datavorous/entrance-exam-dataset huggingface dataset:
Let A=201110101, B=B1B2B3, where B1, B2, B3 are column matrices, and AB1=100, AB2=230,AB3=321
If α=|B| and β is the sum of all the diagonal elements of B, then α3+β3 is equal to

Options:

The correct answer is:
28

Solution:
Given: A=201110101, B=B1B2B3, B1=x1y1z1, B2=x2y2z2 and B3=x3y3z3.

⇒AB1=201110101x1y1z1=100

⇒2x1+0+z1x1+y1+0x1+0+z1=100

⇒2x1+z1=1, x1+y1=0, x1+z1=0

⇒x1=1,y1=-1,z1=-1

⇒AB2=201110101x2y2z2=230

⇒AB2=2x2+z2x2+y2x2+z2=230

⇒2x2+z2=2, x2+y2=3, x2+z2=0

⇒x2=2,y2=1,z2=-2

⇒AB3=201110101x3y3z3=321

⇒AB2=2x3+z3x3+y3x3+z3=321

⇒2x3+z3=3, x3+y3=2, x3+z3=1

⇒x3=2,y3=0,z3=-1

⇒B=122-110-1-2-1

⇒α=|B|=3

⇒β=1

⇒α3+β3=27+1=28

Topic: Mathematics, Matrices, JEE Main, JEE Main 2024 (27 Jan Shift 1)
Let A=2070101-21 and B=-k14k7k010k-4k-2k. If AB=I, where I is an identity matrix of order 3, then the sum of all elements of matrix B is equal to

Options:
A 2
B 37
C 103
D 4

The correct answer is:
4

Solution:
AB=2070101-21  -k14k7k010k-4k-2k=5k00010010k-25k
AB=I
5k00010010k-25k=100010001⇒ k=15
Sum of all element of B=15k+1=4

Topic: Mathematics, Matrices, JEE Main
Let $A=\left[\begin{array}{lll}2 & a & 0 \\ 1 & 3 & 1 \\ 0 & 5 & b\end{array}\right]$. If $A^3=4 A^2-A-21 I$, where $I$ is the identity matrix of order $3 \times 3$, then $2 a+3 b$ is equal to

Options:
A -9
B -13
C -10
D -12

The correct answer is:
-13

Solution:
$\begin{aligned} & A^3-4 A^2+A+21 I=0 \\ & \operatorname{tr}(A)=4=5+6 \Rightarrow b=-1 \\ & |A|=-21 \\ & -16+a=-21 \Rightarrow a=-5 \\ & 2 a+3 b=-13\end{aligned}$

Topic: Mathematics, Matrices, JEE Main, JEE Main 2024 (08 Apr Shift 1)
Let $A=\left[\begin{array}{cc}2 & -1 \\ 1 & 1\end{array}\right]$. If the sum of the diagonal elements of $A^{13}$ is $3^n$, then $n$ is equal to_________

Options:

The correct answer is:
7

Solution:
$A=\left[\begin{array}{cc}2 & -1 \\ 1 & 1\end{array}\right]$
$\begin{aligned} & A^2=\left[\begin{array}{cc}2 & -1 \\ 1 & 1\end{array}\right]\left[\begin{array}{cc}2 & -1 \\ 1 & 1\end{array}\right]=\left[\begin{array}{cc}3 & -3 \\ 3 & 0\end{array}\right] \\ & A^3=\left[\begin{array}{cc}3 & -3 \\ 3 & 0\end{array}\right]\left[\begin{array}{cc}2 & -1 \\ 1 & 1\end{array}\right]=\left[\begin{array}{cc}3 & -6 \\ 6 & -3\end{array}\right] \\ & A^4=\left[\begin{array}{cc}3 & -6 \\ 6 & -3\end{array}\right]\left[\begin{array}{cc}2 & -1 \\ 1 & 1\end{array}\right]=\left[\begin{array}{cc}0 & -9 \\ 9 & -9\end{array}\right] \\ & A^5=\left[\begin{array}{cc}0 & -9 \\ 9 & -9\end{array}\right]\left[\begin{array}{cc}2 & -1 \\ 1 & 1\end{array}\right]=\left[\begin{array}{cc}-9 & -9 \\ 9 & -18\end{array}\right] \\ & A^6=\left[\begin{array}{cc}-9 & -9 \\ 9 & -18\end{array}\right]\left[\begin{array}{cc}2 & -1 \\ 1 & 1\end{array}\right]=\left[\begin{array}{cc}-27 & 0 \\ 0 & -27\end{array}\right] \\ & A^7=\left[\begin{array}{cc}-27 & -0 \\ 0 & -27\end{array}\right]\left[\begin{array}{cc}-54 & 27 \\ -27 & -27\end{array}\right]=\left[\begin{array}{cc}3^6 \times 2 & -27^2 \\ 27^2 & 3^6\end{array}\right] \\ & 3^7=3^n \Rightarrow n=7\end{aligned}$

