datavorous/entrance-exam-dataset huggingface dataset:
Let a→=i^+4j^+2k^,b→=3i^-2j^+7k^ and c→=2i^-j^+4k^. If a vector d→ satisfies d→×b→=c→×b→ and d→·a→=24, then d→2 is equal to

Options:
A 323
B 423
C 313
D 413

The correct answer is:
413

Solution:
We have been given that

d→×b→=c→×b→

⇒(d→-c→)×b→=0

⇒d→-c→=λb→ (∵d→≠c→)

⇒d→=c→+λb→

It is given that a→·d→=24

⇒a→·c→+λb→·a→=24

⇒6+λ(3-8+14)=24

⇒9λ=18

⇒λ=2

∴d→=c→+2b→

⇒d→=8i^-5j^+18k^

|d→|2=64+25+324=413

Hence, this is the correct option.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (13 Apr Shift 1)
Let a→=i^+5j^+αk^, b→=i^+3j^+βk^ and c→=-i^+2j^-3k^ be three vectors such that, |b→×c→|=53 and a→ is perpendicular to b→. Then the greatest amongst the values of |a→|2 is ________.

Options:

The correct answer is:
90

Solution:
a→·b→=0

1+15+αβ=0

αβ=-16 ...(1)

Also,

b→×c→2=75

10+β2·14-5-3β2=75

5β2+30β+40=0

β=-4, -2

α=4, 8

a→max2=26+α2max=90

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2021 (27 Aug Shift 1)
Let A=i-i-ii,i=-1. Then, the system of linear equations A8xy=864 has :

Options:
A A unique solution
B Infinitely many solutions
C No solution
D Exactly two solutions

The correct answer is:
No solution

Solution:
A=i-i-ii

A2=-222-2=2-111-1

A4=222-2-22=81-1-11

A8=642-2-22=1281-1-11

A8xy=864

⇒1281-1-11xy=864

⇒128x-y-x+y=864

⇒x-y=116 ...(1)

and -x+y=12 ...(2)

⇒ From (1) and (2) : No solution.

Topic: Mathematics, Matrices, JEE Main, JEE Main 2021 (16 Mar Shift 1)
Let a→=i^+j^+2k^ and b→=-i^+2j^+3k^. Then the vector product a→+b→×a→×a→-b→×b→×b→ is equal to :

Options:
A 534i^-5j^+3k^
B 734i^-5j^+3k^
C 730i^-5j^+7k^
D 530i^-5j^+7k^

The correct answer is:
734i^-5j^+3k^

Solution:
a→=i^+j^+2k^

b→=-i^+2j^+3k^

a→+b→=3j^+5k^;a→·b→=-1+2+6=7

Now a→×a→-b→×b→×b→

=a→×a→×b→-b→×b→×b→

=a→×a→×b→-0×b→

=a→×a→×b→×b→

=a→·b→a→-a→·a→b→×b→

=a→·b→a→×b→-a→·a→b→×b→

=a→·b→a→×b→

Here, a→×b→=i^j^k^112-123=-i^-5j^+3k^

∴a→·b→a→×b→=7-i^-5j^+3k^

So, a→+b→×7-i^-5j^+3k^

70i^+3j^+5k^×-i^-5j^+3k^

=7i^j^k^035-1-53

=734i^-5j^+3k^

=734i^-5j^+3k^

Topic: Mathematics, Vector Algebra, JEE Main
Let a→=i^+j^+2k^ and b→=-i^+2j^+3k^. Then the vector product a→+b→×a→×a→-b→×b→×b→ is equal to :

Options:
A 534i^-5j^+3k^
B 734i^-5j^+3k^
C 730i^-5j^+7k^
D 530i^-5j^+7k^

The correct answer is:
734i^-5j^+3k^

Solution:
a→=i^+j^+2k^

b→=-i^+2j^+3k^

a→+b→=3j^+5k^;a→·b→=-1+2+6=7

Now a→×a→-b→×b→×b→

=a→×a→×b→-b→×b→×b→

=a→×a→×b→-0×b→

=a→×a→×b→×b→

=a→·b→a→-a→·a→b→×b→

=a→·b→a→×b→-a→·a→b→×b→

=a→·b→a→×b→

Here, a→×b→=i^j^k^112-123=-i^-5j^+3k^

∴a→·b→a→×b→=7-i^-5j^+3k^

So, a→+b→×7-i^-5j^+3k^

70i^+3j^+5k^×-i^-5j^+3k^

=7i^j^k^035-1-53

=734i^-5j^+3k^

=734i^-5j^+3k^

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2021 (27 Jul Shift 1)
Let a→=i^-j^+2k^ and let b→ be a vector such that a→×b→=2i^-k^ and a→·b→=3. Then the projection of b→ on the vector a→-b→ is:

Options:
A 221
B 237
C 2373
D 23

The correct answer is:
221

Solution:
Given,

a→=i^-j^+2k^ , a→×b→=2i^-k^ and a→·b→=3

Now using the formula a→×b→2+a→·b→2=a→2·b→2sin2θ+a→2·b→2cos2θ

⇒a→×b→2+a→·b→2=a→2·b→2sin2θ+cos2θ=a→2·b→2

We get,

⇒2i^-k^2+32=i^-j^+2k^2b→2

⇒b→2=5+96=146=73

Now finding a→-b→=a→2+b→2-2a→b→=6+73-6=73

Now projection of b→ on a→-b→=b→·a→-b→a→-b→

=b→·a→-b→2a→-b→=3-7373

=221

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2022 (25 Jul Shift 2)
Let a→=i^+j^+2k^,b→=2i^-3j^+k^ and c→=i^-j^+k^ be the three given vectors. Let v→ be a vector in the plane of a→ and b→ whose projection on c→ is 23. If v→,j^=7, then v→·i^+k^ is equal to

Options:
A 6
B 7
C 8
D 9

The correct answer is:
9

Solution:
Since vector v→, a→ & b→ are coplanar then, v→=λ1a→+λ2b→, where λ1,λ2∈ℝ.

⇒v→=λ1+2λ2i^+λ1-3λ2j^+2λ1+λ2k^

∵ Projection of v→ on c→ is 23

∴λ1+2λ2-λ1+3λ2+2λ1+λ23=23

∴ λ1+3λ2=1 ...1

and v→·j^=7⇒λ1-3λ2=7 ...2

⇒v→=λ1+2λ2i^+λ1-3λ2j^+2λ1+λ2k^

∴λ1+3λ2=1 ...1

⇒λ1-3λ2=7 2
From equation 1 and 2, we get

λ1=4,λ2=-1

∴v→=2i^+7j^+7k^

∴v→·i^+k^=2+7=9

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2022 (26 Jun Shift 2)
Let a→=i^+j^+2k^, b→=b1i^+b2j^+2k^ and c→=5i^+j^+2k^ be three vectors such that the projection vector of b→ on a→ is a→ . If a→+b→ is perpendicular to c→ , then b→ is equal to:

Options:
A 22
B 32
C 6
D 4

The correct answer is:
6

Solution:
Since a→+b→ is perpendicular to c→, hence

a→+b→·c→=0

⇒a→·c→+b→·c→=0

⇒5+1+2+5b1+b2 +2=0

⇒5b1+b2=-10 ...1

Given,

b→·a→a→=a→

∵p→=ri^+sj^+tk^⇒p→=r2+s2+t2

⇒b1+b2+2=4

⇒b1+b2=2 ...2

Solving 1 and 2, b1=-3, b2=5

∴b→=9+25+2=6

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2019 (09 Jan Shift 2)
Let a→=i^+αj^+3k^ and b→=3i^-αj^+k^. If the area of the parallelogram whose adjacent sides are represented by the vectors a→ and b→ is 83 square units, then a→·b→ is equal to ___ .

Options:

The correct answer is:
2

Solution:
a→=i^+αj^+3k^

b→=3i^-αj^+k^

area of parallelogram =|a→×b→|=83.

a→×b→=i^j^k^1α33-α1=i^4α-j^-8+k^-4α

∴|a→×b→|=64+32α2=83

⇒2+α2=6⇒α2=4

∴a→·b→=3-α2+3=2

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2021 (25 Feb Shift 2)
Let A→=i^ + j^ and B→=(2i^- j^) . The magnitude of a coplanar vector C→ such that A→ . C→=B→. C→= A→ . B→ is given by:

Options:
A 912
B 209
C 59
D 109

The correct answer is:
59

Solution:
If C→=ai^ + bj^ then,
A→ . C→=A→ . B→
a+b=1 .......(i)
B→. C→= A→ . B→
2a-b=1 ......(ii)
solving equations (i) and (ii) we get,
a=13, b=23
C→= 19+49= 59

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2018 (16 Apr Online)
Let a→=-i^-j^+k^,a→·b→=1 and a→×b→=i^-j^. Then a→-6b→ is equal to

