics, Matrices, JEE Main
If $A^2-A+I=0$, then the inverse of $A$ is

Options:
A
$A+I$

B
$A$

C
$A-I$

D
$I-A$

The correct answer is:
$I-A$

Solution:
Given $A^2-A+I=0$
$A^{-1} A^2-A^{-1} A+A^{-1}-I=A^{-1} \cdot 0$ (Multiplying $A^{-1}$ on both sides)
$\Rightarrow A-I+A^{-1}=0$ or $A^{-1}=I-A$

Topic: Mathematics, Matrices, JEE Main, JEE Main 2005
If A2=A, then I+A4 is equal to

Options:
A I+15A
B 1+7A
C 1+8A
D 1+11A

The correct answer is:
I+15A

Solution:
As I and A commute, we can apply binomial theorem to expand 1+A4, we have,
I+A4=I+4A+6A2+4A3+A4
But, A3=A2=A, A4=A3=A (âµââ A 2 =A)
â´ I+A4=I+4A+6A+4A+A=I+15A

Topic: Mathematics, Matrices, JEE Main
If $a^{2}+b^{2}+c^{2}=0$, then what is $\frac{\left(a^{4}-b^{4}\right)^{3}+\left(b^{4}-c^{4}\right)^{3}+\left(c^{4}-a^{4}\right)^{3}}{\left(a^{2}-b^{2}\right)^{3}+\left(b^{2}-c^{2}\right)^{3}+\left(c^{2}-a^{2}\right)^{3}}$
equal to ?

Options:
A $\mathrm{a}^{2} \mathrm{~b}^{2} \mathrm{c}^{2}$
B $-a^{2} b^{2} c^{2}$
C $\mathrm{abc}$
D $3 \mathrm{a}^{2} \mathrm{~b}^{2} \mathrm{c}^{2}$

The correct answer is:
$-a^{2} b^{2} c^{2}$

Solution:
We know that if $\mathrm{x}+\mathrm{y}$ ' Hereboth in nominator and denominator $a^{4}-b^{4}+b^{4}-c^{4}+c^{4}-a^{4}=0$
and $a^{2}-b^{2}+b^{2}-c^{2}+c^{2}-a^{2}=0$
Hence, $\frac{\left(a^{4}-b^{4}\right)^{3}+\left(b^{4}-c^{4}\right)^{3}+\left(c^{4}-a^{4}\right)^{3}}{\left(a^{2}-b^{2}\right)^{3}+\left(b^{2}-c^{2}\right)^{3}+\left(c^{2}-a^{2}\right)^{3}}$
$=\frac{3\left(a^{4}-b^{4}\right)\left(b^{4}-c^{4}\right)\left(c^{4}-a^{4}\right)}{3\left(a^{2}-b^{2}\right)\left(b^{2}-c^{2}\right)\left(c^{2}-a^{2}\right)}$
$=\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)\left(c^{2}+a^{2}\right.$
$=\left(-c^{2}\right)\left(-a^{2}\right)\left(-b^{2}\right)$
$=-a^{2} b^{2} c^{2}$
$\left(\because a^{2}+b^{2}+c^{2}=0\right)$

Topic: Mathematics, Quadratic Equation, JEE Main
If $a^2+b^2+c^2=-2$ and $f(x)=\left|\begin{array}{ccc}1+a^2 x & \left(1+b^2\right) x & \left(1+c^2\right) x \\ \left(1+a^2\right) x & 1+b^2 x & \left(1+c^2\right) x \\ \left(1+a^2\right) x & \left(1+b^2\right) x & 1+c^2 x\end{array}\right|$ then $f(x)$ is $a$ polynomial of degree