Topic: Mathematics, Matrices, JEE Main, JEE Main 2024 (08 Apr Shift 1)
Let A=2-11-23-2-44-x be a matrix. If A2=A, then the value of x is equal to

Options:

The correct answer is:
3

Solution:
A2=2-11-23-2-44-x2-11-23-2-44-x

=2-14-x-232x-84x-1616-4xx2-12
∵A2=A⇒4-x=1⇒x=3
⇒2x-8=-2⇒x=3
⇒-4=4x-16⇒x=3
⇒4=16-4x⇒x=3
⇒-x=x2-12⇒x=-4 or 3
Hence, x=3

Topic: Mathematics, Matrices, JEE Main
Let A=2126211332 and P=120502715. The sum of the prime factors of P-1AP-2I is equal to

Options:
A 26
B 27
C 66
D 23

The correct answer is:
26

Solution:
Let, y=P-1AP-2I

⇒y=P-1AP-2P-1P

⇒y=P-1I( A-2I)P

⇒y=P-1I|A-2I||P|

We know that, A=1A-1.

⇒y=|A-2I|

⇒y=2126211332-200020002

⇒y=0126011330

⇒y=0-0-33+218-0

⇒y=69

Now, prime factor of 69 is 3 and 23.

So, sum =26

Topic: Mathematics, Matrices, JEE Main, JEE Main 2024 (29 Jan Shift 2)
Let A=θ∈-π2,π:3+2i sinθ1-2i sinθ is purely imaginary . Then the sum of the elements in A is:

Options:
A 5π6
B π
C 2π3
D 3π4

The correct answer is:
2π3

Solution:
Given complex number is z=3+2isinθ1-2isinθ
⇒z=(3+2isinθ)(1+2isinθ)(1+4sin2θ)
Since given complex number is purely imaginary
⇒Re(z)=0
⇒3-4sin2θ=0
⇒sinθ=±32
⇒θ=-π3,π3,2π3
⇒ sum of all possible values of θ=2π3 .

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2019 (09 Jan Shift 1)
Let A=2,3,4 and B=8,9,12. Then the number of elements in the relation R=a1,b1,a2,b2∈A×B,A×B:a1 divides b2 and a2 divides b1 is

Options:
A 36
B 24
C 18
D 12

The correct answer is:
36

Solution:
Given,

A=2,3,4, B=8,9,12

And a1∈A, b2∈B

Now finding elements where a1 divides b2 we get,

a1,b2∈2,8,2,12,3,9,3,12,4,8,4,12

Also given, a2∈A,b1∈B

Now finding elements when a2 divides b1, we get,

a2,b1∈2,8,2,12,3,9,3,12,4,8,4,12

Hence, Number of relations = 6 × 6 = 36

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2023 (10 Apr Shift 2)
Let A=2357 and B=a00b where a,bâN. The number of matrices B such that AB=BA, is equal to

Options:
A 0
B 1
C 2
D infinite

The correct answer is:
infinite

Solution:
AB=2357a00b=2a3b5a7b
BA=a00b2357=2a3a5b7b
∴AB=BA⇒a=b (infinite matrices are possible)

Topic: Mathematics, Matrices, JEE Main
Let A(2,3,5) and C(-3,4,-2) be opposite vertices of a parallelogram ABCD if the diagonal BD→=i^+2j^+3k^ then the area of the parallelogram is equal to

Options:
A 12410
B 12474
C 12586
D 12306

The correct answer is:
12474

Solution:
Area of the parallelogram is given by 12AC→×BD→.

