Aspire's Library

A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations
  • CUET
    Previous Year Questions

CUET Previous Year Questions (PYQs)

CUET Vector PYQ

CUET PYQ
List I List II
A. Kailash Satyarthi I. Chemistry
B. Abhijit Banerjee II. Peace
C. Vinkatraman Ramakrishnan III. Physics
D. Subrahmanyan Chandrasekhar IV. Economics






Go to Discussion
CUET Previous Year PYQ CUET CUET 2023 PYQ

Solution

List I List II
A. Kailash Satyarthi II. Peace
B. Abhijit Banerjee IV. Economics
C. Venkatraman Ramakrishnan I. Chemistry
D. Subrahmanyan Chandrasekhar III. Physics
CUET PYQ
Which of the following is true:

A. Two vectors are said to be identical if their difference is zero.
B. Velocity is not a vector quantity.
C. Projection of one vector on another is not an application of dot product.
D. The maximum space rate of change of the function which is increasing direction of line function is known as gradient of scalar function.

Choose the most appropriate answer from the options given below:





Go to Discussion
CUET Previous Year PYQ CUET CUET 2023 PYQ

Solution

Statement A is true because if the difference of two vectors is zero, then both vectors are equal in magnitude and direction.
Statement B is false because velocity is a vector quantity.
Statement C is false because projection of one vector on another is an application of dot product.
Statement D is true because gradient gives the direction and maximum rate of increase of a scalar function.

Correct answer: (3) A and D only
CUET PYQ
The unit vectors orthogonal to the vector $- \hat{i} + 2\hat{j} + 2\hat{k}$ and making equal angles with the x and y axis is (are)





Go to Discussion
CUET Previous Year PYQ CUET CUET 2023 PYQ

Solution

Let required unit vector be $\vec{a}=l\hat{i}+m\hat{j}+n\hat{k}$. Equal angles with x and y axis $\Rightarrow l=m$. Orthogonal condition: $\vec{a}\cdot(-\hat{i}+2\hat{j}+2\hat{k})=0$ $-l+2m+2n=0$ Since $l=m$: $-l+2l+2n=0 \Rightarrow l+2n=0 \Rightarrow n=-\dfrac{l}{2}$ So vector $\propto (l,l,-\dfrac{l}{2}) \Rightarrow (2,2,-1)$ Unit vector: $\pm \dfrac{1}{3}(2\hat{i}+2\hat{j}-\hat{k})$
CUET PYQ
List I List II
A. Dog : Rabies :: Mosquito : I. Bacteria
B. Amnesia : Memory :: Paralysis : II. Liver
C. Meningitis : Brain :: Cirrhosis : III. Movement
D. Influenza : Virus :: Typhoid : IV. Malaria






Go to Discussion
CUET Previous Year PYQCUET CUET 2023 PYQ

Solution

Dog causes Rabies, Mosquito causes Malaria → A-IV
Amnesia affects Memory, Paralysis affects Movement → B-III
Meningitis affects Brain, Cirrhosis affects Liver → C-II
Influenza is caused by Virus, Typhoid is caused by Bacteria → D-I

Correct matching: A-IV, B-III, C-II, D-I
CUET PYQ
$\vec a = 2\hat i + 2\hat j + 3\hat k,; \vec b = -\hat i + 2\hat j + \hat k$ and $\vec c = 3\hat i + \hat j$ are such that $\vec a + \gamma \vec b$ is perpendicular to $\vec c$, then determine the value of $\gamma$.