Options:
A 3i^-j^-k^
B 3i^+j^+k^
C 3i^-j^+k^
D 3i^+j^-k^

The correct answer is:
3i^+j^+k^

Solution:
Let

b→=xi^+yj^+zk^

Then,

a→×b→=i^j^k^-1-11xyz

⇒a→×b→=-y+zi^+x+zj^+x-yk^

So,

-y+zi^+x+zj^+x-yk^=i^-j^+0k^

Hence,

y+z=-1x+z=-1x-y=0⇒x=y

Also,

a→·b→=1

⇒-x-y+z=1

⇒-2x+z=1

⇒z=2x+1

⇒-1-x=2x+1

⇒3x=-2⇒x=-23

So, y=-23; z=-13

Hence,

b→=-23i^-23j^-13k^

⇒6b→=-4i^-4j^-2k^

So,

a→-6b→=3i^+j^+k^

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (25 Jan Shift 2)
Let a→=αi^+j^-k^ and b→=2i^+j^-αk^,α>0. If the projection of a→×b→ on the vector -i^+2j^-2k^ is 30, then α is equal to

Options:
A 152
B 8
C 132
D 7

The correct answer is:
7

Solution:
Given, a→=αi^+j^-k^ and b→=2i^+j^-αk^,α>0. If the projection of a→×b→ on the vector -i^+2j^-2k^ is 30, then α is equal to

Now a→×b→=i^j^k^α1-121-α=1-αi^+α2-2j^+α-2k^

Projection of a→×b→ on -i^+2j^-2k^

=a→×b→·-i^+2j^-2k^12+22+22

=1-αi^+α2-2j^+α-2k^.-i^+2j^-2k^3

=-1+α+2α2-4-2α+43

=2α2-α-13

Now given length of projection is 30

So 2α2-α-13=30

⇒2α2-α-91=0

⇒α=7,-132

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2022 (26 Jul Shift 1)
Let a→=i^+j^-k^ and b→=2i^-j^+k^. If c→ is a non-zero vector perpendicular to both a→ and b→, such that its component along x,y,z axes are rational numbers, then c→ is

Options:
A an integer
B an irrational number
C a rational number
D None of these

The correct answer is:
an irrational number

Solution:
c→=λa→×b→
=λi^j^k^11−12−11=−3λj^+k^
c→=32λ

Hence, an irrational number

Topic: Mathematics, Vector Algebra, JEE Main
Let λ∈ℤ, a→=λi^+j^-k^ and b→=3i^-j^+2k^. Let c→ be a vector such that a→+b→+c→×c→=0→, a→·c→=-17 and b→·c→=-20. Then c→×λi^+j^+k^2 is equal to

Options:
A 46
B 53
C 62
D 49

The correct answer is:
46

Solution:
Given that

a→=λi^+j^-k^

b→=3i^-j^+2k^

Also a→+b→+c→×c→=0→

⇒ka→+b→=c→

a→·c→=-17 and b→·c→=-20

Now, Take a→·c→=-17

⇒kλi^+j^-k^·λi^+j^-k^+3i^-j^+2k^=-17

⇒kλ2+3λ+0-1=-17

⇒kλ2+3λ-1=-17 ....(1)

Similarly, on taking b→·c→=-20, we get,

k3i^-j^+2k^·λi^+j^-k^+3i^-j^+2k^=-20

⇒k3λ+9+2=-20

⇒k3λ+11=-20 ...(2)

Now on solving equation 1 & 2 we get,

⇒20λ2+9λ-207=0

⇒λ=3, -6920

For λ=3, k=-1

⇒c→=-1a→+b→

⇒-λ+3i^+k^=-6i^-k^

⇒c→×λi^+j^+k^=i^j^k^-60-1311=i^-3j^+6k^

⇒c→×λi^+j^+k^2=46

Hence this is the correct option.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (12 Apr Shift 1)
Let a→=i^+j^+k^ and b→=j^-k^. If c→ is a vector such that a→×c→=b→ and a→·c→=3, then a→·(b→×c→) is equal to:

Options:
A 6
B -2
C 2
D -6

The correct answer is:
-2

Solution:
We know that, for vectors a→, b→ and c→
a→·(b→×c→)=a→b→c→=c→a→b→

⇒a→·(b→×c→)=(c→×a→)·b→

=-b→·b→

∵a→×c→=b→⇒c→×a→=-b→

=-b→2

=-12+-122

=-2

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2021 (26 Aug Shift 1)
Let aâ=i^+j^+k^, bâ=2i^-3j^+4k^ and aâÃcâ=bâ+2Î»aâ, then the value of aâbâcâ is equal to

