datavorous/entrance-exam-dataset huggingface dataset:
Let $\overrightarrow{\mathrm{a}}=2 \hat{i}+\hat{j}-\hat{k}, \overrightarrow{\mathrm{b}}=((\overrightarrow{\mathrm{a}} \times(\hat{i}+\hat{j})) \times \hat{i}) \times \hat{i}$. Then the square of the projection of $\overrightarrow{\mathrm{a}}$ on $\overrightarrow{\mathrm{b}}$ is :

Options:
A $\frac{1}{3}$
B $\frac{2}{3}$
C 2
D $\frac{1}{5}$

The correct answer is:
2

Solution:
$\begin{aligned} & \vec{a} \times(\hat{i}+\hat{j})=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 1 & 1 & 0\end{array}\right| \\ & =\hat{i}-\hat{j}+\hat{k} \\ & (\vec{a} \times(\hat{i} \times \hat{j})) \times \hat{i}=\hat{k}+\hat{j} \\ & ((\vec{a} \times(\hat{i} \times \hat{j})) \times i) \times \hat{i}=\hat{j}-\hat{k} \\ & \operatorname{projection} \text { of } \vec{a} \text { on } \hat{b}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \\ & =\frac{1+1}{\sqrt{2}}=\sqrt{2}\end{aligned}$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2024 (06 Apr Shift 2)
Let $\vec{a}=2 \hat{i}-\hat{j}+\hat{k}, \vec{b}=\hat{i}+2 \hat{j}-\hat{k}$ and $\vec{c}=\hat{i}+\hat{j}-2 \hat{k}$ be three vectors. A vector of the type $\vec{b}+\lambda \vec{c}$ for some scalar $\lambda$, whose projection on $\vec{a}$ is of magnitude $\sqrt{\frac{2}{3}}$ is :

Options:
A
$2 \hat{i}+\hat{j}+5 \hat{k}$

B
$2 \hat{i}+3 \hat{j}-3 \hat{k}$

C
$2 \hat{i}-\hat{j}+5 \hat{k}$

D
$2 \hat{i}+3 \hat{j}+3 \hat{k}$

The correct answer is:
$2 \hat{i}+3 \hat{j}-3 \hat{k}$

Solution:
Let $\vec{d}=\vec{b}+\lambda \vec{c}$
$$
\begin{aligned}
& \therefore \vec{d}=\hat{i}+2 \hat{j}-\hat{k}+\lambda(\hat{i}+\hat{j}-2 \hat{k}) \\
& =(1+\lambda) \hat{i}+(2+\lambda) \hat{j}-(1+2 \lambda) \hat{k}
\end{aligned}
$$
If $\theta$ be the angle between $\vec{d}$ and $\vec{a}$, then projection of $\vec{d}$ or $(\vec{b}+\lambda \vec{c})$ on $\vec{a}$
$$
\begin{aligned}
& =|\vec{d}| \cos \theta \\
& =|\vec{d}|\left(\frac{\vec{d} \cdot \vec{a}}{|\vec{d}||\vec{a}|}\right)=\frac{\vec{d} \cdot \vec{a}}{|\vec{a}|} \\
& =\frac{2(\lambda+1)-(\lambda+2)-(2 \lambda+1)}{\sqrt{4+1+1}} \\
& =\frac{-\lambda-1}{\sqrt{6}}
\end{aligned}
$$
But projection of $\vec{d}$ on $\vec{a}=\sqrt{\frac{2}{3}}$
$$
\therefore-\frac{\lambda+1}{\sqrt{6}}=\sqrt{\frac{2}{3}} \Rightarrow \frac{\lambda^2+2 \lambda+1}{6}=\frac{2}{3}
$$
$$
\begin{aligned}
& \Rightarrow \lambda^2+2 \lambda-3=0 \Rightarrow \lambda^2+3 \lambda-\lambda-3=0 \\
& \Rightarrow \lambda(\lambda+3)-1(\lambda+3)=0, \Rightarrow \lambda=1,-3
\end{aligned}
$$
when $\lambda=1$, then $\vec{b}+\lambda \vec{c}=2 \hat{i}+3 \hat{j}-3 \hat{k}$
when $\lambda=-3$, then $\vec{b}+\lambda \vec{c}=-2 \hat{i}-\hat{j}+5 \hat{k}$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2013 (09 Apr Online)
Let $\vec{a}=2 \hat{i}+\alpha \hat{j}+\hat{k}, \vec{b}=-\hat{i}+\hat{k}, \vec{c}=\beta \hat{j}-\hat{k}$, where $\alpha$ and $\beta$ are integers and $\alpha \beta=-6$. Let the values of the ordered pair $(\alpha, \beta)$, for which the area of the parallelogram of diagonals $\vec{a}+\vec{b}$ and $\vec{b}+\vec{c}$ is $\frac{\sqrt{21}}{2}$, be $\left(\alpha_1, \beta_1\right)$ and $\left(\alpha_2, \beta_2\right)$. Then $\alpha_1^2+\beta_1^2-\alpha_2 \beta_2$ is equal to

