datavorous/entrance-exam-dataset huggingface dataset:
Let a→and b→ be two vectors. Let |a→|=1,|b→|=4 and a→·b→=2. If c→=(2a→×b→)-3b→, then the value of b→·c→ is

Options:
A -24
B -48
C -84
D -60

The correct answer is:
-48

Solution:
Given,

a→and b→ be two vectors, where |a→|=1,|b→|=4 and a→·b→=2,

Now given, c→=2a→×b→-3b→

So, b→·c→=b→·2a→×b→-3b→·b→

⇒b→·c→=-3|b|2 as b→2a→b→=0 or b→·2a→×b→=0

⇒b→·c→=-48

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (30 Jan Shift 2)
Let a→ and b→ be two vectors such that |2a→+3b→|=|3a→+b→| and the angle between a→ and b→ is 60°. If 18a→ is a unit vector, then |b→| is equal to :

Options:
A 8
B 4
C 6
D 5

The correct answer is:
5

Solution:
Given that

18a→ is a unit vector.

|a→|=8
|2a→+3b→|2=|3a→+b→|2

We know thata→ · b→ = a→ b→ cos θ ,
⇒4|a→|2+9|b→|2+12|a→||b→|cos60°=9|a→|2+|b→|2+6|a→||b→|cos60°
⇒5|a→|2=8|b→|2+6|a→|||b→∣cos60°
⇒564=8|b→|2+68·|b→|12
⇒|b→|2+3|b→|-40=0⇒b→+8|b→|-5=0
⇒|b→|=5

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2021 (31 Aug Shift 1)
Let a→ and b→ be two vectors such that a→=1, b→=4 and a→⋅b→=2. If c→=2a→×b→−3b→ and the angle between b→ andc→ is α, then 192sin2α is equal to _________

Options:

The correct answer is:
48

Solution:
Given,

a→ and b→ be two vectors such that a→=1, b→=4 and a→⋅b→=2.

Also, given

c→=2a→×b→−3b→

Taking dot product with b→ both side we get,

⇒b→⋅c→=2a→×b→⋅b→−3b2

⇒b→c→cosα=0−3b→2

⇒c→cosα=−12 {as b→=4}

Now, solving a→⋅b→=2

⇒a→b→cosθ=2

⇒cosθ=12⇒θ=π3

Now, solving

c→2=2a→×b→−3b→2

⇒c→2=64×34+144=192

Now, squaring both side of c→cosα=−12 we get,

c→2cos2α=144

⇒192cos2α=144

⇒1921-sin2α=144

⇒192sin2α=48

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2024 (31 Jan Shift 1)
Let a→ and b→ be two vectors such that a→+b→2=a→2+2b→2,a→·b→=3 and a→×b→2=75. Then a→2 is equal to ______.

Options:

The correct answer is:
14

Solution:
Given,

a→+b→2=a→2+2b→2 & a→·b→=3

We know that

a→+b→2=a→2+b→2+2a→·b→

Now using the above formula in a→+b→2=a→2+2b→2 we get,

a→2+b→2+2a→·b→=a→2+2b→2

Now on comparing both side we get, b→2=2a→·b→=6 .........1

Also given a→×b→2=75

⇒a→2b→2sin2θ=75

⇒a→2b→21-cos2θ=75

⇒a→2b→2-a→2b→2cos2θ=75

⇒a→2b→2-a→·b→2=75

Now putting the value of b→2=2a→·b→=6 we get, 6a→2-9=75⇒a→2=14

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2022 (29 Jul Shift 2)
Let $A$ and $B$ denote the statements
A: $\cos \alpha+\cos \beta+\cos \gamma=0$
B: $\sin \alpha+\sin \beta+\sin \gamma=0$
If $\cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha-\beta)=-\frac{3}{2}$, then

Options:
A
$A$ is true and $B$ is false

B
$A$ is false and $B$ is true

C
both $A$ and $B$ are true

D
both $A$ and $B$ are false

The correct answer is:
both $A$ and $B$ are true

Solution:
$$
\begin{aligned}
& \cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha-\beta)=-\frac{3}{2} \\
& \Rightarrow 2[\cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha-\beta)]+3=0 \\
& \Rightarrow 2[\cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha-\beta)]+\sin ^2 \alpha+\cos ^2 \alpha+\sin ^2 \beta+\cos ^2 \beta+\sin ^2 \gamma+\cos ^2 \gamma=0 \\
& \Rightarrow(\sin \alpha+\sin \beta+\sin \gamma)^2+(\cos \alpha+\cos \beta+\cos \gamma)^2=0
\end{aligned}
$$