Options:
A 12
B 26
C -26
D 3

The correct answer is:
-26

Solution:
aâÃcâ=bâ+2Î»aâ
Taking dot product with aâ
0=bââaâ+2Î»aââaââ0=3+6Î»âÎ»=-12

i.e. aâÃcâ=bâ-aâ
Taking dot product with bâ
aâÃcââbâ=bââbâ-aââbâ
âaâcâbâ=29-3âaâbâcâ=-26

Topic: Mathematics, Vector Algebra, JEE Main
Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}+4 \hat{j}-5 \hat{k}$ and $\vec{c}=x \hat{i}+2 \hat{j}+3 \hat{k}, x \in \mathbb{R}$.
If $\vec{d}$ is the unit vector in the direction of $\vec{b}+\vec{c}$ such that $\vec{a} \cdot \vec{d}=1$, then $(\vec{a} \times \vec{b}) \cdot \vec{c}$ is equal to

Options:
A 11
B 3
C 9
D 6

The correct answer is:
11

Solution:
$\begin{aligned} & \overrightarrow{\mathrm{d}}=\lambda(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}) \\ & \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{d}}=\lambda(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}) \\ & 1=\lambda(1+\mathrm{x}+5)\end{aligned}$
$1=\lambda(x+6)$ ...(1)
$\begin{aligned}
& |\overrightarrow{\mathrm{d}}|=1 \\
& |\lambda(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})|=1 \\
& |\lambda((\mathrm{x}+2) \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-2 \hat{\mathrm{k}})|=1 \\
& \lambda^2\left((\mathrm{x}+2)^2+6^2+2^2\right)=1 \\
& \mathrm{x}^2+4 \mathrm{x}+4+36+4=(\mathrm{x}+6)^2 \\
& \mathrm{x}^2+4 \mathrm{x}+44=\mathrm{x}^2+12 \mathrm{x}+36 \\
& 8 \mathrm{x}=8, \mathrm{x}=1 \\
& \left|\begin{array}{ccc}
1 & 1 & 1 \\
2 & 4 & -5 \\
\mathrm{x} & 2 & 3
\end{array}\right|=(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}) \cdot \overrightarrow{\mathrm{c}} \\
& \left|\begin{array}{ccc}
0 & 0 & 1 \\
-2 & 9 & -4 \\
\mathrm{x}-2 & -1 & 3
\end{array}\right|=2-9(\mathrm{x}-2) \\
& =20-9 \mathrm{x} \\
& \text { at } \mathrm{x}=1 \\
& 20-9=11
\end{aligned}$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2024 (04 Apr Shift 2)
Let a→=i^-αj^+βk^,b→=3i^+βj^-αk^ and c→=-αi^-2j^+k^, where α and β are integers. If a→·b→=-1 and b→·c→=10, then a→×b→·c→ is equal to ______.

Options:

The correct answer is:
9

Solution:
We have, a→=1,-α,β,b→=3,β,-α and

c→=-α,-2,1;α,β∈I

a→·b→=-1⇒3-αβ-αβ=-1

⇒αβ=2 ...1

Now, b→·c→=10

⇒-3α-2β-α=10

⇒2α+β+5=0 ...2

From equations 1 and 2,

⇒2α+2α+5=0

⇒2α2+2+5α=0

⇒2α2+5α+2=0

⇒2α2+4α+α+2=0

⇒2αα+2+1α+2=0

⇒2α+1α+2=0

α=-12 is neglected since α is an integer.

∴ α=-2

Put α=-2 in equation 1, we get β=-1.

Now, a→b→c→=12-13-122-21

=1-1+4-23-4-1-6+2

=3+2+4=9.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2021 (27 Jul Shift 2)
Let a→=i^+j^+k^,b→ and c→=j^-k^ be three vectors such that a→×b→=c→ and a→·b→=1. If the length of projection vector of the vector b→ on the vector a→×c→ is l, then the value of 3l2 is equal to _____.

Options:

The correct answer is:
2

Solution:
Given,

a→×b→=c→

⇒a→×b→·c→=c→2=12+122=2

Here, a→×c→=i^j^k^11101-1=-2i^+j^+k^

Projection of b→ on a→×c→=ℓ

⇒b→·a→×c→a→×c→=ℓ

⇒a→×b→·c→22+1+1=ℓ

∴ ℓ=26⇒ℓ2=46

So, 3ℓ2=2

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2021 (27 Jul Shift 1)
Let aâ=i^+j^+k^, bâ=i^+4j^-k^ and câ=i^+j^+2k^. If Sâ be a unit vector, then the magnitude of the vector aââSâbâÃcâ+bââSâcâÃaâ+cââSâaâÃbâ is equal to