⇒Area=12-5i^+j^-7k^×i^+2j^+3k^

⇒Area=12i^j^k^-51-7123

⇒Area=1217i^+8j^-11k^

⇒Area=12289+64+121

⇒Area=4742 square units.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2024 (30 Jan Shift 1)
Let $\mathrm{A}(2,3,5), \mathrm{B}(-1,3,2)$ and $\mathrm{C}(\lambda, 5, \mu)$ be the vertices of a $\triangle \mathrm{ABC}$. If the median through $\mathrm{A}$ is equally inclined to the coordinate axes, then:

Options:
A
$5 \lambda-8 \mu=0$

B
$8 \lambda-5 \mu=0$

C
$10 \lambda-7 \mu=0$

D
$7 \lambda-10 \mu=0$

The correct answer is:
$10 \lambda-7 \mu=0$

Solution:
If D be the mid-point of $B C$, then
$$
D=\left(\frac{\lambda-1}{2}, 4, \frac{\mu+2}{2}\right)
$$
Direction ratios of $\mathrm{AD}$ are $\frac{\lambda-5}{2}, 1, \frac{\mu-8}{2}$
Since median AD is equally inclined with coordinate axes, therefore direction ratios of $\mathrm{AD}$ will be equal, i.e,
$$
\begin{aligned}
&\frac{\left(\frac{\lambda-5}{2}\right)^2}{\left(\frac{\lambda-5}{2}\right)^2+1+\left(\frac{\mu-8}{2}\right)^2} \\
&=\frac{1}{\left(\frac{\lambda-5}{2}\right)^2+1+\left(\frac{\mu-8}{2}\right)^2} \\
&=\frac{\left(\frac{\mu-8}{2}\right)^2}{\left(\frac{\lambda-5}{2}\right)^2+1+\left(\frac{\mu-8}{2}\right)^2} \\
&\Rightarrow\left(\frac{\lambda-5}{2}\right)^2=1=\left(\frac{\mu-8}{2}\right)^2
\end{aligned}
$$
$$
\begin{aligned}
&\Rightarrow\left(\frac{\lambda-5}{2}\right)^2=1=\left(\frac{\mu-8}{2}\right)^2 \\
&\Rightarrow \lambda=7,3 \text { and } \mu=10,6
\end{aligned}
$$
If $\lambda=7$ and $\mu=10$
Then $\frac{\lambda}{\mu}=\frac{7}{10} \Rightarrow 10 \lambda-7 \mu=0$

Topic: Mathematics, Properties of Triangles, JEE Main, JEE Main 2014 (11 Apr Online)
Let $A=\{2,3,6,8,9,11\}$ and $B=\{1,4,5,10,15\}$. Let $R$ be a relation on $A \times B$ defined by $(a, b) R(c, d)$ if and only if $3 a d-7 b c$ is an even integer. Then the relation $R$ is

Options:
A an equivalence relation.
B reflexive and symmetric but not transitive.
C transitive but not symmetric.
D reflexive but not symmetric.

The correct answer is:
reflexive and symmetric but not transitive.

Solution:
$\begin{array}{lc}\mathrm{A}=\{2,3,6,8,9,11\} & (\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{c}, \mathrm{d}) \\ \mathrm{B}=\{1,4,5,10,15\} & 3 \mathrm{ad}-7 \mathrm{bc}\end{array}$
Reflexive : $(\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{a}, \mathrm{b})$
$\Rightarrow 3 \mathrm{ab}-7 \mathrm{ba}=-4 \mathrm{ab}$ always even so it is reflexive.
Symmetric : If $3 \mathrm{ad}-7 \mathrm{bc}=$ Even
Case-I : odd odd
Case-II : even even
(c, d) $\mathrm{R}(\mathrm{a}, \mathrm{b}) \Rightarrow 3 \mathrm{bc}-3 \mathrm{ab}$
Case-I : odd odd
Case-II : even even
so symmetric relation
Transitive :
Set $(3,4) R(6,4)$ Satisfy relation
Set $(6,4) R(3,1)$ Satisfy relation
but $(3,4) \mathrm{R}(3,1)$ does not satisfy relation so not transitive.

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2024 (08 Apr Shift 2)
Let A=23a0, a∈R be written as P+Q where P is a symmetric matrix and Q is skew symmetric matrix. If detQ=9, then the modulus of the sum of all possible values of determinant of P is equal to:

Options:
A 36
B 24
C 45
D 18

The correct answer is:
36

Solution:
We have,

A=23a0, a∈R

Then,

AT=2a30

Now,

P=A+AT2

⇒P=1223a0+2a30

⇒P=23+a2a+320
And,

Q=A-AT2

⇒Q=1223a0-2a30

⇒Q=03-a2a-320

As,

det(Q)=9

⇒a-32=36

⇒a=3±6

∴ a=9,-3

Now,

detP=-3+a22

For a=-3, detP=0

For a=9, detP=-3+922=-36
∴ Modulus of the sum of all possible values of detP

=-36+0=36

Topic: Mathematics, Matrices, JEE Main, JEE Main 2021 (20 Jul Shift 1)
Let $A(2,-3)$ and $B(-2,1)$ be vertices of a triangle $A B C$. If the centroid of this triangle moves on the line $2 x+3 y=1$, then the locus of the vertex $C$ is the line