Topic: Mathematics, Determinants, JEE Main, JEE Main 2009
Let a and b respectively be the semi-transverse and semi-conjugate axes of a standard hyperbola whose eccentricity satisfies the equation 9e2-18e+5=0. If S5, 0 is a focus and 5x=9 is the corresponding directrix of this hyperbola, then a2-b2 is equal to

Options:
A -7
B -5
C 5
D 7

The correct answer is:
-7

Solution:
Given that focus of hyperbola is S5, 0

⇒ae=5......i
Therefore, Directrix x=ae

⇒ ae=95 ......ii
From i and ii, we get ⇒ a2=9

⇒a=3 and e=53

(It satisfies 9e2-18e+5=0 )

We also know that for an ellipse,
b2=a2e2-1

⇒b2=16
⇒ a2-b2=9-16= -7

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2016 (09 Apr Online)
Let A and E be any two events with positive probabilities
Statement I: PE/A≥PA/EPE.
Statement II: PA/E≥PA∩E.

Options:
A Both the statements are false
B Both the statements are true
C Statement - I is false, Statement - II is true
D Statement - I is true, Statement - II is false

The correct answer is:
Both the statements are true

Solution:
We know that PE/A=PE∩APA

⇒PE∩A=PE/A·PA ...i

Also, PA/E=PE∩APE

⇒PE∩A=PA/E·PE ii

From the equations i and ii, we get

PE/A·PA=PA/E·PE

Now, since 0<PA≤1

⇒PE/A≥PA/E·PE

Again, we have PA/E=PE∩APE

⇒PA/E·PE=PE∩A

Now, since 0<PE≤1

PA/E≥PA∩E

⇒Both the statements are true.

Topic: Mathematics, Probability, JEE Main, JEE Main 2014 (19 Apr Online)
Let A and \(\mathrm{B}\) the two gases and given :
\(\frac{T_A}{M_A}=4 \cdot \frac{T_B}{M_B}\) Where \(T\) is the temperature and \(M\) is molecular mass. If \(C_A\) and \(C_B\) are the r.m.s. speed, then the ratio \(\frac{C_A}{C_B}\) will be equal to:

Options:
A $1$
B $0.5$
C $2$
D $4$

The correct answer is:
$2$

Solution:
Let \(A\) and \(B\) two gases, then,
\(\frac{T_A}{M_A}=\frac{4 T_B}{M_B}\)
Now,
\(\begin{aligned}
& \frac{C_A}{C_B}=\frac{\sqrt{\frac{3 R T_A}{M_A}}}{\sqrt{\frac{3 R T_B}{M_B}}} \\
& \frac{C_A}{C_B}=2
\end{aligned}\)

Topic: Physics, Kinetic Theory of Gases, JEE Main
Let A, B and C are n×n matrices such that A=2, B=3 and C=5. If 2A23B5C-1=1728125, then the value of n is equal to

Options:
A 2
B 3
C 4
D 6

The correct answer is:
2

Solution:
2A23B5C-1=2A23B5C=22n3n5nA2BC
⇒125n+1=1253⇒n=2

Topic: Mathematics, Determinants, JEE Main
Let a, b and c are the roots of the equation x3-7x2+9x-13=0 and A and B are two matrices given by A=abcbcacab and B=bc-a2ca-b2ab-c2ca-b2ab-c2bc-a2ab-c2bc-a2ca-b2, then the value of AB is equal to

Options:
A -154
B -1543
C -22
D -223

The correct answer is:
-1543

Solution:
a+b+c=7
ab+bc+ca=9
abc=13
B is the cofactor matrix of A⇒B=A2
⇒A⋅B=A3….i
A=abcbcacab=abc-a2-bb2-ac+cab-c2=3abc-a3-b3-c3
=-a+b+ca+b+c2-3ab+bc+ca
=-749-27
=-154
From i
BA=A3=-1543

Topic: Mathematics, Determinants, JEE Main
Let aâ, bâ and câ are three non-collinear vectors in a plane such that aâ=2, bâ=5 and câ=29. If the angle between aâ and câ is Î¸1 and the angle between bâ and câ is Î¸2, where Î¸1,Î¸2âÏ2,Ï, then the value of Î¸1+Î¸2 is equal to