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:
Let r→=a→⋅S→b→×c→+b→⋅S→c→×a→+c→⋅S→a→×b→
S→ can be written as a linear combination of b→×c→,c→×a→,a→×b→
S→=xb→×c→+yc→×a→+za→×b→
Taking dot with a→
S→⋅a→=xa→b→c→⇒x=S→·a→[a→b→c→]
Taking dot with b→
S→⋅b→=ya→b→c→⇒x=S→·b→[a→b→c→]
Similarly z=S→·c→a→ b→ c→
Hence S→=S→⋅a→b→×c→+S→⋅b→c→×a→+S→⋅c→a→×b→a→ b→ c→
⇒r→=S→a→ b→ c→

⇒r→=S→a→ b→ c→
=111114-1112
=19-13+1-3
=3

Topic: Mathematics, Vector Algebra, JEE Main
Let $\overline{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overline{\mathrm{b}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$ and $\overline{\mathrm{c}}=\mathrm{x} \hat{\mathrm{i}}+(\mathrm{x}-2) \hat{\mathrm{j}}-\hat{\mathrm{k}}$. If the vector $\overline{\mathrm{c}}$ lies in the plane of $\bar{a}$ and $\bar{b}$, then $x$ equals

Options:
A
$0$

B
$1$

C
$-4$

D
$-2$

The correct answer is:
$-2$

Solution:
$\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=\hat{i}-\hat{j}+2 \hat{k}$ and $\vec{c}=x \hat{i}+(x-2) \hat{j}-\hat{k}$
$\left|\begin{array}{ccr}x & x-2 & -1 \\ 1 & 1 & 1 \\ 1 & -1 & 2\end{array}\right|=0$
$3 x+2-x+2=0$
$2 x=-4$
$x=-2$.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2007
Let a→=i^+j^+k^, c→=j^-k^ and a vector b→ be such that a→×b→=c→ and a→⋅b→=3. Then b→ equals

Options:
A 113
B 113
C 113
D 113

The correct answer is:
113

Solution:
Given, a→×b→=c→
⇒a→×a→×b→=a→×c→
⇒a→⋅b→a→-a→⋅a→b→=a→×c→

a→×c→=i^j^k^11101-1=-2i^+j^+k^
⇒3a→-3b→=-2i^+j^+k^
⇒3i^+3j^+3k^-3b→=-2 i^+j^+k^
⇒b→=135i^+2j^+2k^
∴ b→=25+4+43
⇒b→=113

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2018 (16 Apr Online)
Let $\vec{a}=\hat{i}-\hat{k}, \vec{b}=x \hat{i}+\hat{j}+(1-x) \hat{k}$ and $\vec{c}=y \hat{i}+x \hat{j}+(1+x-y) \hat{k}$. Then $[\vec{a}, \vec{b}, \vec{c}]$ depends on

Options:
A
only $y$

B
only $\mathrm{x}$

C
both $x$ and $y$

D
neither $x$ nor $y$

The correct answer is:
neither $x$ nor $y$

Solution:
$\vec{a}=\hat{i}-\hat{k}, \vec{b}=x \hat{i}+\hat{j}+(1-x) \hat{k}$ and $\vec{c}=y \hat{i}+x \hat{j}+(1+x-y) \hat{k}$
$[\vec{a} \vec{b} \vec{c}]=\vec{a} \cdot(\vec{b} \times \vec{c})$
$$
\begin{aligned}
& \vec{b} \times \vec{c}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
x & 1 & 1-x \\
y & x & 1+x-y
\end{array}\right|=\hat{i}\left(1+x-x-x^2\right)-\hat{j}\left(x+x^2-x y-y+x y\right)+\hat{k}\left(x^2-y\right) \\
& \vec{a} .(\vec{b} \times \vec{c})=1
\end{aligned}
$$
which does not depend on $\mathrm{x}$ and $\mathrm{y}$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2005
Let $a=\operatorname{Im}\left(\frac{1+z^2}{2 i z}\right)$, where $z$ is any non-zero complex number. The set $\mathrm{A}=\{a:|z|=1$ and $z \neq \pm 1\}$ is equal to:

Options:
A
$(-1,1)$

B
$[-1,1]$

C
$[0,1)$

D
$(-1,0]$

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

Solution:
Let $z=x+i y \Rightarrow z^2=x^2-y^2+2 i x y$ Now,
$$
\begin{aligned}
& \frac{1+z^2}{2 i z}=\frac{1+x^2-y^2+2 i x y}{2 i(x+i y)}=\frac{\left(x^2-y^2+1\right)+2 i x y}{2 i x-2 y} \\
& =\frac{\left(x^2-y^2+1\right)+2 i x y}{-2 y+2 i x} \times \frac{-2 y-2 i x}{-2 y-2 i x} \\
& =\frac{y\left(x^2+y^2-1\right)+x\left(x^2+y^2+1\right) i}{2\left(x^2+y^2\right)} \\
&
\end{aligned}
$$
$$
a=\frac{x\left(x^2+y^2+1\right)}{2\left(x^2+y^2\right)}
$$
Since, $|z|=1 \Rightarrow \sqrt{x^2+y^2}=1$
$$
\begin{aligned}
& \Rightarrow x^2+y^2=1 \\
& \therefore a=\frac{x(1+1)}{2 \times 1}=x \\
& \text { Also } z \neq 1 \Rightarrow x+i y \neq 1 \\
& \therefore \mathrm{A}=(-1,1)
\end{aligned}
$$

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2013 (23 Apr Online)
Let A is a matrix of order 3×3 defined as A=aij3×3, where aij=limx→01-cosixsinixtanjx∀1≤i,j≤3, then A2 is equal to

Options:
A A
B 32A
C 23A
D 14A

The correct answer is:
32A

Solution:
Applying L' Hospital rule, we get,

aij=12ij
A=12141611213323412=121121321233321
A2=1411213212333211121321233321
=143323363639923=32A

Topic: Mathematics, Matrices, JEE Main
Let $\vec{a}=\hat{j}-\hat{k}$ and $\vec{c}=\hat{i}-\hat{j}-\hat{k}$. Then vector $\vec{b}$ satisfying $\vec{a} \times \vec{b}+\vec{c}=\overrightarrow{0}$ and $\vec{a} \cdot \vec{b}=3$ is

Options:
A
$2 \hat{i}-\hat{j}+2 \hat{k}$

B
$\hat{i}-\hat{j}-2 \hat{k}$

C
$\hat{i}+\hat{j}-2 \hat{k}$

D
$-\hat{i}+\hat{j}-2 \hat{k}$

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

Solution:
$\overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}$
$\Rightarrow \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=0$
$\Rightarrow\left(\mathrm{b}_1 \hat{\mathrm{i}}+\mathrm{b}_2 \hat{j}+\mathrm{b}_3 \hat{\mathrm{k}}\right) \cdot(\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}})=0$
$\mathrm{~b}_1-\mathrm{b}_2-\mathrm{b}_3=0$
and $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=3$
$\Rightarrow \mathrm{b}_2-\mathrm{b}_3=3$
$\mathrm{~b}_1=\mathrm{b}_2+\mathrm{b}_3=3+2 \mathrm{~b}_3$
$\overrightarrow{\mathrm{b}}=\left(3+2 \mathrm{~b}_3\right) \hat{\mathrm{i}}+\left(3+\mathrm{b}_3\right) \hat{\mathrm{j}}+\mathrm{b}_3 \hat{\mathrm{k}}$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2010
Let ak=10Ck for 0≤k≤10, k∈N.If Ak=ak-100ak for 0≤k≤10 and ∑ k=1 9 A k ⋅ A k+1 = a 0 0 b , then the value of a-b is equal to

Options:
A 1
B 20C11
C 0
D 20C11-10

The correct answer is:
0

Solution:
Ak=10Ck-10010Ck, Ak+1=10Ck0010Ck+1
AkAk+1=10Ck-1·10Ck0010Ck·10Ck+1
∑k=19AkAk+1=C010C110+C110C210−−+C810.C91000C110C210+C210C310+−−C910C1010
=20C11-100020C11-10=a00b
a=20C11-10
b=20C11-10

Topic: Mathematics, Matrices, JEE Main
Let A lies on 3x-4y+1=0, B lies on 4x+3y-7=0 and C is -2,5. If ABCD is a rhombus, then the locus of D is a conic whose length of the latus rectum is equal to

Options:
A 10 units
B 15 units
C 5 units
D 20 units

The correct answer is:
10 units

Solution:
Let, Dh,k
C-2,5 lies on 4x+3y-7=0
⇒BC is perpendicular to 3x-4y+1=0
⇒DA is perpendicular to 3x-4y+1=0
⇒ Because DA=DC so, the distance of D from 3x-4y+1=0 is equal to DC
⇒h+22+k-52=3h-4k+15
⇒25x+22+y-52=3x-4y+12
Which represents parabola with focus -2,5 and directrix is 3x-4y+1=0
⇒ Length of the latus rectum is 2-6-20+15=10 units