Options:
A
1

B
0

C
3

D
2

The correct answer is:
2

Solution:
$$
\begin{aligned}
& f(x)=\left|\begin{array}{ccc}
1+\left(a^2+b^2+c^2+2\right) x & \left(1+b^2\right) x & \left(1+c^2\right) x \\
1+\left(a^2+b^2+c^2+2\right) x & 1+b^2 x & \left(1+c^2\right) x \\
1+\left(a^2+b^2+c^2+2\right) x & \left(1+b^2\right) x & 1+c^2 x
\end{array}\right| \text {, Applying } C_1 \rightarrow C_1+C_2+C_3 \\
& =\left|\begin{array}{ccc}
1 & \left(1+b^2\right) x & \left(1+c^2\right) x \\
1 & 1+b^2 x & \left(1+c^2\right) x \\
1 & \left(1+b^2\right) x & 1+c^2 x
\end{array}\right| \because a^2+b^2+c^2+2=0 \\
& f(x)=\left|\begin{array}{ccc}
0 & x-1 & 0 \\
0 & 1-x & x-1 \\
1 & \left(1+b^2\right) x & 1+c x
\end{array}\right| ; \text { Applying } R_1 \rightarrow R_1-R_2, R_2 \rightarrow R_2-R_3 \\
& f(x)=(x-1)^2
\end{aligned}
$$
Hence degree $=2$.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2005
If a2b2c2a+λ2b+λ2c+λ2a-λ2b-λ2c-λ2=kλa2b2c2abc111,λ≠0, then k is equal to

Options:
A 4λabc
B -4λ2
C 4λ2
D -4λabc

The correct answer is:
4λ2

Solution:
R2→R2-R1,R1→R1-R3
λ2a-λλ2b-λλ2c-λ4aλ4bλ4cλa-λb-λ2c-λ2
R3→R3+R1,R1→R1-12R2
=-λ2-λ2-λ24aλ4bλ4cλa2b2c2
=4λ3111abca2b2c2
=4λ3a2b2c2abc111
∴k=4λ2

Topic: Mathematics, Determinants, JEE Main
If a2b2c2a+λ2b+λ2c+λ2a-λ2b-λ2c-λ2=kλa2b2c2abc111,λ≠0, then k is equal to:

Options:
A 4λabc
B -4λ2
C 4λ2
D -4λabc

The correct answer is:
4λ2

Solution:
R2→R2-R1,R1→R1-R3
λ2a-λλ2b-λλ2c-λ4aλ4bλ4cλa-λb-λ2c-λ2
R3→R3+R1,R1→R1-12R2
=-λ2-λ2-λ24aλ4bλ4cλa2b2c2
=4λ3111abca2b2c2
=4λ3a2b2c2abc111
∴K=4λ2

Topic: Mathematics, Determinants, JEE Main
If a→=2, b→=3 and 2a→-b→=5, then 2a→+b→ equals :

Options:
A 5
B 7
C 17
D 1

The correct answer is:
5

Solution:
a→=2, b→=3, 2a→-b→=5
2 a→-b→2=2 a→-b→·2 a→-b→
=4a→2-4a→·b→+b→2 ...i
2a→+b→2=4a→2+4a→·b→+b→2 ...ii

Adding equation (i) and equation (ii), we get
2 a→+b→2+2 a→-b→2=8a→2+2b→2
⇒ 2 a→+b→2+52=822+232
⇒ 2a→+b→2=32+18-25 = 2 5
⇒ 2a→+b→=5

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2014 (09 Apr Online)
If $|\mathbf{a}|=2,|\mathbf{b}|=3$ and $\mathbf{a}, \mathbf{b}$ are mutually perpendicular, then the area of the triangle whose vertices are $\mathbf{0}, \mathbf{a}+\mathbf{b}, \mathbf{a}-\mathbf{b}$ is

Options:
A 5
B 1
C 6
D 8

The correct answer is:
6

Solution:
Let the position vectors of the points $A, B, C$ are $\mathbf{0}, \mathbf{a + b}, \mathbf{a}-\mathbf{b}$ and $\theta=90^{\circ}$.
$\begin{aligned}
\text { Area of triangle } & =\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{A C}| \\
& =\frac{1}{2}|(\mathbf{a}+\mathbf{b}) \times(\mathbf{a}-\mathbf{b})|=\frac{1}{2}|2 \mathbf{b} \times \mathbf{a}| \\
& =b a \sin \theta=3 \times 2 \sin 90^{\circ}=6 .
\end{aligned}$

Topic: Mathematics, Vector Algebra, JEE Main
If $|\overrightarrow{\mathbf{a}}|=2,|\overrightarrow{\mathbf{b}}|=3$ and $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$ are mutually
perpendicular, then the area of the triangle whose vertices are $\overrightarrow{\mathbf{0}}, \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}$ is