Options:
A 7π6
B 4π6
C 3π2
D 7π4

The correct answer is:
3π2

Solution:
a→2+b→2=c→2 (property of triangles)
From the diagram,
π-θ1+π-θ2=π2
⇒θ1+θ2=3π2

Topic: Mathematics, Vector Algebra, JEE Main
Let a,b and c be any real numbers. Suppose that there are real numbers x, y and 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 1
B 2
C −1
D 0

The correct answer is:
1

Solution:
System of equations
x - c y - b z = 0 c x - y + a z = 0 b x + a y - z = 0
has non trivial solution if the determinant of coefficient matrix is zero.
⇒ 1 - c - b c - 1 a b a - 1 = 0
⇒ 1 1 - a 2 + c - c - a b - b c a + b = 0
⇒ a 2 + b 2 + c 2 + 2 a b c = 1

Topic: Mathematics, Determinants, JEE Main
Let $a, b$ and $c$ be distinct non-negative numbers. If the vectors $a \hat{i}+a \hat{j}+c \hat{k}, \hat{i}+\hat{k}$ and $c \hat{i}+c \hat{j}+b \hat{k}$ lie in a plane, then $c$ is

Options:
A
the Geometric Mean of $a$ and $b$

B
the Arithmetic Mean of $a$ and $b$

C
equal to zero

D
the Harmonic Mean of $a$ and $b$

The correct answer is:
the Geometric Mean of $a$ and $b$

Solution:
Vector $a \hat{i}+a \hat{j}+c \hat{k}, \hat{i}+\hat{k}$ and $c \hat{i}+c \hat{j}+b \hat{k}$ are coplanar
$$
\begin{aligned}
& \left|\begin{array}{lll}
a & a & c \\
1 & 0 & 1 \\
c & c & b
\end{array}\right|=0 \Rightarrow c^2=a b \\
& \therefore a, b, c \text { are in G.P. }
\end{aligned}
$$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2005
Let a,b and c be distinct positive numbers. If the vectors ai^+aj^+ck^,i^+k^ and ci^+cj^+bk^ are co-planar, then c is equal to:

Options:
A 21a+1b
B a+b2
C 1a+1b
D ab

The correct answer is:
ab

Solution:
Given vectors ai^+aj^+ck^,i^+k^ and ci^+cj^+bk^ are co-planar, then

aac101ccb=0

⇒a0-c-ab-c+cc-0=0

⇒c2=ab

⇒c=ab.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2021 (25 Jul Shift 2)
Let a, b and c be in G.P. with common ratio r, where a≠0 and 0<r≤12. If 3a, 7b and 15c are the first three terms of an A.P., then the 4th term of this A.P. is :

Options:
A a
B 73a
C 5a
D 23a

The correct answer is:
a

Solution:
We know that, the nth term of a G.P. with first term a and common ratio r is tn=arn-1.

Then, b=ar and c=ar2

Now, 3a, 7b and 15c are in A.P.

⇒14b=3a+15c

⇒14ar=3a+15(ar2)

⇒14r=3+15r2

⇒15r2-14r+3=0

⇒3r-15r-3=0

⇒r=13 or r=35

The only acceptable value is r=13, because r∈0, 12

∴ the first term of the A.P. is A=3a and the common difference is d=7b-3a=7ar-3a=73a-3a=-23a

Also, we know that the nth term of an A.P. with first term A and common difference d is Tn=A+n-1d

∴ T4=3a+4-1-23a= 3a-2a=a.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2019 (10 Apr Shift 2)
Let $\overline{\mathrm{a}}, \overline{\mathrm{b}}$ and $\overline{\mathrm{c}}$ be non-zero vectors such that $(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}=\frac{1}{3}|\overline{\mathrm{b}}||\overline{\mathrm{c}}| \overline{\mathrm{a}}$. If $\theta$ is the acute angle between the vectors $\overline{\mathrm{b}}$ and $\overline{\mathrm{c}}$, then $\sin \theta$ equals