Options:
A
$2 x+3 y=9$

B
$2 x-3 y=7$

C
$3 x+2 y=5$

D
$3 x-2 y=3$

The correct answer is:
$2 x+3 y=9$

Solution:
If $\mathrm{C}$ be $(\mathrm{h}, \mathrm{k})$ then centroid is $(\mathrm{h} / 3,(\mathrm{k}-2) / 3)$ it lies on $2 x+3 y=1$. $\Rightarrow$ locus is $2 x+3 y=9$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2004
Let A+2B=1206-33-531 and 2A-B=2-152-16012. If TrA denotes the sum of all diagonal elements of the matrix A, then Tr(A)-Tr(B) has value equal to

Options:
A 1
B 2
C 0
D 3

The correct answer is:
2

Solution:
A+2B=1206-33-531 …1

2A-B=2-152-16012

⇒4A-2B=4-2104-212024 …2

1+2⇒5 A=501010-515-555

A=1022-13-111 and 2A=2044-26-222

∴B=2044-26-222-2-152-16012

B=01-12-10-210

tr(A)=1-1+1=1

tr(B)=-1

tr(A)=1 and tr(B)=-1

∴tr(A)-tr(B)=2

Topic: Mathematics, Matrices, JEE Main, JEE Main 2021 (18 Mar Shift 1)
Let A=2b1bb2+1b1b2, where b>0. Then the minimum value of detAb is:

Options:
A 23
B -23
C 3
D -3

The correct answer is:
23

Solution:
A=2b1bb2+1b1b2
=22b2+2-b2-b2b-b+1b2-b2-1
=2b2+2-b2-1
=2b2+4-b2-1
=b2+3
Ab=b+3b=b-3b2+23
Hence, the minimum value of Ab is 23. ∵b-3b2≥0

Topic: Mathematics, Determinants, JEE Main, JEE Main 2019 (10 Jan Shift 2)
Let A+2B= 2 40 6-33-5 35 and 2A-B=6-246 156 34, then trA-trB is equal to
(where, trA= trace of matrix A i.e. sum of the principal diagonal elements of matrix A )

Options:

The correct answer is:
5.80

Solution:
Let, x=2406-33-535
y=6-24615234
A+2B=x
2A-B=y
A=15x+2y, B=152x-y
A-B=15-x+3y=1516-101212612167
trA-trB=trA-B
=295

Topic: Mathematics, Matrices, JEE Main
Let |a→|=2,|b→|=3 and the angle between the vectors a→ and b→ be π4. Then |(a→+2b→)×(2a→-3b→)|2 is equal to

Options:
A 441
B 482
C 841
D 882

The correct answer is:
882

Solution:
Given,

|a→|=2,|b→|=3 and the angle between the vectors a→ and b→ be π4,

Now solving,

|(a→+2b→)×(2a→-3b→)|2

=|-3(a→×b→)+4(b→×a→)|2 as a→×a→=0 & b→×b→=0

=|7(b→×a→)|2 as b→×a→=-a→×b→

=49|a|2|b|2sin2π4

=49×4×9×12

=882

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (13 Apr Shift 2)
Let Aα,-2,Bα,6 and Cα4,-2 be vertices of a ∆ABC. If 5,α4 is the circumcentre of ∆ABC, then which of the following is NOT correct about ∆ABC

Options:
A ares is 24
B perimeter is 25
C circumradius is 5
D inradius is 2

The correct answer is:
perimeter is 25

Solution:
Given,

Aα,-2, Bα,6 & Cα4,-2

Now slope of AC=0 and slope of AB=80→∞

Now we can see AC is perpendicular to AB.

So, ∆ABC is right angled at A.