Options:
A 19
B 17
C 24
D 21

The correct answer is:
19

Solution:
$\begin{aligned} & \text { Area of parallelogram } \left.=\frac{1}{2} \right\rvert\, \overrightarrow{\mathrm{d}}_1 \times \overrightarrow{\mathrm{d}}_2 \\ & \mathrm{~A}=\frac{1}{2}|(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})|=\frac{\sqrt{21}}{2} \\ & \text { so, } \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+\alpha \hat{\mathrm{j}}+2 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=-\hat{\mathrm{i}}+\beta \hat{\mathrm{j}} \\ & (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & \alpha & 2 \\ -1 & \beta & 0\end{array}\right|\end{aligned}$
$\begin{aligned}
& =\hat{\mathrm{i}}(-2 \beta)-\hat{\mathrm{j}}(2)+\hat{\mathrm{k}}(\beta+\alpha) \\
& |(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})|=\sqrt{4 \beta^2+4+(\alpha+\beta)^2}=\sqrt{21} \\
& 4 \beta^2+4+\alpha^2+\beta^2+2 \alpha \beta=21 \\
& \alpha^2+5 \beta^2-12=17 \\
& \alpha^2+5 \beta^2=29
\end{aligned}$
and $\alpha \beta=-6$
and given $\alpha_{\mathrm{i}} \beta$ are integers
so, $\alpha=-3, \beta=2$
or
$\begin{aligned}
& \alpha=3, \beta=-2 \\
& \left(\alpha_1, \beta_1\right)=(-3,2) \\
& \left(\alpha_2, \beta_2\right)=(3,-2) \\
& \alpha_1^2+\beta_1^2-\alpha_2 \beta_2=9+4+6=19
\end{aligned}$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2024 (09 Apr Shift 2)
Let $A(-3,2)$ and $B(-2,1)$ be the vertices of a triangle $\mathrm{ABC}$. If the centroid of this triangle lies on the line $3 x+4 y+2=0$, then the vertex $C$ lies on the line :

Options:
A
$4 x+3 y+5=0$

B
$3 x+4 y+3=0$

C
$4 x+3 y+3=0$

D
$3 x+4 y+5=0$

The correct answer is:
$3 x+4 y+3=0$

Solution:
Let $\mathrm{C}=\left(x_1, y_1\right)$
$$
\mathrm{A}(-3,2)
$$
Centroid, $\mathrm{E}=\left(\frac{x_1-5}{3}, \frac{y_1+3}{3}\right)$
Since centroid lies on the line
$$
\begin{aligned}
& 3 x+4 y+2=0 \\
\therefore \quad & 3\left(\frac{x_1-5}{3}\right)+4\left(\frac{y_1+3}{3}\right)+2=0
\end{aligned}
$$
$$
\Rightarrow \quad 3 x_1+4 y_1+3=0
$$
Hence vertex $\left(x_1, y_1\right)$ lies on the line
$$
3 x+4 y+3=0
$$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2013 (25 Apr Online)
Let A0,3 and B0,12 be two vertices of a âABC where C=x,0. If the circumcircle of âABC touches the x-axis, then the value of cos 2Î¸ is (where, Î¸ is angle ACB)