Topic: Mathematics, Parabola, JEE Main
Let a line having direction ratios 1,-4,2 intersect the lines x-73=y-1-1=z+21 and x2=y-73=z1 at the points A and B. Then AB2 is equal to

Options:

The correct answer is:
84

Solution:
Let A be 3λ+7,-λ+1,λ-2 and B be 2μ,3μ+7,μ

DR's of AB is

3λ-2μ+7,-λ-3μ-6,λ-μ-2

Now,

3λ-2μ+71=-λ-3μ-6-4=λ-μ-22

i.e. -12λ+8μ-28=-λ-3μ-6

⇒λ=μ-2

Also

-2λ-6μ-12=-4λ+4μ+8

⇒λ=5μ+10

After solving above two equations, we get λ=-5,μ=-3

i.e. A=-8,6,-7 and B=-6,-2,-3

Hence, AB2=4+64+16=84

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2022 (24 Jun Shift 1)
Let a line L1 be tangent to the hyperbola x216-y24=1 and let L2 be the line passing through the origin and perpendicular to L1. If the locus of the point of intersection of L1 and L2 is x2+y22= αx2+βy2, then α+β is equal to ______.

Options:

The correct answer is:
12

Solution:
The equation of tangent to the given hyperbola is y=mx±16m2-4 ...i

Hence, l1:y=mx±16m2−4

Given that, l2 is a straight line passing through origin and perpendicular to l1.

So, l2:y=-1mx ⇒m=-xy ...ii

On solving equations i & ii, we get

y=-xyx±16-xy2-4 ⇒y=-x2y±16x2-4y2y

⇒y2+x22=16x2-4y2

On comparing the above equation with x2+y22=αx2+βy2, we get α=16, β=−4

∴ α+β=12

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2022 (26 Jun Shift 2)
Let a line L:2x+y=k, k>0 be a tangent to the hyperbola x2-y2=3. If L is also a tangent to the parabola y2=αx, then α is equal to:

Options:
A 12
B -12
C 24
D -24

The correct answer is:
-24

Solution:
A line y=mx+c is a tangent to the hyperbola x2a2-y2b2=1 if c2=a2m2-b2.

Given that, tangent to hyperbola x23-y23=1 is 2x+y=k or y=-2x+k

Thus, we have, slope m=-2, c= k & a2=b2=3

k2=3-22-3

⇒ k2=9

Given k>0, ⇒ k=3.

Thus, the equation of the tangent to the hyperbola is y=-2x+3.

Given this line is also a tangent to the parabola, y2=αx and a line y=mx+c is tangent to the parabola y2=4Ax, if c=Am

Thus, we have 3=α4-2

⇒ α-8=3

⇒ α=-24.

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2021 (22 Jul Shift 1)
Let a line L pass through the origin and be perpendicular to the lines L1:r→=i^-11j^-7k^+λi^+2j^+3k^, λ∈ℝ and
L2:r→=-i^+k^+μ2i^+2j^+k^, μ∈ℝ. If P is the point of intersection of L and L1, and ,Qα,β,γ is the foot of perpendicular from P on L2, then 9α+β+γ is equal to ________.

Options:

The correct answer is:
5

Solution:
Given,

A line L pass through the origin and be perpendicular to the lines L1:r→=i^-11j^-7k^+λi^+2j^+3k^, λ∈ℝ and
L2:r→=-i^+k^+μ2i^+2j^+k^, μ∈ℝ

So, direction ratio of line will be,
i^j^k^123221

=i^-4-j^-5+k^-2

=-4i^+5j^-2k^

Hence, the equation of line L which passes through origin will be,

L:r→=σ(-4i^+5j^-2k^)

Now finding the intersection point P we get,

1+λ=-4σ ....1

-11+2λ=5σ ....2

-7+3λ=-2σ ....3

So, from equation 1 & 3 we get, 1+λ=-14+6λ⇒λ=3, σ=-1

Hence, point P≡4,-5,2

Now finding the point Q which is foot of perpendicular from point P4,-5,2 on line L2: -i^+k^+μ2i^+2j^+k^ we get,

Now any point on line L2 will be Q2μ-1i^+2μj^+μ+1k^

Now direction ratio of PQ will be -5+2μi^+2μ+5j^+μ-1k^=0

Now using the perpendicular condition we get,

2(-5+2μ)+2(2μ+5)+1(μ-1)=0

⇒μ=19

Hence, α+β+γ=2μ-1+2μ+μ+1=5μ=59

⇒9α+β+γ=5

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (11 Apr Shift 1)
Let a line L pass through the point P(2,3,1) and be parallel to the line x+3y-2z-2=0=x-y+2z. If the distance of L from the point (5,3,8) is α, then 3α2 is equal to ________