Go to Discussion
CUET Previous Year PYQ CUET CUET 2023 PYQ

Solution

Perpendicular condition: $(\vec a + \gamma \vec b)\cdot \vec c = 0$ $\vec a + \gamma \vec b = (2-\gamma)\hat i + (2+2\gamma)\hat j + (3+\gamma)\hat k$ $\vec c = 3\hat i + \hat j$ Dot product: $3(2-\gamma) + 1(2+2\gamma) = 0$ $6 - 3\gamma + 2 + 2\gamma = 0$ $8 - \gamma = 0 \Rightarrow \gamma = 8$
CUET PYQ
If $\vec{a}$ and $\vec{b}$ are two unit vectors such that $\vec{a}+2\vec{b}$ and $5\vec{a}-4\vec{b}$ are perpendicular to each other, then the angle between $\vec{a}$ and $\vec{b}$ is:





Go to Discussion
CUET Previous Year PYQ CUET CUET 2022 PYQ

Solution

Solution:

Given two unit vectors \( \vec{a} \) and \( \vec{b} \), and the vectors \( \vec{a} + 2\vec{b} \) and \( 5\vec{a} - 4\vec{b} \) are perpendicular, we use the condition for perpendicular vectors: $$ (\vec{a} + 2\vec{b}) \cdot (5\vec{a} - 4\vec{b}) = 0 $$ Expanding the dot product: $$ (\vec{a} + 2\vec{b}) \cdot (5\vec{a} - 4\vec{b}) = \vec{a} \cdot 5\vec{a} + \vec{a} \cdot (-4\vec{b}) + 2\vec{b} \cdot 5\vec{a} + 2\vec{b} \cdot (-4\vec{b}) $$ Using properties of dot products and knowing \( \vec{a} \) and \( \vec{b} \) are unit vectors (\( \vec{a} \cdot \vec{a} = 1 \) and \( \vec{b} \cdot \vec{b} = 1 \)): $$ 5(\vec{a} \cdot \vec{a}) - 4(\vec{a} \cdot \vec{b}) + 10(\vec{b} \cdot \vec{a}) - 8(\vec{b} \cdot \vec{b}) = 0 $$ Simplifying: $$ 5(1) - 4(\vec{a} \cdot \vec{b}) + 10(\vec{a} \cdot \vec{b}) - 8(1) = 0 $$ $$ 5 - 8 + 6(\vec{a} \cdot \vec{b}) = 0 $$ $$ -3 + 6(\vec{a} \cdot \vec{b}) = 0 $$ $$ 6(\vec{a} \cdot \vec{b}) = 3 $$ $$ \vec{a} \cdot \vec{b} = \frac{1}{2} $$ The dot product \( \vec{a} \cdot \vec{b} = \cos \theta \), where \( \theta \) is the angle between \( \vec{a} \) and \( \vec{b} \): $$ \cos \theta = \frac{1}{2} $$ Therefore, the angle \( \theta \) is: $$ \theta = \cos^{-1} \left( \frac{1}{2} \right) = 60^\circ $$
Final Answer:
$$ \boxed{60^\circ} $$
CUET PYQ
Let $\vec{a}=\hat{i}-\hat{j}$ and $\vec{b}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{c}$ be a vector such that $(\vec{a} \times \vec{c})+\vec{b}=0$ and $\vec{a}.\vec{c}=4$, then $|\vec{c}|^2$ is equal to 





Go to Discussion
CUET Previous Year PYQ CUET CUET 2022 PYQ

Solution

Solution:

Given vectors:
  • \(\vec{a} = \hat{i} - \hat{j}\)
  • \(\vec{b} = \hat{i} + \hat{j} + \hat{k}\)
  • And \((\vec{a} \times \vec{c}) + \vec{b} = 0\)
  • \(\vec{a} \cdot \vec{c} = 4\)
From \((\vec{a} \times \vec{c}) + \vec{b} = 0\), we get: \[ \vec{a} \times \vec{c} = -\vec{b} \] Let \(\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}\). The cross product \(\vec{a} \times \vec{c}\) is: \[ \vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 0 \\ x & y & z \end{vmatrix} \] Expanding this determinant: \[ \vec{a} \times \vec{c} = (z \hat{i} + z \hat{j} + (x + y) \hat{k}) \] Setting \(\vec{a} \times \vec{c} = -\vec{b}\), we get: \[ z = -1, \quad z = -1, \quad x + y = -1 \] Therefore: \[ x + y = -1 \] Now, from \(\vec{a} \cdot \vec{c} = 4\): \[ \vec{a} \cdot \vec{c} = 1 \cdot x + (-1) \cdot y = 4 \] Simplifying: \[ x - y = 4 \] Solving the system of equations: \[ x + y = -1 \] \[ x - y = 4 \] Adding the two equations: \[ 2x = 3 \quad \Rightarrow \quad x = \frac{3}{2} \] Substituting into \(x + y = -1\): \[ \frac{3}{2} + y = -1 \quad \Rightarrow \quad y = -\frac{5}{2} \] Now, \(\vec{c} = \frac{3}{2} \hat{i} - \frac{5}{2} \hat{j} - \hat{k}\). To find \(|\vec{c}|^2\), we compute: \[ |\vec{c}|^2 = \left( \frac{3}{2} \right)^2 + \left( -\frac{5}{2} \right)^2 + (-1)^2 = \frac{9}{4} + \frac{25}{4} + 1 \] \[ |\vec{c}|^2 = \frac{9 + 25 + 4}{4} = \frac{38}{4} = 9.5 \]
Final Answer:
$$ \boxed{9.5} $$
CUET PYQ
If the unit vectors $\vec a$ and $\vec b$ are inclined at an angle $2\theta$ such that $|\vec a - \vec b| < 1$ and $0 \le \theta \le \pi$, then $\theta$ lies in the interval





Go to Discussion
CUET Previous Year PYQ CUET CUET 2023 PYQ

Solution

$|\vec a - \vec b| = \sqrt{2-2\cos 2\theta}$

Given:
$\sqrt{2-2\cos 2\theta} < 1$

Squaring:
$2-2\cos 2\theta < 1$

$\cos 2\theta > \dfrac{1}{2}$

So,
$0 \le 2\theta < \dfrac{\pi}{3}$

$\Rightarrow 0 \le \theta < \dfrac{\pi}{6}$

From given options, valid interval is included in
$\left[0,\dfrac{\pi}{2}\right]$
CUET PYQ
If $\vec{a}$, $\vec{b}$, $\vec{c}$ and $\vec{d}$ are the unit vectors such that $(\vec{a} \times \vec{b}).(\vec{c} \times \vec{d})=1$ and $(\vec{a}.\vec{c})=\frac{1}{2}$, then 





Go to Discussion
CUET Previous Year PYQ CUET CUET 2022 PYQ

Solution

CUET PYQ
The moment of the couple formed by the forces $5\hat i+\hat k$ and $-5\hat i-\hat k$ acting at the points $(9,-1,2)$ and $(3,-2,1)$ respectively is





Go to Discussion
CUET Previous Year PYQ CUET CUET 2023 PYQ

Solution

Moment of a couple $=\vec r \times \vec F$ $\vec r=(9,-1,2)-(3,-2,1)=(6,1,1)$ $\vec F=5\hat i+\hat k$ $\vec M=\begin{vmatrix} \hat i & \hat j & \hat k \ 6 & 1 & 1 \ 5 & 0 & 1 \end{vmatrix} = \hat i(1-0)-\hat j(6-5)+\hat k(0-5)$ $=\hat i-\hat j-5\hat k$
CUET PYQ
If $\oplus$ and $\odot$ denote exclusive OR and exclusive NOR operations respectively, then which one of the following is not correct?





Go to Discussion
CUET Previous Year PYQ CUET CUET 2023 PYQ

Solution

XOR and XNOR are complements of each other. Statement (1) claims them equal, which is false.
CUET PYQ
Each of the angle between vectors $\vec a$, $\vec b$ and $\vec c$ is equal to $60^\circ$. If $|\vec a|=4$, $|\vec b|=2$ and $|\vec c|=6$, then the modulus of $\vec a+\vec b+\vec c$ is