Options:
A 13
B 12
C 815
D 725

The correct answer is:
725

Solution:
OA×OB=OC2
⇒ 3×12=x2
⇒ x=6
Centre C'=6,152
∠AC'B=2∠ACB=2θ
⇒cos⁡2θ=1522+1522-922×152×152
=2252-812252=225-162225=63225
=725

Topic: Mathematics, Circle, JEE Main
Let a→=3, b→=4, c→=5 and a→×a→×c→+4b→=0, then the value of a→×c→2 equals to

Options:
A 9256
B 163
C 2569
D 316

The correct answer is:
2569

Solution:
a→×a→×c→=-4b→⇒a→⋅c→a→-a→⋅a→c→=-4b→

Let, a→⋅c→=a→c→cosθ
⇒15cos⁡θa→-9c→=-4b→
⇒15cos⁡θa→-9c→2=16b→2
⇒225cos2θ×9+81×25-270×15cos2θ=16×16
⇒2025cos2θ=1769
⇒cos2θ=17692025
⇒sin2θ=2562025
a→×c→2=a→2c→2sin2θ
⇒a→×c→2=2569

Topic: Mathematics, Vector Algebra, JEE Main
Let a→=3i^+2j^+2k^ and b→=i^+2j^-2k^ be two vectors. If a vector perpendicular to both the vectors a→+b→ and a→-b→ has the magnitude 12 then one such vector is:

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

The correct answer is:
4(2i^-2j^-k^)

Solution:
a→=3i^+2j^+2k^,b→=i+2j^-2k^
∵ α→=a→+b→=4i^+4j^
β→=a→-b→=2i^+4k^
∴ α→×β→=i^j^k^440204=i^16-j^16+k^-8=16i^-16j^-8k^
∴ Unit vector perpendicular to  α→& β→ is
n^=α→×β→α→×β→=2i^-2j^-k^3
∴ Required vector =122i^-2j^-k^3=42i^-2j^-k^

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2019 (12 Apr Shift 1)
Let a→=3i^+2j^+xk^ and b→=i^-j^+k^, for some real x. Then the condition for a→×b→ =r to follow

Options:
A 0<r≤ 32
B r≥ 532
C 32<r≤332
D 3 32<r<532

The correct answer is:
r≥ 532

Solution:
As, We know that a→×b→=i^j^k^32x1-11
=2+xi^+x-3j^-5k^

a→×b→=4+x2+4x+x2+9-6x+25

=2x2-2x+38

Therefore, we have
⇒a→×b→≥752

⇒a→×b→≥532

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2019 (08 Apr Shift 2)
Let a→=3i^+j^−2k^, b→=4i^+j^+7k^ and c→=i^−3j^+4k^ be three vectors. If a vectors p→ satisfies p→×b→=c→×b→ and p→⋅a→=0, then p→⋅i^−j^−k^ is equal to

Options:
A 24
B 36
C 28
D 32

The correct answer is:
32

Solution:
Given: p→×b→=c→×b→

⇒p→-c→×b→=0

⇒p→-c→||b→

⇒p→-c→=kb→

⇒p→=kb→+c→

⇒p→=4k+1i^+k-3j^+7k+4k^

Now, p→.a→=0

⇒34k+1+k-3-27k+4=0

⇒k=-8

⇒p→=-31i^-11j^-52k^

⇒p→.i^-j^-k^=-31+11+52

⇒p→.i^-j^-k^=32

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2024 (31 Jan Shift 1)
Let a→=3i^+j^ and b→=i^+2j^+k^. Let c→ be a vector satisfying a→×b→×c→=b→+λc→. If b→ and c→ are non-parallel, then the value of λ is

Options:
A -5
B 5
C 1
D -1

The correct answer is:
-5

Solution:
Given,

a→=3i^+j^ and b→=i^+2j^+k^

Let c→ be a vector satisfying a→×b→×c→=b→+λc→. If b→ and c→ are non-parallel, then the value of λ is

Now putting the value of a→=3i^+j^ & b→=i^+2j^+k^ in a→×b→×c→=b→+λc→ we get,

⇒a→×b→×c→=b→+λc→

⇒a→·c→b→-a→·b→c→=b→+λc→

On comparing both side we get,

⇒a→·c→=1 & a→·b→=-λ

⇒3i^+j^·i^+2j^+k^=-λ

⇒λ=-5

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2022 (29 Jul Shift 1)
Let a→=3i^+j^-k^ and c→=2i^-3j^+3k^. If b→ is a vector such that a→=b→×c→ and |b→|2=50, then 72-b→+c→2 is equal to __________.