Options:

The correct answer is:
158

Solution:
Given,

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

Now direction ratio is given by,

i^j^k^13-21-12=4i^-4j^-4k^

So, Equation of line will be x-24=y-3-4=z-1-4

⇒x-21=y-3-1=z-1-1=k

Let Q be (5,3,8) and foot of ⊥ from Q on this line be R

Now, R≡(k+2,-k+3,-k+1)

So, direction ratio of QR will be (k-3,-k,-k-7),

Now using the perpendicular condition we get,

(1)(k-3)+(-1)(-k)+(-1)(-k-7)=0

⇒k=-43

So, α2=1332+432+1732=4749

⇒3α2=158

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (30 Jan Shift 2)
Let a line passing through the point (-1,2,3) intersect the lines L1:x-13=y-22=z+1-2 at M(α,β,γ) and L2:x+2-3=y-2-2=z-14 at N(a,b,c). Then the value of (α+β+γ)2(a+b+c)2 equals ________________.

Options:

The correct answer is:
196

Solution:
Let,

L1:x-13=y-22=z+1-2=λ

⇒M(3λ+1,2λ+2,-2λ-1)≡Mα,β,γ

⇒α+β+γ=3λ+2

Also, L2:x+2-3=y-2-2=z-14=μ

⇒N(-3μ-2,-2μ+2,4μ+1)≡Na,b,c

⇒a+b+c=-μ+1

Now, from the above diagram, we can see MN lie on the same line,

So, 3λ+2-3μ-1=2λ-2μ=-2λ-44μ-2

Now, solving 3λ+2-3μ-1=2λ-2μ

⇒3λμ+2μ=3λμ+λ

⇒2μ=λ ......i

Now, solving 2λ-2μ=-2λ-44μ-2 we get,

⇒2λμ-λ=λμ+2μ

⇒λμ=λ+2μ

⇒λμ=2λ as 2μ=λ

⇒μ=2 (λ≠0)

⇒λ=4 from equation i

Hence, α+β+γ=14 and a+b+c=-1

So, (α+β+γ)2(a+b+c)2=196

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2024 (30 Jan Shift 2)
Let a line perpendicular to the line $2 x-y=10$ touch the parabola $y^2=4(x-9)$ at the point $P$. The distance of the point $P$ from the centre of the circle $x^2+y^2-14 x-8 y+56=0$ is __________

Options:

The correct answer is:
10

Solution:
$y^2=4(x-9)$
slope of tangent $=\frac{-1}{2}$
Point of contact $\mathrm{P}\left(9+\frac{1}{\left(-\frac{1}{2}\right)^2}, \frac{2 \times 1}{\frac{-1}{2}}\right)$
$\begin{aligned} & P(13,-4) \\ & \text { center of circle } C(7,4) \\ & \text { distance } C P=\sqrt{(13-7)^2+(-4-4)^2} \\ & =10\end{aligned}$

Topic: Mathematics, Parabola, JEE Main, JEE Main 2024 (05 Apr Shift 2)
Let a line with direction ratios a,-4 a,-7 be perpendicular to the lines with direction ratios 3,-1,2b and b, a,-2. If the point of intersection of the line x+1a2+b2=y-2a2-b2=z1 and the plane x-y+z=0 is α,β,γ, then α+β+γ is equal to ________.

Options:

The correct answer is:
10

Solution:
Given a line with direction ratios a,-4 a,-7 be perpendicular to the lines with direction ratios 3,-1,2b and b, a,-2. If the point of intersection of the line x+1a2+b2=y-2a2-b2=z1 and the plane x-y+z=0 is α,β,γ, then α+β+γ is equal to ________.

Now given a,-4a,-7⊥ to 3,-1,2b

So, 3a+4a-14b=0

⇒a=2b ⋯i

Also a,-4a,-7⊥ to b,a,-2

So, ab-4a2+14=0 ⋯ii

From Equations i and ii we get,

2b2-16b2+14=0

⇒b2=1

So, a2=4b2=4

So, line becomes x+15=y-23=z1=k

Let points be α=5k-1,β=3k+2,γ=k

As α,β,γ satisfies plane x-y+z=0

So, 5k-1-3k+2+k=0

⇒k=1

∴ α+β+γ=9k+1=10

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2022 (29 Jul Shift 1)