So, Circumcentre =mid point of BC=5α8,2

Now given circumcentre ≡5,α4

So on comparing we get,5α8=5 & α4=2

⇒α=8

Now plotting the diagram and finding the sides with distance formula we get,

Area =1268=24

Perimeter =24

Circumradius =102=5

Inradius =Δs=2412=2

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2022 (29 Jul Shift 2)
Let a-2b+c=1.
If fx=x+ax+2x+1x+bx+3x+2x+cx+4x+3, then:

Options:
A f-50=501
B f-50=-1
C f50=-501
D f50=1

The correct answer is:
f50=1

Solution:
Apply R1=R1+R3-2R2
⇒fx=a-2b+c100x+bx+3x+2x+cx+4x+3

⇒fx=1x+32-x+2x+4

⇒fx=1

⇒f50=1

Topic: Mathematics, Determinants, JEE Main, JEE Main 2020 (09 Jan Shift 2)
Let a→=2i^+λ1j^+3k^, b→=4i^+3-λ2j^+6k^ and c→=3i^+6j^+λ3-1k^ be three vectors such that b→=2a→ and a→ is perpendicular to c→. Then a possible value of λ1, λ2, λ3 is

Options:
A -12, 4, 0
B 1,5,1
C 12, 4,-2
D 1,3,1

The correct answer is:
-12, 4, 0

Solution:
Since b→=2a→, so 3-λ2=2λ1

λ2=3-2λ1 ...1

We know that if two vectors a1i^+b1j^+c1k^ and a2i^+b2j^+c2k^ are perpendicular, then a1a2+b1b2+c1c2=0

Since, a→ is perpendicular to c→ so

6+6λ1+3λ3-1=0

⇒6+6λ1+3λ3-3=0

⇒λ3=-1-2λ1 ...2

From equations 1 and 2, we get

λ1, λ2, λ3=λ1, 3-2λ1, -1-2λ1 where λ1∈R

⇒ -12, 4, 0 satisfies the above triplet.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2019 (10 Jan Shift 1)
Let $\mathbf{a}=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}=-\hat{\mathbf{j}}+\hat{\mathbf{k}}$. If $\mathbf{c}$ is a vector such that $\mathbf{a} \cdot \mathbf{c}=|\mathbf{c}|,|\mathbf{c}-\mathbf{a}|=2 \sqrt{2}$ and the angle between $\mathbf{a} \times \mathbf{b}$ and $\mathbf{c}$ is $\frac{\pi}{3}$, then $|(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}|=$

Options:
A $3 \sqrt{3}$
B $\frac{3}{2}$
C $\frac{3 \sqrt{3}}{2}$
D 0

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

Solution:
Given, $\mathbf{a}=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$, then $|\mathbf{a}|=\sqrt{4+4+1}=3$
Now given $|\mathbf{c}-\mathbf{a}|=2 \sqrt{2}$
Squaring on both sides,
$|\mathbf{c}-\mathbf{a}|^2=8 \Rightarrow|\mathbf{c}|^2+|\mathbf{a}|^2-2 \mathbf{a} \cdot \mathbf{c}=8$
$|\mathbf{c}|^2+9-2|\mathbf{c}|=8 \quad[\because|\mathbf{a}|=3$ and $\mathbf{a} \cdot \mathbf{c}=|\mathbf{c}|]$
$\therefore \quad|\mathbf{c}|^2-2|\mathbf{c}|+1=0$
$(|\mathbf{c}|-1)^2=0 \Rightarrow|\mathbf{c}|-1=0 \Rightarrow|\mathbf{c}|=1$
Now, we find $(\mathbf{a} \times \mathbf{b})=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & -2 & 1 \\ 0 & -1 & 1\end{array}\right|$
$(\mathbf{a} \times \mathbf{b})=\hat{\mathbf{i}}(-2+1)-\hat{\mathbf{j}}(2-0)+\hat{\mathbf{k}}(-2)$
$(\mathbf{a} \times \mathbf{b})=-\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$
$\therefore \quad|\mathbf{a} \times \mathbf{b}|=\sqrt{1+4+4}=3$
Now, given angle between $\mathbf{a} \times \mathbf{b}$ and $\mathbf{c}$ is $\frac{\pi}{3}$.
$\therefore|(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}|=|\mathbf{a} \times \mathbf{b}||\mathbf{c}| \cdot \sin \frac{\pi}{3}=3 \times 1 \times \frac{\sqrt{3}}{2}=\frac{3 \sqrt{3}}{2}$

Topic: Mathematics, Vector Algebra, JEE Main
Let $\vec{a}=2 \hat{i}-3 \hat{j}+4 \hat{k}, \vec{b}=3 \hat{i}+4 \hat{j}-5 \hat{k}$ and a vector $\vec{c}$ be such that $\vec{a} \times(\vec{b}+\vec{c})+\vec{b} \times \vec{c}=\hat{i}+8 \hat{j}+13 \hat{k}$. If $\vec{a} \cdot \vec{c}=13$, then $(24-\vec{b} \cdot \vec{c})$ is equal to_______

Options:

The correct answer is:
46

Solution:
$\begin{aligned} & \vec{a} \times \vec{b}+\vec{a} \times \vec{c}+\vec{b} \times \vec{c}=(1,8,13) \\ & \vec{a} \times(\vec{a} \times \vec{b})+\vec{a} \times(\vec{a} \times \vec{c})+\vec{a} \times(\vec{b} \times \vec{c}) \\ & =\vec{a} \times(\hat{i}+8 \hat{j}+13 \hat{k})\end{aligned}$
$\begin{aligned} & (\vec{a} \cdot \vec{b}) \vec{a}-a^2 \vec{b}+(\vec{a} \cdot \vec{c}) \vec{a}-a^2 \vec{c}+(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}=\vec{a} \times(\hat{i}+8 \hat{j}+13 \hat{k}) \\ & \Rightarrow-26 \vec{a}-29 \vec{b}+13 \vec{a}-29 \vec{c}+13 \vec{b}+26 \vec{c}=\vec{a} \times(\hat{i}+8 \hat{j}+13 \hat{k})\end{aligned}$
$\begin{aligned}
& \Rightarrow \quad-13 \overrightarrow{\mathrm{a}}-16 \overrightarrow{\mathrm{b}}-3 \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}} \times(\hat{\mathrm{i}}+8 \hat{\mathrm{j}}+13 \hat{\mathrm{k}}) \\
& \Rightarrow \quad-13 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}-16 \mathrm{~b}^2-3 \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=\{\overrightarrow{\mathrm{a}} \times(\hat{\mathrm{i}}+8 \hat{\mathrm{j}}+13 \hat{\mathrm{k}})\} \cdot \overrightarrow{\mathrm{b}} \\
& \Rightarrow \quad(-13)(-26)-16(50)-3 \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=\left|\begin{array}{ccc}
2 & -3 & 4 \\
1 & 8 & 13 \\
3 & 4 & -5
\end{array}\right| \\
& \Rightarrow \quad-462-3 \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=-396 \\
& \Rightarrow \quad \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=-22
\end{aligned}$
Hence $24-\vec{b} \cdot \vec{c}=46$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2024 (06 Apr Shift 1)
Let a→=2i^+3j^+4k^, b→=i^-2j^-2k^ and c→=-i^+4j^+3k^. If d→ is a vector perpendicular to both b→ and c→, and a→·d→=18, then |a→×d→|2 is equal to

Options:
A 640
B 680
C 720
D 760

The correct answer is:
720

Solution:
Given that,

a→=2i^+3j^+4k^, b→=i^-2j^-2k^ and c→=-i^+4j^+3k^.

Let us find b→×c→

⇒b→×c→=i^j^k^1-2-2-143

=-6+8i^-3-2j^+4-2k^

⇒b→×c→=2i^-j^+2k^

Since d→ is perpendicular to both b→ and c→

∴d→=λ(2i^-j^+2k^)

Also a→·d→=18

⇒λ2i^+3j^+4k^.(2i^-j^+2k^)=18

⇒9λ=18

⇒λ=2

Now let us apply LaGrange's identity which is as follows,

∴|a→×d→|2=|a→|2·|d→|2-(a→·d→)2

=2i^+3j^+4k^24i^-2j^+4k^2-182

=720

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (06 Apr Shift 1)
Let aâ=2i^+3j^+4k^, bâ=i^-2j^+k^ and câ=i^+j^-k^. If râÃaâ=bâ and rââcâ=3, then the value of râ is equal to

Options:
A 155
B 17
C 217
D 3

The correct answer is:
155

Solution:
Given, r→×a→=b→
Taking right cross product with c→, we get,
r→×a→×c→=b→×c→⇒r→⋅c→a→-a→⋅c→r→=b→×c→
⇒3a→-r→=b→×c→⇒r→=32i^+3j^+4k^-i^+2j^+3k^
⇒r→=5i^+7j^+9k^⇒r→=155

Topic: Mathematics, Vector Algebra, JEE Main
Let a→=2i^-7j^+5k^ , b→=i^+k^ and c→=i^+2j^-3k^ be three given vectors. If r→ is a vector such that r→×a→=c→×a→ and r→·b→=0, then |r→| is equal to:

Options:
A 1172
B 117
C 1152
D 9147

The correct answer is:
1172

Solution:
Given data as below:

a→=2i^-7j^+5k^

b→=i^+k^

c→=i^+2j^-3k^

Given, r→×a→=c→×a→

⇒(r→-c→)×a→=0

It means (r→-c→) is parallel to a→,

∴r→=c→+λa→.......(1)