Options:

The correct answer is:
66

Solution:
Given,

a→=3i^+j^-k^ and c→=2i^-3j^+3k^.

And b→ is a vector such that a→=b→×c→ and |b→|2=50,

Now solving, |a→|=|b→×c→|

⇒32+12+12=|b→|·22+32+32·sinθ

⇒11=50·22·sinθ

∴sinθ=110 or cosθ=9910

Now solving,72-b→+c→2 we get,

72-b→+c→2

=72-b→2+c→2+2b→c→cosθ

=72-50+22+2×52.229910

=72-72+2×5×2×11×310

=66=66

Topic: Mathematics, Vector Algebra, JEE Main
Let a→=3i^+j^-k^ and c→=2i^-3j^+3k^. If b→ is a vector such that a→=b→×c→ and |b→|2=50, then 72-b→+c→2 is equal to __________.

Options:

The correct answer is:
66

Solution:
Given,

a→=3i^+j^-k^ and c→=2i^-3j^+3k^.

And b→ is a vector such that a→=b→×c→ and |b→|2=50,

Now solving, |a→|=|b→×c→|

⇒32+12+12=|b→|·22+32+32·sinθ

⇒11=50·22·sinθ

∴sinθ=110 or cosθ=9910

Now solving,72-b→+c→2 we get,

72-b→+c→2

=72-b→2+c→2+2b→c→cosθ

=72-50+22+2×52.229910

=72-72+2×5×2×11×310

=66=66

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (13 Apr Shift 1)
Let $\mathrm{A}=\left(\begin{array}{ccc}3-\mathrm{t} & 1 & 0 \\ -1 & 3-\mathrm{t} & 1 \\ 0 & -1 & 0\end{array}\right)$ and $\operatorname{det} \mathrm{A}=5$, then

Options:
A $\mathrm{t}=1$
B $\mathrm{t}=2$
C $t=-1$
D $\mathrm{t}=-2$

The correct answer is:
$\mathrm{t}=-2$

Solution:
Hint:
$|\mathrm{A}|=\left|\begin{array}{ccc}3-\mathrm{t} & 1 & 0 \\ -1 & 3-\mathrm{t} & 1 \\ 0 & -1 & 0\end{array}\right|=3-\mathrm{t}=5, \mathrm{t}=-2$

Topic: Mathematics, Determinants, JEE Main
Let $A=(4,0), B=(0,12)$ be two points in the plane. The locus of a point $C$ such that the area of triangle $A B C$ is 18 sq. units is -

Options:
A $(y+3 x+12)^{2}=81$
B $(y+3 x+81)^{2}=12$
C $(y+3 x-12)^{2}=81$
D $(y+3 x-81)^{2}=12$

The correct answer is:
$(y+3 x-12)^{2}=81$

Solution:
$\frac{1}{2}\left|\begin{array}{ccc}1 & x & y \\ 1 & 0 & 12 \\ 1 & 4 & 0\end{array}\right|=\pm 18$
$1(-48)-x(-12)+y(4)=\pm 36$
$12 x+4 y-48=\pm 36$
$3 x+y-12=\pm 9$
$(3 x+y-12)^{2}=81$

Topic: Mathematics, Straight Lines, JEE Main
Let A=4-2αβ. If A2+γA+18I=O, then detA is equal to _______.

Options:
A -18
B 18
C -50
D 50

The correct answer is:
18

Solution:
Given A=4-2αβ ⇒detA=4β+2α

Characteristic equation of the matrix is

4-λ-2αβ-λ=0

⇒4β+λ2-β+4λ+2α=0

⇒λ2-β+4λ+2α+4β=0

Comparing with given equation A2+γA+18I=0, we get

⇒γ=-β+4 & 2α+4β=18

Hence detA=4β+2α=18

Topic: Mathematics, Matrices, JEE Main, JEE Main 2022 (27 Jul Shift 2)
Let A={-4,-3,-2,0,1,3,4} and R={(a,b)∈A×A : b=|a| or b2=a+1 be a relation on A. Then the minimum number of elements, that must be added to the relation R so that it becomes reflexive and symmetric, is

Options:

The correct answer is:
7

Solution:
Given,

A={-4,-3,-2,0,1,3,4} and R={(a,b)∈A×A : b=|a| or b2=a+1 be a relation on A,

So, the relation is given by,

R={(-4,4),(-3,3),(0,0),(1,1),(3,3),(4,4),(0,1),(3,-2)}

Now, relation to be reflexive (a,a)∈R ∀ a∈A

⇒(-4,-4),(-3,-3),(-2,-2) also should be added in R.

