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Phrases Vector PYQ

Phrases 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






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Phrases Previous Year PYQ Phrases 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
Phrases 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:





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Phrases Previous Year PYQ Phrases 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
Phrases 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)





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Phrases Previous Year PYQ Phrases 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: $\pm \dfrac{1}{3}(2\hat{i}+2\hat{j}-\hat{k})$
Phrases 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






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Phrases Previous Year PYQPhrases 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
Phrases 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 $\gamma$.





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Phrases Previous Year PYQ Phrases 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: $3(2-\gamma) + 1(2+2\gamma) = 0$ $6 - 3\gamma + 2 + 2\gamma = 0$ $8 - \gamma = 0 \Rightarrow \gamma = 8$
Phrases 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:





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Phrases Previous Year PYQ Phrases 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:
$$ \boxed{60^\circ} $$
Phrases 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 





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Phrases Previous Year PYQ Phrases 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:
$$ \boxed{9.5} $$
Phrases 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





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Phrases Previous Year PYQ Phrases 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
$\left[0,\dfrac{\pi}{2}\right]$
Phrases 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 





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Phrases Previous Year PYQ Phrases CUET 2022 PYQ

Solution

Phrases 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





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Phrases Previous Year PYQ Phrases CUET 2023 PYQ

Solution

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





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Phrases Previous Year PYQ Phrases CUET 2023 PYQ

Solution

XOR and XNOR are complements of each other. Statement (1) claims them equal, which is false.
Phrases 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





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Phrases Previous Year PYQ Phrases 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, $=16+4+36+2\left(\dfrac{4\cdot2+2\cdot6+6\cdot4}{2}\right)$ $=56+44=100$ $\Rightarrow |\vec a+\vec b+\vec c|=10$
Phrases 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





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Phrases Previous Year PYQ Phrases CUET 2022 PYQ

Solution

Phrases 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 





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Phrases Previous Year PYQ Phrases CUET 2022 PYQ

Solution

Phrases 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:






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Phrases Previous Year PYQ Phrases 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
Phrases 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. $60^\circ$





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Phrases Previous Year PYQ Phrases 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
Phrases 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





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Phrases Previous Year PYQ Phrases 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,
$\cos\dfrac{\theta}{2}=\dfrac12|\hat n_1+\hat n_2|$

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