Given, r→·b→=0

From equation (1),

c→+λa→·b→=0

⇒c→·b→+λb→·a→=0

⇒-2+λ(7)=0⇒λ=27

On putting values in equation (1) we get,

∴r→=c→+27a→=17(11i^-11k^)

|r→|=1127

Topic: Mathematics, Vector Algebra, JEE Main
Let a→=2i^-7j^+5k^ , b→=i^+k^ and c→=i^+2j^-3k^ be three given vectors. If r→ is a vector such that r→×a→=c→×a→ and r→·b→=0, then |r→| is equal to:

Options:
A 1172
B 117
C 1152
D 9147

The correct answer is:
1172

Solution:
Given data as below:

a→=2i^-7j^+5k^

b→=i^+k^

c→=i^+2j^-3k^

Given, r→×a→=c→×a→

⇒(r→-c→)×a→=0

It means (r→-c→) is parallel to a→,

∴r→=c→+λa→.......(1)

Given, r→·b→=0

From equation (1),

c→+λa→·b→=0

⇒c→·b→+λb→·a→=0

⇒-2+λ(7)=0⇒λ=27

On putting values in equation (1) we get,

∴r→=c→+27a→=17(11i^-11k^)

|r→|=1127

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (01 Feb Shift 2)
Let a→=2i^+7j^-k^, b^=3i^+5k^ and c→=i^-j^+2k^ Let d→ be a vector which is perpendicular to both a→ and b→, and c→·d→=12. Then-i^+j^-k^·c→×d→ is equal to

Options:
A 24
B 44
C 42
D 48

The correct answer is:
44

Solution:
Given,

a→=2i^+7j^-k^, b^=3i^+5k^ and c→=i^-j^+2k^

And d→ be a vector which is perpendicular to both a→ and b→,

So, d→=λa→×b→

Now finding, a→×b→=i^j^k^27-1305

⇒a→×b→=35i^-13j^-21k^

So, d→=λ35i^-13j^-21k^

Also given,

c→·d→=12

⇒c→·d→=i^-j^+2k^·λ35i^-13j^-21k^

⇒λ(35+13-42)=12

⇒ λ = 2

So, c→×d→=2×i^j^k^1-1235-13-21

⇒c→×d→=247i^+91j^+22k^

Now finding,

-i^+j^-k^·c→×d→=2-i^+j^-k^47i^+91j^+22k^

⇒-i^+j^-k^·c→×d→=2-47+91-22=44

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (10 Apr Shift 2)
Let a→=2i^-j^+2k^ and b→=i^+2j^-k^. Let a vector v→ be in the plane containing a→ and b→. If v→ is perpendicular to the vector 3i^+2j^-k^ and its projection on a→ is 19 units, then |2v→|2 is equal to _____.

Options:

The correct answer is:
1494

Solution:
v→,a→,b→ are coplanar

v→=λa→+μb→
v→=2λ+μi^+2μ-λj^+2λ-μk^

given v→⊥3i^+2j^-k^

v→·3i^+2j^-k^=0

⇒(2λ+μi^+2μ-λj^+2λ-μk^).(3i^+2j^-k^)=0

⇒3(2λ+μ)+2(2μ-λ)-(2λ-μ)=0

⇒λ=-4μ ...(i)

and also v→·a^=19⇒9λ-2μ=57 ...ii

solving (i) and (ii) we get

λ=6, μ=-32
⇒2v→=21i^-18j^+27k^
⇒|2v→|2=441+324+729
=1494

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2021 (01 Sep Shift 2)
Let a→=2i^+j^-2k^ and b→=i^+j^. If c→ is a vector such that a→·c→=|c→|,|c→-a→|=22 and the angle between a→×b→ and c→ is π6, then the value of a→×b→×c→ is:

Options:
A 23
B 4
C 3
D 32

The correct answer is:
32

Solution:
We have,

a→=3=a

a→·c→=c→=c

Now,

c→-a→=22

⇒ c→-a→2=8

⇒ c2+a2-2c→·a→=8

⇒ c2+9-2c=8

⇒ c2-2c+1=0

⇒ c-12=0

⇒ c=1=c→

Now,

a→×b→=i^j^k^21-2110

⇒ a→×b→=2i^-2j^+k^

⇒ a→×b→=22+-22+12=3

Now,

a→×b→×c→=a→×b→c→sinπ6

⇒a→×b→×c→=3·1·12=32

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2021 (20 Jul Shift 1)
Let $\vec{a}=2 \hat{i}+\hat{j}-2 \hat{k}, \vec{b}=\hat{i}+\hat{j}$. If $\vec{c}$ is a vector such that $\vec{a} \bullet \vec{c}=|\vec{c}|,|\vec{c}-\vec{a}|=2 \sqrt{2}$ and the angle between $\vec{a} \times \vec{b}$ and $\vec{c}$ is $30^{\circ}$, then $|(\vec{a} \times \vec{b}) \times \vec{c}|$ equals:

Options:
A
$\frac{1}{2}$

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

C
3

D
$\frac{3}{2}$

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

Solution:
$\vec{a}=2 \hat{i}+\hat{j}-2 \hat{k}, \vec{b}=\hat{i}+\hat{j}$
$\Rightarrow|\vec{a}|=3$
and $\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 1 & 0\end{array}\right|=2 \hat{i}-2 \hat{j}+\hat{k}$
$|\vec{a} \times \vec{b}|=\sqrt{4+4+1}=3$
Now, $|\vec{c}-\vec{a}|=2 \sqrt{2} \Rightarrow|\vec{c}-\vec{a}|^2=8$
$\Rightarrow|\vec{c}-\vec{a}| \cdot(\vec{c}-\vec{a})=8$
$\Rightarrow|\vec{c}|^2+|\vec{a}|^2-2 \vec{c} \cdot \vec{a}=8$
$\Rightarrow|\vec{c}|^2+9-2|\vec{c}|=8$
$\Rightarrow(|\vec{c}|-1)^2=0 \Rightarrow|\vec{c}|=1$
$\therefore|(\vec{a} \times \vec{b}) \times \vec{c}|=|\vec{a} \times \vec{b}||\vec{c}| \sin 30^{\circ}=3 \times 1 \times \frac{1}{2}=\frac{3}{2}$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2013 (25 Apr Online)
Let aâ=2i^-j^+3k^, bâ=i^+j^-4k^ and non-zero vector câ are such that aâÃbâÃcâ=aâÃbâÃcâ, then vector câ may be

Options:
A 4i^-2j^+6k^
B 4i^+2j^+6k^
C i^+j^-k^
D i^-4j^+k^

The correct answer is:
4i^-2j^+6k^

Solution:
aâÃbâÃcâ=aâÃbâÃcâ
a → ⋅ c → b → − b → ⋅ c → a → = a → ⋅ c → b → − a → ⋅ b → c →
âcâ must be parallel to aâ

Topic: Mathematics, Vector Algebra, JEE Main
Let a→=2i^-j^+5k^ and b→=αi^+βj^+2k^. If a→×b→×i^·k^=232, then b→×2j^ is equal to

Options:
A 4
B 5
C 21
D 17

The correct answer is:
5

Solution:
Given, a→=2i^-j^+5k^,b→=αi^+βj^+2k^

Also given a→×b→×i^·k^=232, then b→×2j^ is

Now using triple cross product we have, a→·i^b→-b→·i^a→·k^=232

⇒2.b→-α·a→.k^=232

⇒2×2-α×5=232

⇒5α=4-232

⇒α=-32

Now finding, b→×2j^=i^j^k^αβ2020=-4i^+2αk^

∴ b→×2j^=16+4α2=16+4×94=5

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2022 (27 Jul Shift 1)
Let a→=2i^+j^+k^, and b→ and c→ be two nonzero vectors such that a→+b→+c→=a→+b→-c→ and b→·c→=0. Consider the following two statement:

A a→+λc→≥a→ for all λ∈ℝ.

B a→ and c→ are always parallel

Options:
A only (B) is correct
B neither (A) nor (B) is correct
C only (A) is correct
D both (A) and (B) are correct.

The correct answer is:
only (A) is correct

Solution:
Given,

a→+b→+c→=a→+b→-c→

Now squaring both side we get,

⇒a→+b→+c→2=a→+b→-c→2

⇒a→2+b→2+c→2+2a→·b→+2b→·c→+2c→·a→=a→2+b→2+c→2+2a→·b→-2b→·c→-2c→·a→

⇒2b→·c→+2c→·a→=-2b→·c→+2c→·a→

⇒4b→·c→+c→·a→=0

Now here given b→·c→=0, so

c→·a→=0 which means c→ & a→ are perpendicular vector, so statement B is false,

Now statementA: a→+λc→≥a→

⇒a→+λc→2≥a→2

⇒a→2+λc→2+2a→·c→≥a→2

⇒λc→2+2a→·c→≥0

⇒λc→2≥0

This is not always true, since λ∈R.

Hence, we can say that statementA is true and statement B is false.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (31 Jan Shift 1)