Now relation to be symmetric if (a,b)∈R, then (b,a)∈R ∀ a,b∈A

⇒(4,-4),(3,-3),(1,0),(-2,3) also should be added in R

Hence, minimum number of elements to be added to
R=3+4=7

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2023 (13 Apr Shift 2)
Let A=-4-3-31a14b3 and A=A-1, then a+2b is equal to

Options:
A 0
B 4
C 8
D 5

The correct answer is:
8

Solution:
â´A=A-1âA2=I
A 2 =A⋅A= −4 −3 −3 1 a 1 4 b 3 −4 −3 −3 1 a 1 4 b 3
=112-3a-3b0a-3+a2+bab-4-12+ab+3bb-3
â´A2=Iâa=0,b=4âa+2b=8

Topic: Mathematics, Determinants, JEE Main
Let A4,-4 and B9,6 be points on the parabola, y2=4x. Let C be chosen on the arc AOB of the parabola, where O is the origin, such that the area of ΔACB is maximum. Then, the area (in sq. units) of ΔACB , is:

Options:
A 32
B 3134
C 3012
D 3114

The correct answer is:
3114

Solution:
Let the coordinates of C be t2, 2t.
Area of ∆ABC=12 t22t19614-41
∆=56+t-t2, t∈-2, 3
d∆dt=1-2t=0⇒t=12

d2△dt2=-2, hence area is maximum at t=12
∴ Δmax=1254=3114

Topic: Mathematics, Parabola, JEE Main, JEE Main 2019 (09 Jan Shift 2)
Let a→=4i^+3j^ and b→=3i^-4j^+5k^ and c→ is a vector such that c→·(a→×b→)+25=0,c→·(i^+j^+k^)=4 and projection of c→ on a→ is 1 , then the projection of c→ on b→ equals:

Options:
A 52
B 15
C 12
D 32

The correct answer is:
52

Solution:
Given:

a→=4i^+3j^

b→=3i^-4j^+5k^

So,

a→×b→=i^j^k^4303-45

⇒a→×b→=15i^-20j^-25k^

Let c→=xi^+yj^+zk^

Then,

c→·(a→×b→)+25=0

⇒15x-20y-25z+25=0

⇒3x-4y-5z=-5 ...1

Also,

c→·i^+j^+k^=4

⇒x+y+z=4 ....2

And projection of c→ on a→ is

c→·a→|a→|=1

⇒xi^+yj^+zk^·4i^+3j^16+9=1

⇒4x+3y=5 ....3

Solving 1, 2 & 3, we get

x=2, y=-1, z=3

So,

c→=2i^-j^+3k^

Projection of c→ on b→ is

c→·b→b→=2i^-j^+3k^·3i^-4j^+5k^9+16+25=2552=52

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (29 Jan Shift 2)
Let $\vec{a}=4 \hat{i}-\hat{j}+\hat{k}, \vec{b}=11 \hat{i}-\hat{j}+\hat{k}$ and $\vec{c}$ be a vector such that $(\vec{a}+\vec{b}) \times \vec{c}=\vec{c} \times(-2 \vec{a}+3 \vec{b})$. If $(2 \vec{a}+3 \vec{b}) \cdot \vec{c}=1670$, then $|\vec{c}|^2$ is equal to :