Options:
A
$\frac{1}{3}$

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

C
$\frac{2}{3}$

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

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

Solution:
$(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=\frac{1}{3}|\overrightarrow{\mathrm{b}}||\overline{\mathrm{c}}| \overline{\mathrm{a}} \Rightarrow(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}) \overrightarrow{\mathrm{b}}-(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}) \overrightarrow{\mathrm{a}}=\frac{1}{3}|\overline{\mathrm{b}}||\overline{\mathrm{c}}| \overline{\mathrm{a}}$
$\Rightarrow(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}) \overrightarrow{\mathrm{b}}=\left(\frac{1}{3}|\overline{\mathrm{b}}||\overline{\mathrm{c}}|+(\overline{\mathrm{b}} \cdot \overline{\mathrm{c}})\right) \overrightarrow{\mathrm{a}} \Rightarrow \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=0$ and $\frac{1}{3}|\overline{\mathrm{b}}||\overline{\mathrm{c}}|+(\overline{\mathrm{b}} \cdot \overline{\mathrm{c}})=0$
$\Rightarrow|\overline{\mathrm{b}}||\overline{\mathrm{c}}|\left(\frac{1}{3}+\cos \theta\right)=0 \Rightarrow \cos \theta=-1 / 3 \Rightarrow \sin \theta=\frac{2 \sqrt{2}}{3}$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2004
Let $a, b$ and $c$ be positive real numbers. If $\frac{x^2-b x}{a x-c}=\frac{m-1}{m+1}$ has two roots which are numerically equal but opposite in sign, then the value of $m$ is

Options:
A $c$
B $\frac{1}{c}$
C $\frac{a+b}{a-b}$
D $\frac{a-b}{a+b}$

The correct answer is:
$\frac{a-b}{a+b}$

Solution:
$$
\begin{aligned}
& \text { Given, } \frac{x^2-b x}{a x-c}=\frac{m-1}{m+1} \\
& \Rightarrow\left(x^2-b x\right)(m+1)=(m-1)(a x-c) \\
& \Rightarrow x^2(m+1)-x(b(m+1)+a(m-1)+c(m-1)=0
\end{aligned}
$$
Let $p$ and $-p$ are roots of Eq. (i), then
$$
\begin{aligned}
& \text { Sum of roots } \Rightarrow p+(-p)=\frac{b(m+1)+a(m-1)}{m+1} \\
& 0=\frac{(b+a) m+(b-a)}{m+1} \\
& \therefore m(b+a)+(b-a)=0 \\
& \Rightarrow \quad m=\frac{a-b}{a+b}
\end{aligned}
$$

Topic: Mathematics, Quadratic Equation, JEE Main
Let A, B and C be sets such that ϕ≠A∩B⊆C. Then which of the following statements is not true?

Options:
A B∩C≠ϕ
B (C∪A)∩(C∪B)=C
C If (A-B)⊆C,  then A⊆C
D If (A-C)⊆B, then A⊆B

The correct answer is:
If (A-C)⊆B, then A⊆B

Solution:
For A=C,A-C=ϕ
⇒ϕ⊆B
But A⊈B
Hence, If (A-C)⊆B, then A⊆B is not true
Let x∈ C, ⇒x∈C∪A∩C∪B
⇒x∈C∪A  and x∈C∪B
⇒(x∈C  or x∈A) and (x∈C or  x∈B)
⇒ x∈C or  x∈A∩B
⇒x∈C or x∈C  as A∪B⊆C
⇒x∈C
⇒C∪A∩C∪B⊆C .... (i)
Now x∈C⇒x∈C∪A and x∈C∪B
⇒x∈(C∪A)∩C∪B
⇒C⊆(C∪A)∩C∪B ..... (ii)
⇒ From (i) and (ii)
C=C∪A∩C∪B
Hence, (C∪A)∩(C∪B)=C is true
Let x∈A  and x∉B
⇒x∈A-B
⇒x∈C  as A-B⊆C
Let x∈A and x∈B
⇒x∈A∩B
⇒x∈C as A∩B⊆C
Hence x∈A ⇒x∈C
⇒A⊆C
Hence, If (A-B)⊆C,  then A⊆C is true
as  C⊇A∩B
⇒B∩C⊇A∩B
as A∩B≠ϕ
⇒B∩C≠ϕ
Hence, B∩C≠ϕ is true

Alternate solution

Given

ϕ≠A∩B⊆C

Let answer that

C=1,2,3,4,5,6

A=1,2,3

B=3,4,5

A∩B=3

Option (a)

B∩C=3,4,5≠ϕ

Option (b) C∪A=1,2,3,4,5,6

C∪B=1,2,3,4,5,6

C∪A∩C∪B=1,2,3,4,5,6=C

Option (c) A−B=1,2,3⊆CA⊆C True

Option (d) A−C=ϕ⊆B  A⊆C Not correct

∴ Option (d) is correct

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2019 (12 Apr Shift 2)
Let $a, b$, and $c$ be such that
$\frac{1}{(1-x)(1-2 x)(1-3 x)}=\frac{a}{1-x}+\frac{b}{1-2 x}+\frac{c}{1-3 x}$
then $\frac{a}{1}+\frac{b}{3}+\frac{c}{5}$ is equal to