Options:
A 5
B 1
C 6
D 8

The correct answer is:
6

Solution:
Let the position vectors of the points $A, B, C$ are
$$
\begin{array}{l}
\mathbf{0}, \mathbf{a}+\mathbf{b}, \mathbf{a}-\mathbf{b} \text { and } \theta=90^{\circ} . \\
\therefore \text { Area of triangle }=\frac{1}{2}|\overrightarrow{\mathbf{A B}} \times \overrightarrow{\mathbf{A C}}| \\
\left.\quad=\frac{1}{2} \mid \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}\right) \times(\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}) \mid \\
=\frac{1}{2}|2 \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{a}}| \\
=|\overrightarrow{\mathbf{b}}||\overrightarrow{\mathbf{a}}| \sin \theta \\
=3 \times 2 \sin 90^{\circ} \\
=6
\end{array}
$$

Topic: Mathematics, Vector Algebra, JEE Main
If a→+2b→+3c→×b→+2c→+3a→⋅c→+2a→+3b→=54, where a→, b→ and c→ are 3 non-coplanar vectors, then the value of a→⋅a→a→⋅b→a→⋅c→b→⋅a→b→⋅b→b→⋅c→c→⋅a→c→⋅b→c→⋅c→ is equal to

Options:
A 9
B 3
C 6
D 12

The correct answer is:
9

Solution:
a→+2b→+3c→b→+2c→+3a→c→+2a→+3b→=54
123312231a→b→c→=54
1-5-2-1+37a→b→c→=54
a→b→c→=3
Also a→⋅a→a→⋅b→a→⋅c→b→⋅a→b→⋅b→b→⋅c→c→⋅a→c→⋅b→c→⋅c→=a→b→c→2=9

Topic: Mathematics, Vector Algebra, JEE Main
If $|\mathbf{a}|=2,|\mathbf{b}|=5$ and $|\mathbf{a} \times \mathbf{b}|=8$, then $\mathbf{a} \cdot \mathbf{b}$ is equal to

Options:
A 0
B 2
C 4
D 6

The correct answer is:
6

Solution:
$\begin{aligned}
& \mathbf{a} \cdot \mathbf{b}=|\mathbf{a}||\mathbf{b}| \cos \theta ; B u t \\
& \Rightarrow \sin \theta=\frac{|\mathbf{a} \times \mathbf{b}|}{|\mathbf{a}||\mathbf{b}|}=\frac{4}{5} \Rightarrow \cos \theta=\frac{3}{5} \\
& \text {Therefore, } \\
& \mathbf{a} \cdot \mathbf{b}=2 \times 5 \times \frac{3}{5}=6
\end{aligned}$

Topic: Mathematics, Vector Algebra, JEE Main
If a→=2,b→=5 and a→×b→=8, then a→·b→ is equal to:

Options:
A 6
B 4
C 3
D 5

The correct answer is:
6

Solution:
We have, a→=2,b→=5

and a→×b→=8

Now, a→×b→=a→b→sinθ=8

⇒sinθ=±45

∴ a→·b→=a→b→cosθ

=10.±35=±6

⇒a→·b→=6.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2021 (25 Jul Shift 2)
If a+2bcosx)(a-2bcosy=a2-b2, where a>b>0, thendxdy at π4,π4 is:

Options:
A a-2ba+2b
B a-ba+b
C a+ba-b
D 2a+b2a-b

The correct answer is:
a+ba-b

Solution:
a+2bcosxa−2bcosy=a2−b2

a−2bcosy−2bsinxdxdy+2bsinya+2bcosx=0

dxdy=2bsinya+2bcosx2bsinxa−2bcosy=sinya+2bcosxsinxa−2bcosy

dxdyπ4,π4=a+ba−b.