Options:
A 1609
B 1618
C 1600
D 1627

The correct answer is:
1618

Solution:
$\begin{aligned} & (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{c}} \times(-2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}})=0 \\ & (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}+(-2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=0 \\ & \Rightarrow(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})-2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=0 \\ & \Rightarrow \overrightarrow{\mathrm{c}}=\lambda(4 \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}) \\ & \Rightarrow=\lambda(44 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}-4 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}) \\ & =\lambda(40 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})\end{aligned}$
Now
$\begin{aligned}
& (8 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}+33 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \cdot \lambda(40 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})=1670 \\
& \Rightarrow(41 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}) \cdot(40 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \times \lambda=1670) \\
& \Rightarrow(1640+15+15) \lambda=1670 \Rightarrow \lambda=1 \\
& \text { so } \overrightarrow{\mathrm{c}}=40 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}-3 \hat{\mathrm{k}} \\
& \Rightarrow|\overrightarrow{\mathrm{c}}|^2=1600+9+9=1618
\end{aligned}$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2024 (08 Apr Shift 2)
Let $A=\left[\begin{array}{ccc}5 & 5 \alpha & \alpha \\ 0 & \alpha & 5 \alpha \\ 0 & 0 & 5\end{array}\right]$. If $\left|A^2\right|=25$, then $|\alpha|$ equals

Options:
A
$5^2$

B
$1$

C
$1 / 5$

D
$5$

The correct answer is:
$1 / 5$

Solution:
$\mathrm{A}^2=\left[\begin{array}{ccc}5 & 5 \alpha & \alpha \\ 0 & \alpha & 5 \alpha \\ 0 & 0 & 5\end{array}\right]\left[\begin{array}{ccc}5 & 5 \alpha & \alpha \\ 0 & \alpha & 5 \alpha \\ 0 & 0 & 5\end{array}\right]$
$\mathrm{A}^2=\left[\begin{array}{ccc}25 & 25 \alpha+5 \alpha^2 & 5 \alpha+25 \alpha^2+5 \alpha \\ 0 & \alpha^2 & 5 \alpha^2+25 \alpha \\ 0 & 0 & 25\end{array}\right]$
$625 \alpha^2=25$
$\Rightarrow|\alpha|=\frac{1}{5}$.

Topic: Mathematics, Matrices, JEE Main, JEE Main 2007
Let a→=5i^-j^-3k^ and b→=i^+3j^+5k^ be two vectors. Then which one of the following statements is TRUE?

Options:
A Projection of a→ on b→ is -1335 and the direction of the projection vector is opposite to the direction of b→
B Projection of a→ on b→ is -1735 and the direction of the projection vector is opposite to the direction of b→
C Projection of a→ on b→ is 1735 and the direction of the projection vector is opposite to the direction of b→
D Projection of a→ on b→ is 1335 and the direction of the projection vector is opposite to the direction of a→

The correct answer is:
Projection of a→ on b→ is -1335 and the direction of the projection vector is opposite to the direction of b→

Solution:
Given

a→=5i^-j^-3k^

So, a→=52+12+32=35

b→=i^+3j^+5k^

So, 52+12+32=35

So, a→·b→=5-3-15=-13

Now finding the projection of a→ on b→ using the formula we get,

a→·b→b→=-1335

And projection of b→ on a→ is given by,

a→·b→a→=-1335

Note- The original answer key was dropped by NTA in Jee mains 1st feb 2023, here we have modified the options.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (01 Feb Shift 2)
Let a→=−5i^+j^−3k^, b→=i^+2j^−4k^ and c→=a→×b→×i^×i^×i^. Then c→⋅−i^+j^+k^ is equal to

Options:
A -12
B -10
C -13
D -15

The correct answer is:
-12

Solution:
Given:

a→=−5i^+j−3k^ and b→=i^+2j^−4k^

⇒a→×b→×i^=a→⋅i^b→−b→⋅i^a→

⇒a→×b→×i^=−5b→−a→

⇒a→×b→×i^=−5b→−a→×i^×i^

⇒a→×b→×i^=−11j^+23k^×i^×i^

⇒a→×b→×i^=11k^+23j^×i^

⇒a→×b→×i^=11j^−23k^

⇒c→.−i^+j^+k^=11−23

⇒c→.−i^+j^+k^=−12

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2024 (01 Feb Shift 1)
Let a→=6i^+9j^+12k^, b→=αi^+11j^-2k^ and c→ be vectors such that a→×c→=a→×b→ If a→·c→=-12, and c→·i^-2j^+k^=5 then c→·i^+j^+k^ is equal to _______