Options:
A $\frac{1}{15}$
B $\frac{1}{6}$
C $\frac{1}{5}$
D $\frac{1}{3}$

The correct answer is:
$\frac{1}{15}$

Solution:
We have,
$1=a(1-2 x)(1-3 x)+b(1-x)(1-3 x)$$+c(1-x)(1-2 x)$
Put $x=\frac{1}{2}$
$1=0+b\left(1-\frac{1}{2}\right)\left(1-\frac{3}{2}\right)+0$
$\Rightarrow \quad 1=b\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right) \Rightarrow b=-4$
Put $x=1$,
$1=a(1-2)(1-3)+0+0$
$\Rightarrow \quad 1=a(-1)(-2) \Rightarrow a=\frac{1}{2}$
Put $x=\frac{1}{3}$,
$1=0+0+c\left(1-\frac{1}{3}\right)\left(1-\frac{2}{3}\right)$
$\Rightarrow \quad 1=c\left(\frac{2}{3}\right)\left(\frac{1}{3}\right) \Rightarrow c=\frac{9}{2}$
Now, $\frac{a}{1}+\frac{b}{3}+\frac{c}{5}=\frac{1}{2}+\frac{(-4)}{3}+\frac{9}{5 \cdot 2}$
$=\frac{1}{2}-\frac{4}{3}+\frac{9}{10}$
$=\frac{15-40+27}{30}=\frac{2}{30}=\frac{1}{15}$

Topic: Mathematics, Functions, JEE Main
Let a, b and c be the 7th, 11th and 13th terms respectively of a non-constant A.P. . If these are also the three consecutive terms of a G.P. , then ac is equal to:

Options:
A 2
B 713
C 12
D 4

The correct answer is:
4

Solution:
Let the first terms of A.P. be A and common difference be d, then
a=A+6d
b=A+10d
c=A+12d
Given a, b, c are in G. P. ⇒b2=ac
⇒A+10d2=A+6dA+12d
⇒A2+100d2+20Ad=A2+18Ad+72d2
⇒A=-14d
Now ac=A+6dA+12d=-14d+6d-14d+12d=-8d-2d=4

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2019 (09 Jan Shift 2)
Let \( a, b \) and \( c \) be the three distinct non-zero real numbers satisfying the system of equations \( \frac{1}{a}+\frac{1}{a-1}+\frac{1}{a-2}=1 \), \( \frac{1}{b}+\frac{1}{b-1}+\frac{1}{b-2}=1 \) and \( \frac{1}{c}+\frac{1}{c-1}+\frac{1}{c-2}=1 \). Then, \( a b c \) is equal to

Options:
A \( 1 \)
B \( 2 \)
C \( 3 \)
D \( 4 \)

The correct answer is:
\( 2 \)

Solution:
Given,

1a+1a-1+1a-2=1

1b+1b-1+1b-2=1

1c+1c-1+1c-2=1

We can see that a, b and c satisfy the equation of the form given as,

1x+1(x-1)+1x-2=1

Thus, the roots of the above equation are a, b and c.

⇒x(x-1)(x-2)=(x-1)(x-2)+x(x-1)+x(x-2)

x3-3x2+2x=3x2-6x+2

x3-6x2+8x-2=0

The product of the roots of abc is 2.

Topic: Mathematics, Quadratic Equation, JEE Main
Let A, B and C be three events, which are pair-wise independent and E¯ denotes the complement of an event E. If PA∩B∩C=0 and PC>0, then PA¯∩B¯|C is equal to

Options:
A PA¯-PB
B PA¯-PB¯
C PA¯+PB¯
D PA+PBÂ¯

The correct answer is:
PA¯-PB

Solution:
PA¯∩B¯|C=PA¯∩B¯∩CPC

=PC-PA∩C-PB∩C+PA∩B∩CPC

=PC-PAPC-PBPCPC

=1-PA-PB

=PA¯-PB or PB¯-PA

Topic: Mathematics, Probability, JEE Main, JEE Main 2018 (16 Apr Online)
Let a→,b→ and c→ be three non zero vectors such that b→·c→=0 and a→×(b→×c→)=b→-c→2. If d→ be a vector such that b→·d→=a→·b→, then (a→×b→)·(c→×d→) is equal to