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2020 (04 Sep Shift 1)
If A=θ:2cos2⁡θ+sin⁡θ≤2 and B=θ:π2≤θ≤3π2, then A∩B is equal to

Options:
A Î¸:Ï2â¤Î¸â¤5Ï6
B Î¸:Ïâ¤Î¸â¤3Ï2
C θ:π2≤θ≤5π6 or θ:π≤θ≤3π2
D Î¸:3Ï4â¤Î¸â¤7Ï4

The correct answer is:
θ:π2≤θ≤5π6 or θ:π≤θ≤3π2

Solution:
2-2sin2⁡θ+sin⁡θ≤2
2sin2⁡θ-sin⁡θ≥0
sin⁡θ2sin⁡θ-1≥0
Case I: sin⁡θ≥0, sin⁡θ≥12
⇒sin⁡θ≥12⇒θ∈π2,5π6
Case II: sin⁡θ≤0, sin⁡θ≤12
⇒sin⁡θ≤0
⇒θ∈π,3π2
⇒A∩B=θ:π2≤θ≤5π6 or θ:π≤θ≤3π2

Topic: Mathematics, Trigonometric Equations, JEE Main
If aâ=2i^+3j^+4k^,aââbâ=2 and aâÃbâ=2i^-k^, then bâ is

Options:
A i^-2j^+k^
B 4i^-4j^+2k^
C 123i^+7j^+9k^
D 1297i^-4j^+14k^

The correct answer is:
1297i^-4j^+14k^

Solution:
a→×b→=2i^-k^
a→×b→×a→=2i^-k^×2i^+3j^+4k^
29b→-2a→=3i^-10j^+6k^
b→=1297i^-4j^+14k^

Topic: Mathematics, Vector Algebra, JEE Main
If a → =2 i ^ −3 j ^ +4 k ^ ,  a → ⋅ b → =2 and a→×b→ =i^+2j^+k^, then b→ is equal to

Options:
A 15i^-8j^+k^
B 12915i^-8j^+k^
C 152i^+j^+k^
D 2i^+j^+k^

The correct answer is:
12915i^-8j^+k^

Solution:
a→×b→=i^+2j^+k^

Taking cross product with a→ on both the sides, we get,
a→×b→×a→=i^+2j^+k^×a→
⇒a→⋅a→b→−b→⋅a→a→=i^j^k^1212−34
⇒29b→-2a→=i^11-j^2+k^-7
⇒29b→=15i^-8j^+k^⇒b→=12915i^-8j^+k^

Topic: Mathematics, Vector Algebra, JEE Main
If $\bar{a}=2 i+3 j-\hat{k}, \bar{b}=-i+2 j-4 \hat{k}$ and $\bar{c}=i+j+\hat{k}$, then $(\bar{a} \times \bar{b}) \cdot(\bar{a} \times \bar{c})=$

Options:
A $-74$
B 64
C $-64$
D 74

The correct answer is:
$-74$

Solution:
$\bar{a} \times \bar{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ -1 & 2 & -4\end{array}\right|=\hat{i}(-12+2)-\hat{j}(-8-1)+\hat{k}(4+3)=-10 \hat{i}+9 \hat{j}+7 \hat{k}$
$\begin{aligned} \bar{a} \times \bar{c} &=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 1 & 1 & 1\end{array}\right|=\hat{i}(3+1)-\hat{j}(2+1)+\hat{k}(2-3)=4 \hat{i}-3 \hat{j}-\hat{k} \\ \text { Here } \quad(\bar{a} \times \bar{b}) \cdot(\bar{a} \times \bar{c}) &=(-10 \hat{i}+9 \hat{j}+7 \hat{k}) \cdot(4 \hat{i}-3 \hat{j}-\hat{k}) \\ &=(-10)(4)+9(-3)+7(-1) \\ &=-40-27-7=-74 \end{aligned}$

Topic: Mathematics, Vector Algebra, JEE Main
If $\mathbf{a}=2 \mathbf{i}+3 \mathbf{j}-\mathbf{k}, \boldsymbol{\beta}=-\mathbf{i}+2 \mathbf{j}-4 \mathbf{k}$ and $\mathbf{Y}=\mathbf{i}+\mathbf{j}+\mathbf{k}$, then $(\boldsymbol{\alpha} \times \boldsymbol{\beta}) \cdot(\boldsymbol{a} \times \mathbf{Y})$ is equal to