Options:

The correct answer is:
11

Solution:
Given,

a→=6i^+9j^+12k^, b→=αi^+11j^-2k^ and c→ be vectors such that a→×c→=a→×b→,

a→×c→-a→×b→=0

⇒a→×c→-b→=0

So, a→ & c→-b→ are parallel vectors,

Hence, λa→=c→-b→

⇒c→=b→+λa→

⇒a→·c→=a→·b→+λa→2

⇒-12 =6α+75+λ261

⇒2α +87λ=-29 ....(i)

Now again using c→=b→+λa→ we get,

c→=i^α+6λ+j^(11+9λ)+k^(-2+12λ)

Also given c→·i^-2j^+k^=5

⇒α+6λ-2(11+9λ)+(-2+12λ)=5

⇒α=29

So, 2α +87λ=-29

⇒λ=-1

Hence, c→=23i^+2j^-14k^

So, the value of c→·(i^+j^+k^)=23+2-14=11

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2023 (08 Apr Shift 1)
Let $\overrightarrow{\mathrm{a}}=6 \hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{\mathrm{b}}=\hat{i}+\hat{j}$. If $\overrightarrow{\mathrm{c}}$ is a is vector such that $|\overrightarrow{\mathrm{c}}| \geq 6, \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=6|\overrightarrow{\mathrm{c}}|,|\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}|=2 \sqrt{2}$ and the angle between $\vec{a} \times \vec{b}$ and $\vec{c}$ is $60^{\circ}$, then $|(\vec{a} \times \vec{b}) \times \vec{c}|$ is equal to:

Options:
A $\frac{9}{2}(6-\sqrt{6})$
B $\frac{3}{2} \sqrt{6}$
C $\frac{9}{2}(6+\sqrt{6})$
D $\frac{3}{2} \sqrt{3}$

The correct answer is:
$\frac{9}{2}(6+\sqrt{6})$

Solution:
$\begin{aligned} & |(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})|=|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}||\overrightarrow{\mathrm{c}}| \frac{\sqrt{3}}{2} \\ & |\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}|=2 \sqrt{2} \\ & |\mathrm{c}|^2+|\mathrm{a}|^2-2 \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=8 \\ & |\mathrm{z}|^2+38-12|\mathrm{z}|=8 \\ & |\mathrm{z}|^2-12|\mathrm{z}|+30=0 \\ & |\mathrm{z}|=\frac{12 \pm \sqrt{144-120}}{2} \\ & =\frac{12 \pm 2 \sqrt{6}}{2}\end{aligned}$
$\begin{aligned} & |z|=6+\sqrt{6} \\ & \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\left|\begin{array}{ccc}\hat{\ell} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 6 & 1 & -1 \\ 1 & 1 & 0\end{array}\right| \\ & \hat{\ell}-\hat{\mathrm{j}}+5 \hat{\mathrm{k}} \\ & |\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=\sqrt{27} \\ & |(\overrightarrow{\mathrm{a}} \times \mathrm{b}) \times \mathrm{z}|=\sqrt{27}(6+\sqrt{6}) \frac{\sqrt{3}}{2} \\ & \frac{9}{2}(6+\sqrt{6})\end{aligned}$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2024 (06 Apr Shift 2)
Let $\mathrm{A}=\left(\begin{array}{ccc}12 & 24 & 5 \\ \mathrm{x} & 6 & 2 \\ -1 & -2 & 3\end{array}\right) .$ The value of $\mathrm{x}$ for which the matrix $\mathrm{A}$ is not invertible is

Options:
A 6
B 12
C 3
D 2

The correct answer is:
3

Solution:
Hint:
If matrix is not invertible $\Rightarrow|\mathrm{A}|=0$
$\therefore|\mathrm{A}|=\left|\begin{array}{ccc}12 & 24 & 5 \\ \mathrm{x} & 6 & 2 \\ -1 & -2 & 3\end{array}\right|=0 \Rightarrow \mathrm{x}=3$

Topic: Mathematics, Matrices, JEE Main
Let A2,0 and B-2,0 are two fixed vertices of âABC. If the vertex C moves in the first quadrant in such a way that cotâ¡A+cotâ¡B=2, then the locus of the point C is