Options:
A 34

B 12

C -14
D 14

The correct answer is:
14

Solution:
Given,

a→×b→×c→=b→-c→2

Using the vector triple cross product we have,

a→×b→×c→=a→·c→b→-a→·b→c→

So, (a→·c→)b→-(a→·b→)c→=b→-c→2

On comparing both sides we get,

a→·c→=12,a→·b→=12

∵ b→·d→ = a→·b→

∴b→·d→=12

Using vector cross-product formulae we have

(a→×b→)·(c→×d→)=a→·(b→×(c→×d→))

=a→·((b→·d→)c→-(b→·c→)d→) ∵ b→·c→=0

=(a→·c→)(b→·d→)=12×12=14

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (25 Jan Shift 1)
Let a→, b→ and c→ be three non - zero vectors such that no two of them are collinear and a→×b→×c→=13b→c→a→. If θ is the angle between vectors b→ and c→, then a value of sinθ is

Options:
A - 2 3 3
B 2 2 3
C - 2 3
D 2 3

The correct answer is:
2 2 3

Solution:
Given a→ × b→ × c→= 13 b→ c→a→
a→⋅c→ b→- b→⋅c→ a→=13 b→c→ a→
⇒ a→⋅c→=0 and b→⋅c→= -13 b→c→
b→c→cosθ= -13 b→c→
cosθ= -13
sinθ= 1-19= 89 = 223

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2015 (04 Apr)
Let a→, b→ and c→ be three non-zero non-coplanar vectors and p→, q→ and r→ be three vectors defined as p→=a→+b→−2c→; q→=3a→-2b→+c→ and r→=a→-4b→+2c→

If the volume of parallelepiped determined by a→, b→ and c→ is V1 and that of the parallelepiped determined by p→, q→ and r→ is V2 then, V2=KV1 implies that the value of K is

Options:

The correct answer is:
15

Solution:
Given, [a⃗b⃗c⃗]=V1

[p⃗q⃗r⃗]=|11−23−2+11−42|[a⃗b⃗c⃗]=[1(−4+4)−1(6−1)−2(−12+2)]V1

V2=(−5+20)V1=15V1 ⇒K=15

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Let a→,b→ and c→ be three non-zero non-coplanar vectors. Let the position vectors of four points A, B, C and D be a→-b→+c→, λa→-3b→+4c→, -a→+2b→-3c→ and 2a→-4b→+6c→ respectively. If AB→, AC→ and AD→ are coplanar, then λ is :

Options:

The correct answer is:
2

Solution:
Given,

a→,b→ and c→ be three non-zero non-coplanar vectors,

Also the position vectors of four points A, B, C and D be a→-b→+c→, λa→-3b→+4c→, -a→+2b→-3c→ and 2a→-4b→+6c→ respectively,

So, AB→=(λ-1)a→-2b→+3c→

AC→=-2a→+3b→-4c→

AD→=a→-3b→+5c→

Now using the condition of coplanar we get,

λ-1-23-23-41-35=0

⇒(λ-1)(15-12)+2(-10+4)+3(6-3)=0

⇒(λ-1)=1⇒λ=2

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (29 Jan Shift 1)
Let a→, b→ and c→ be three non-zero vectors such that b→ and c→ are non-collinear if a→+5b→ is collinear with c→,b→+6c→ is collinear with a→ and a→+αb→+βc→=0→, then α+β is equal to

Options:
A 35
B 30
C -30
D -25

The correct answer is:
35

Solution:
Given: a→+5b→ is collinear with c→, b→+6c→ is collinear with a→ and a→+αb→+βc→=0→.

⇒a→+5b→=λc→ ...i

⇒b→+6c→=μa→ ...ii

⇒a→=λc→-5b→=b→+6c→μ

⇒λc→-5b→=6μc→+1μb→

Comparing both the sides,

⇒μ=-15, λ=-30

Putting these values in equation i

⇒a→+5b→=-30c→

⇒a→+5b→+30c→=0

Comparing this equation with a→+αb→+βc→=0→.