Options:
A 60
B 64
C 74
D -74

The correct answer is:
-74

Solution:
$\begin{aligned}
& \mathbf{a} \cdot[\boldsymbol{\gamma} \times(\boldsymbol{\alpha} \times \boldsymbol{\beta})]=\mathbf{a} \cdot[(\boldsymbol{y} \cdot \boldsymbol{\beta}) \mathbf{a}-(\boldsymbol{y} \cdot \boldsymbol{a}) \boldsymbol{\beta}] \\
& =(\mathbf{a})^2(\boldsymbol{\gamma} \cdot \boldsymbol{\beta})-(\mathbf{y} \cdot \mathbf{a})(\boldsymbol{\beta} \cdot \mathbf{a})=14(-3)-(4)(\boldsymbol{)})=-74
\end{aligned}$

Topic: Mathematics, Vector Algebra, JEE Main
If A→=2i^+3j^-k^ m and B→=i^+2j^+2k^ m. The magnitude of component of vector A→ along vector B→ will be _____ m.

Options:

The correct answer is:
2

Solution:
GIven vector are A→=2i^+3j^-k^ m and B→=i^+2j^+2k^ m

Component of A→ along B→=A→·B^

=A→·B→B→=2+6-212+22+22=63=2

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2022 (26 Jul Shift 2)
If $\vec{A}=2 \hat{i}+4 \hat{j}-5 \hat{k}$ the direction of cosines of the vector $\vec{A}$ are

Options:
A $\frac{2}{\sqrt{45}}, \frac{4}{\sqrt{45}}$ and $\frac{-5}{\sqrt{45}}$
B $\frac{1}{\sqrt{45}}, \frac{2}{\sqrt{45}}$ and $\frac{3}{\sqrt{45}}$
C $\frac{4}{\sqrt{45}}, 0$ and $\frac{4}{\sqrt{45}}$
D $\frac{3}{\sqrt{45}}, \frac{2}{\sqrt{45}}$ and $\frac{5}{\sqrt{45}}$

The correct answer is:
$\frac{2}{\sqrt{45}}, \frac{4}{\sqrt{45}}$ and $\frac{-5}{\sqrt{45}}$

Solution:
$\vec{A}=2 \hat{i}+4 \hat{j}-5 \hat{k} \therefore|\vec{A}|=\sqrt{(2)^2+(4)^2+(-5)^2}=\sqrt{45}$
$\therefore \cos \alpha=\frac{2}{\sqrt{45}}, \quad \cos \beta=\frac{4}{\sqrt{45}}, \quad \cos \gamma=\frac{-5}{\sqrt{45}}$

Topic: Physics, Mathematics in Physics, JEE Main
If $\mathbf{a}=2 \mathbf{i}+5 \mathbf{j}$ and $\mathbf{b}=2 \mathbf{i}-\mathbf{j}$, then the unit vector along $\mathbf{a}+\mathbf{b}$ will be

Options:
A $\frac{\mathbf{i}-\mathbf{j}}{\sqrt{2}}$
B $\mathbf{i}+\mathbf{j}$
C $\sqrt{2}(\mathbf{i}+\mathbf{j})$
D $\frac{\mathbf{i}+\mathbf{j}}{\sqrt{2}}$

The correct answer is:
$\frac{\mathbf{i}+\mathbf{j}}{\sqrt{2}}$

Solution:
$\quad \mathbf{a}+\mathbf{b}=4 \mathbf{i}+4 \mathbf{j}$ therefore unit vector
$\frac{4(\mathbf{i}+\mathbf{j})}{\sqrt{32}}=\frac{\mathbf{i}+\mathbf{j}}{\sqrt{2}}$

Topic: Mathematics, Vector Algebra, JEE Main
If a→=2i^+j^+2k^, then, the value of i^×a→×i^2+j^×a→×j^2+k^×a→×k^2, is equal to :

Options:

The correct answer is:
18

Solution:
Let a→=xi^+yj^+zk^, and using vector triple product, we have,
i^×a→×i^=i^.i^a→-i^.a→i^=yj^+zk^
Similarly,

j^×a→×j^=xi^+zk^

and k^×a→×k^=xi^+yk^

Thus,

i^×a→×i^2+j^×a→×j^2+k^×a→×k^2
=yj^+zk^2+xi^+zk^2+xi^+yj^2

=2|a→|2=2(9)=18

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2020 (04 Sep Shift 2)
If a→=2i^+j^+3k^, b→=3i^+3j^+k^ and c→=c1i^+c2j^+c3k^ are coplanar vectors and a→·c→=5,b→⊥c→, then 122c1+c2+c3 is equal to ______.