Options:
A y=2
B x=4
C x=2
D y=1

The correct answer is:
y=2

Solution:
Given that, coordinates of two fixed points A and B are 2,0 and -2,0 respectively
Let variable point C is h,k.
From the adjoining figure,
cotâ¡A=DACD=2-hk
cotâ¡B=BDCD=2+hk
But cotâ¡A+cotâ¡B=2, so we have
2-hk+2+hk=2â2k=1
Hence, locus of C is y=2

Topic: Mathematics, Straight Lines, JEE Main
Let A6,7,B2,3 and C-2,1 be the vertices of a triangle. The point P nearer to the point A such that ΔPBC is an equilateral triangle is

Options:
A -3,2+23
B 3,2+23
C 3,2-23
D -3,2-23

The correct answer is:
-3,2+23

Solution:
We have, BC=25.
Since, ÎPBC is an equilateral triangle and the point P lies in the interior of ÎABC.

Therefore, P lies on the perpendicular bisector of BC at a distance of 32BC=15 from the mid-point of BC such that P and A are on the same side of BC.
The equation of BC is x-2y+4=0
The coordinates of the mid-point of BC are D0,2
The slope of a line perpendicular to BC is -2.

So, if it makes an angle Î¸ with the x -axis. Then,
tanâ¡Î¸=-2âsinâ¡Î¸=25 and cosâ¡Î¸=-15
The parametric form of the perpendicular bisectors of BC is
x-0-15=y-225
So, the coordinates of P are given by
x-0-15=y-225=Â±15
or, x=Â±3,y=2Â±23
Thus, the coordinates of P are 3,2-23 or -3,2+23.
Clearly, A6,7 and -3,2+23 lie on the same side of BC. Hence, the coordinates of P are -3,2+23.

Topic: Mathematics, Straight Lines, JEE Main
Let A=a030b5-130c, where a,b,c are positive integers. If trA=7, then the greatest value of A is (where trA denotes the trace of matrix A i.e. the sum of principal diagonal elements of matrix A)

Options:
A 6
B 12
C 15
D 10

The correct answer is:
15

Solution:
trA=a+b+c=7
Ordered triplets of a,b,c are 1,1,5,1,2,4,1,3,3,2,1,4,2,2,3,2,3,2,2,4,1,3,1,3,3,2,2,3,3,1,4,1,2,4,2,1,5,1,1
A=a030b5-130c=abc+3b3=abc+b
A is greatest for 2,3,2 and equal to 15

Topic: Mathematics, Determinants, JEE Main
Let A(a, 0), B(b, 2b+1) and C(0,b), b≠0, |b|≠1, be points such that the area of triangle ABC is 1 sq. unit, then the sum of all possible values of a is:

Options:
A -2bb+1
B 2b2b+1
C -2b2b+1
D 2bb+1

The correct answer is:
-2b2b+1

Solution:
Given coordinates of triangle ABC, A(a,0),B(b,2b+1),C(0,b)

Area=12a01b2b+110b1

12a(b+1)+b2=±1

ab+a+b2=±2

If 2 is positive

a=2−b2b+1

If 2 is negative

a=−2−b2b+1

Sum of possible values of a=2-b2b+1+-2-b2b+1

=-2b2b+1

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2021 (27 Aug Shift 2)
Let A=a1a2 and B=b1b2 be two 2×1matrices with real entries such that A=XB, where X=131-11k, and k∈R. If a12+a22=23b12+b22 and k2+1b22≠-2 b1b2, then the value of k is __________.

Options:

The correct answer is:
1

Solution:
A=XB

a1a2=131-11kb1 b2

3a13a2=b1-b2 b1+kb2

b1-b2=3a1 …1

b1+kb2=3a2 …2

Given, a12+a22=23b12+b22

(1)2+(2)2

b1+b22+b1+kb22=3a12+a22

a12+a22=23b12+1+k23b22+23b1b2k-1

Given, a12+a22=23b12+23b22

On comparing we get

k2+13=23⇒k2+1=2

⇒k=±1 …3

23(k-1)=0⇒k=1 …4

From both we get k=1

Topic: Mathematics, Matrices, JEE Main, JEE Main 2021 (16 Mar Shift 2)