⇒α=5, β=30

⇒α+β=35

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2024 (29 Jan Shift 1)
Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-zero vectors such that no two of these are collinear. If the vector $\vec{a}+2 \vec{b}$ is collinear with $\vec{c}$ and $\vec{b}+3 \vec{c}$ is collinear with $\vec{a}$ ( $\lambda$ being some non-zero scalar) then $\vec{a}+2 \vec{b}+6 \vec{c}$ equals

Options:
A
$\lambda \vec{a}$

B
$\lambda \vec{b}$

C
$\lambda \overrightarrow{\mathbf{c}}$

D
0

The correct answer is:
0

Solution:
$(\vec{a}+2 b)=t_1 \vec{c}$
and $\vec{b}+3 \vec{c}=t_2 \vec{a}$
$(1)-2 \times(2) \Rightarrow \vec{a}\left(1+2 t_2\right)+\vec{c}\left(-t_1-6\right)=0 \Rightarrow 1+2 t_2=0 \Rightarrow t_2=-1 / 2 \& t_1=-6$.
Since $\vec{a}$ and $\vec{c}$ are non-collinear.
Putting the value of $t_1$ and $t_2$ in (1) and (2), we get $\vec{a}+2 \vec{b}+6 \vec{c}=\overrightarrow{0}$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2004
Let a→, b→ and c→ be three unit vectors, out of which vectors b→ and c→ are non-parallel. If α and β are the angles which vector a→ makes with vectors b→ and c→ respectively and a→×b→×c→=12b→, then α-β is equal to :

Options:
A 90o
B 60o
C 45o
D 30o

The correct answer is:
30o

Solution:
a→×b→×c→=b→2⇒a→.c→b→-a→.b→c→=b→2

Since, b→ is not parallel to c→⇒a→.c→=12⇒accosβ=12⇒cosβ=12 and

a→.b→=0⇒abcosα=0⇒cosα=0

⇒β=π3 and α=π2

∴α-β=π2-π3=π6

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2019 (12 Jan Shift 2)
Let a→, b→ and c→ be three unit vectors such that a→-b→2+a→-c→2=8. Then a→+2b→2+a→+2c→2 is equal to

Options:

The correct answer is:
2

Solution:
a→=b→=c→=1

a→-b→2+a→-c→2=8

⇒ 1+1-2a→·b→+1+1-2a→·c→=8

⇒ a→·b→+a→·c→=-2

Now a→+2b→2+a→+2c→2

=a→2+4b→2+4a→·b→+a→2+4c→2+4a→·c→

=2a→2+4b→2+4c→2+4a→·b→+a→·c→

=2+4+4+4-2

=2

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2020 (02 Sep Shift 1)
Let a→,b→ and c→, be three unit vectors such that a→+b→+c→=0→. If λ=a→⋅b→+b→⋅c→+c→⋅a→ and d→=a→×b→+b→×c→+c→×a→, then the order pair, λ,d→, is equal to.

Options:
A 32,3a→×c→
B -32,3c→×b→
C 32,3b→×c→
D -32,3a→×b→

The correct answer is:
-32,3a→×b→

Solution:
a→+b→+c→2=0
3+2a→·b→+b→·c→+c→·a→=0
a→·b→+b→·c→+c→·a→=-32⇒λ=-32
d→=a→×b→+b→×-a→-b→+-a→-b→×a→
=a→×b→+a→×b→+a→×b→
d→=3a→×b→.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2020 (07 Jan Shift 2)
Let a→, b→ and c→ be three unit vectors such that a→ ×b→ × c→=32b→ + c→. If b→ is not parallel to c→ , then the angle between a→ and b→ is

Options:
A 2Ï3
B 5π6
C 3Ï4
D Ï2

The correct answer is:
5π6

Solution:
Given,

a→×(b→×c→ )=32 (b→+c→ )

⇒a→·c→b→-a→·b→c→=32 b→+32c→

On comparing, we get

-a→·b→=32

⇒a→·b→=-32

⇒a→b→cosθ=-32

Given a→,b→ are unit vector,

a→=b→=1

From equation i

cosθ=-32

⇒cosθ=cos5π6

⇒θ=5π6

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2016 (03 Apr)
Let aâ, bâ and câ be three vectors such that aâ=2, bâ=1 and câ=3. If the projection of bâ along aâ is double of the projection of câ along aâ and bâ, câ are perpendicular to each other, then the value of aâ-bâ+2câ22 is equal to

Options:
A 41
B 14
C 14
D 20.5

The correct answer is:
20.5

Solution:
Given, b→⋅c→=0

b→⋅a^=2c→⋅a^⇒b→⋅a→=2c→⋅a→
So, a→-b→+2c→2=a→2+b→2+4c→2-2a→⋅b→-4b→⋅c→+4a→⋅c→
a→-b→+c→22=4+1+362=20.5