Options:

The correct answer is:
150

Solution:
As the given vectors are coplanar, so 213331c1c2c3=0

⇒8c1-7c2-3c3=0 ...i

Also, a→·c→=5⇒2c1+c2+3c3=5 ...ii

and b→·c→=0⇒3c1+3c2+c3=0 ...iii

Solving the above three equations using Cramer's rule, we get,

∆=8-7-3213331=-122

∆1=0-7-3513031=-10

∆2=80-3253301=85

∆3=8-70215330=-225

Hence, 122c1+c2+c3=12210122-85122+225122=150

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2022 (28 Jun Shift 1)
If a→=2i^+j^+3k^,b→=i^+3j^-k^ and c→=3i^-j^-2k^ then, the value of a→·a→a→·b→a→·c→b→·a→b→·b→b→·c→c→·a→c→·b→c→·c→ is

Options:
A 2020
B 2025
C 2030
D 1849

The correct answer is:
2025

Solution:
Given that, If a→=2i^+j^+3k^,b→=i^+3j^-k^ and c→=3i^-j^-2k^

⇒a→·a→=14, b→·b→=11, c→·c→=14 and a→·b→=b→·a→=2, a→·c→=c→·a→=-1, b→·c→=c→·b→=2

Putting these values in a→·a→a→·b→a→·c→b→·a→b→·b→b→·c→c→·a→c→·b→c→·c→, we get 142-12112-1214

=14154-4-228+2-14+11

=2025

Topic: Mathematics, Vector Algebra, JEE Main
If $\mathbf{a}=2 \hat{\mathbf{i}}+\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{c}=4 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}$, then the vector $r$ satisfying $r \times b=c \times b$ and $r . a=0$ is

Options:
A $\hat{\mathbf{i}}+8 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$
B $\hat{\mathbf{i}}-8 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$
C $\hat{\mathbf{i}}-8 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$
D $\hat{\mathbf{-i}}-8 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$

The correct answer is:
$\hat{\mathbf{-i}}-8 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$

Solution:
(d) Three vectors $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$ are given as, $a=2 \hat{i}+\hat{k}, b=\hat{i}+\hat{j}+\hat{k}$ and $c=4 \hat{i}-3 \hat{j}+7 \hat{k}$
Given condition, $\mathrm{r} \times \mathrm{b}=\mathrm{c} \times \mathrm{b}$
$\begin{aligned}
& \mathrm{r} \times \mathrm{b}-\mathrm{c} \times \mathrm{b}=0 \\
& (\mathrm{r}-\mathrm{c}) \times \mathrm{b}=0
\end{aligned}$
It means, $(\mathrm{r}-\mathrm{c}) \| \mathrm{b}$
So, $r-c=\lambda b$
Also given, $\mathrm{r} \cdot \mathrm{a}=0$
$(c+\lambda b) \cdot a=0$ [using Eq. (i)]
$\begin{aligned}
& (4 \hat{i}-3 \hat{j}+7 \hat{k}+\lambda \hat{i}+\lambda \hat{j}+\lambda \hat{k}) \cdot(2 \hat{i}+\hat{k})=0 \\
& {[(4+\lambda) \cdot \hat{i}+(-3+\lambda) \hat{j}+(7+\lambda) \hat{k}] \cdot(2 \hat{i}+\hat{k})=0} \\
& (4+\lambda) \cdot 2+(7+\lambda) \cdot 1=0 \\
& 8+2 \lambda+7+\lambda=0 \\
& 3 \lambda=-15 \\
& \therefore \quad \lambda=-5
\end{aligned}$
Put the value of $\lambda$ in Eq. (i), we get
$\begin{aligned}
& r=4 \hat{i}-3 \hat{j}+7 \hat{k}-5(\hat{i}+\hat{j}+\hat{k}) \\
& =4 \hat{i}-3 \hat{j}+7 \hat{k}-5 \hat{i}-5 \hat{j}-5 \hat{k}=-\hat{i}-8 \hat{j}+2 \hat{k}
\end{aligned}$

Topic: Mathematics, Vector Algebra, JEE Main