Go to Discussion
CUET Previous Year PYQ CUET CUET 2023 PYQ

Solution

$|\vec a+\vec b+\vec c|^2 = a^2+b^2+c^2 +2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)$ Since angle $=60^\circ$, $\vec a\cdot\vec b=|a||b|\cos60^\circ=\dfrac{ab}{2}$ So, $=16+4+36+2\left(\dfrac{4\cdot2+2\cdot6+6\cdot4}{2}\right)$ $=56+44=100$ $\Rightarrow |\vec a+\vec b+\vec c|=10$
CUET PYQ
If $\vec{a}$, $\vec{b}$ and $\vec{c}$ are unit vectors, then $|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2 $ does not exceed





Go to Discussion
CUET Previous Year PYQ CUET CUET 2022 PYQ

Solution

CUET PYQ
If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$, $\vec{a}.\vec{b}=1$ and $\vec{a} \times \vec{b}=\hat{j}-\hat{k}$, then $\vec{b}$ is equal to 





Go to Discussion
CUET Previous Year PYQ CUET CUET 2022 PYQ

Solution

CUET PYQ
Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R.

Assertion A:
If dot product and cross product of vectors $\vec A$ and $\vec B$ are zero, it implies that one of the vectors $\vec A$ or $\vec B$ must be a null vector.

Reason R:
Null vector is a vector with zero magnitude.

Choose the correct answer:






Go to Discussion
CUET Previous Year PYQ CUET CUET 2023 PYQ

Solution

$\vec A\times\vec B=0$ ⇒ vectors are parallel

A non-zero vector cannot be both parallel and perpendicular
⇒ one vector must be zero

✔ Assertion A is true
✔ Reason R is true
✔ R explains A
CUET PYQ
LIST I LIST II
A. $|\vec A+\vec B|=|\vec A-\vec B|$ I. $45^\circ$
B. $|\vec A\times\vec B|=\vec A\cdot\vec B$ II. $30^\circ$
C. $|\vec A\cdot\vec B|=\dfrac{AB}{2}$ III. $90^\circ$
D. $|\vec A\times\vec B|=\dfrac{AB}{2}$ IV. $60^\circ$





Go to Discussion
CUET Previous Year PYQ CUET CUET 2023 PYQ

Solution

A:
$|\vec A+\vec B|=|\vec A-\vec B| \Rightarrow \vec A\cdot\vec B=0$
$\Rightarrow$ angle $=90^\circ$ → III

B:
$|\vec A\times\vec B|=\vec A\cdot\vec B$
$AB\sin\theta=AB\cos\theta \Rightarrow \theta=45^\circ$ → I

C:
$|\vec A\cdot\vec B|=AB\cos\theta=\dfrac{AB}{2}$
$\Rightarrow \cos\theta=\dfrac12 \Rightarrow \theta=60^\circ$ → IV

D:
$|\vec A\times\vec B|=AB\sin\theta=\dfrac{AB}{2}$
$\Rightarrow \sin\theta=\dfrac12 \Rightarrow \theta=30^\circ$ → II

Correct Matching:
A-III, B-I, C-IV, D-II
CUET PYQ
If $\hat n_1,\hat n_2$ are two unit vectors and $\theta$ is the angle between them, then $\cos\dfrac{\theta}{2}$ is equal to





Go to Discussion
CUET Previous Year PYQ CUET CUET 2023 PYQ

Solution

$|\hat n_1+\hat n_2|^2=2(1+\cos\theta)$

So,
$|\hat n_1+\hat n_2|=2\cos\dfrac{\theta}{2}$

Hence,
$\cos\dfrac{\theta}{2}=\dfrac12|\hat n_1+\hat n_2|$

CUET

Online Test Series,
Information About Examination,
Syllabus, Notification
and More.

Click Here to
View More

CUET

Online Test Series,
Information About Examination,
Syllabus, Notification
and More.

Click Here to
View More
Limited Seats
× Aspire MCA Promotion

Game Changer NIMCET Test Series 2026

Boost your preparation with mock tests, analysis and rank-focused practice.

JOIN NOW
Ask Your Question or Put Your Review.

loading...