Topic: Mathematics, Vector Algebra, JEE Main
Let a→, b→ and c→ be three vectors such that a→=2, b→=4, c→=1 and b→×c→=15. If b→=2c→+λa→, then the value of λ is equal to (here, the angle between b→ and c→ is acute)

Options:
A ±2
B ±1
C ±22
D ±4

The correct answer is:
±22

Solution:
b→-2c→=λa→
⇒b→-2c→2=λa→2
⇒b→2+4c→2-4b→.c→=λ22
⇒16+4-4b→.c→=2λ2⇒λ2=10-2b→.c→ ... 1
Now, b→×c→2+b→.c→2=b→2c→2
15+b→.c→2=16
⇒b→.c→=±1
The angle between b→ and c→ is acute
Hence, b→.c→=1
from 1

λ2=8⇒λ=±22

Topic: Mathematics, Vector Algebra, JEE Main
Let a→,b→ and c→ be three vectors such that a→=b→×b→×c→. If magnitudes of the vectors a→,b→ and c→ are 2,1 and 2 respectively and the angle between b→ and c→ is θ0<θ<π2, then the value of 1+tanθ is equal to :

Options:
A 3+1
B 2
C 1
D 3+13

The correct answer is:
2

Solution:
We have, a→=b→·c→b→-b→·b→c→

=1.2cosθb→-c→

⇒a→=2cosθb→-c→

a→2=2cosθb→2+c→2-2.2cosθb→·c→

a→2=2cosθ2b→2+c→2-2.2cosθb→·c→

Given that, a→=2, b→=1 and c→=2

22=2cosθ2+22-2.2cosθb→·c→

⇒2=4cos2θ+4-4cosθ·2cosθ

∵b→·c→=b→c→cosθ

⇒-2=-4cos2θ

⇒cos2θ=12

⇒cosθ=±12

⇒cosθ=12

⇒θ=π4

∵0<θ<π2

Hence, 1+tanθ=2.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2021 (27 Jul Shift 2)
Let $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ be three vectors. Then scalar triple product $[\mathbf{a} \mathbf{b} \mathbf{c}]$ is equal to

Options:
A $\left[\begin{array}{lll}\mathbf{b} & \mathbf{a} & \mathbf{c}\end{array}\right]$
B $\left[\begin{array}{lll}\mathbf{a} & \mathbf{c} & \mathbf{b}\end{array}\right]$
C $\left[\begin{array}{lll}\mathbf{c} & \mathbf{b} & \mathbf{a}\end{array}\right]$
D $\left[\begin{array}{lll}\mathbf{b} & \mathbf{c} & \mathbf{a}\end{array}\right]$

The correct answer is:
$\left[\begin{array}{lll}\mathbf{b} & \mathbf{c} & \mathbf{a}\end{array}\right]$

Solution:
From the properties of the scalar triple product:
$\Rightarrow [\mathbf{a b c}]=-[\mathbf{b a c}]$
$\Rightarrow [\mathbf{a b c}]=-[\mathbf{a c b}]$
$\Rightarrow [\mathbf{a b c}]=[\mathbf{b c a}]$
$\Rightarrow [\mathbf{a b c}]=-[\mathbf{c b a}]$
The correct option that equals $[\mathbf{a b c}]$ is:[bca]

Topic: Mathematics, Vector Algebra, JEE Main
Let a, b and c satisfy the system of equations a+2b+3c=6, 4a+5b+6c=12 and 6a+9b=4. If the roots of the equation a+b+cx2-abcx+a-1+b-1+c-1=0 are α and β, then 1α+1β is equal to

Options:
A 243
B 100
C 24312
D 100243

The correct answer is:
100243

Solution:
Using cramer’s rule
∆=123456690=1-54-2-36+36
=-54+72+18=36
∆1=6231256490=6-54-2-24+388
=-324+48+264=-12
∆2=1634126640=1-24-6-36+3-56
=-24+216-168=24
∆3=1264512694=1-88-2-56+66
=-88+112+36=60
⇒a=∆1∆=-13,b=∆2∆=23,c=∆3∆=53
⇒ equation is 2x2+1027x-910=0
⇒540x2+100x-243=0
α+β=-527,αβ=-920
1α+1β=α+βαβ=-527-920=100243

Topic: Mathematics, Determinants, JEE Main