\(\overrightarrow{v_{w}}=10 \mathrm{~m} / \mathrm{s}=\overrightarrow{\mathrm{OA}}\)
\(\vec{v}_{r}=30 \mathrm{~m} / \mathrm{s}=\overrightarrow{\mathrm{OB}}\)
\(\therefore \quad \overrightarrow{v_{r w}}=\sqrt{v_{r}^{2}+v_{w}^{2}}\)
\(=\sqrt{(10)^{2}+(30)^{2}}=31.6 \mathrm{~m} / \mathrm{s}\)
\(\tan \theta=\frac{\mathrm{BD}}{\mathrm{OB}}=\frac{10}{30}=0.33(\mathrm{BD}=\mathrm{OA})\)
θ = 18°16' with vertical direction.

Work and current are the scalar quantities in the given list

Here, r = 1 km = 1000 m,
v = 900 kmh^-1 = 900 x (1000m(/(60 x 60s) = 250 ms^-1
Centripetal acceleration, a = \(\frac { { v }^{ 2 } }{ r } =\frac { { (250) }^{ 2 } }{ 1000 } \)
Now,  \(\frac { a }{ g } =\frac { { (250) }^{ 2 } }{ 1000 } \times\frac { 1 }{ 9.8 } =6.38\)

Here, r = 80 cm = 0.8 m, f = 14/25 s^{-1
\(\therefore \omega =2\pi f=2\times\frac { 22 }{ 7 } \times\frac { 14 }{ 25 } =\frac { 88 }{ 25 } rad/s\)}
The cantripetal acceleration
\(a={ \omega }^{ 2 }r=(\frac { 88 }{ 25 } { ) }^{ 2 }\times0.80=9.90m{ s }^{ -1 }\)
The direction of centripetal acceleration is along the string directed towards the centre of circular path.

(i) 12s,
(ii) 4.1 m /s ^2,
(iii) down and
(iv) up

Here, actual path length travelled, s = 23 km; Displacement = 10 km;
Time taken, t  = 28 min = \(\frac{28}{60} h\)
(a) Average speed of taxi = \(\frac{\text { actual path length }}{\text { time taken }}=\frac{23}{\frac{28}{60}} k \frac{m}{h}=49.3 \mathrm{~km} / \mathrm{h}\)
(b) Magnitude of average velocity = \(=\frac{\text { displacement }}{\text { time taken }}=\frac{10}{\frac{28}{60}} \mathrm{~km} / \mathrm{h}=21.4 \mathrm{~km} / \mathrm{h}\)
The average speed is not equal to the magnitude of average velocity. The two are equal for the motion of taxi along a straight path in one direction.

We know that \(A.B=ABcos\theta \)  when vectors are orthogonal \(\theta ={ 90 }^{ o }\)
So,    \(A.B=ABcos90^{ o }=0\)  when vectors are parallel   \(\theta ={ 0 }^{ o }\)   so,   \(A.B=ABcos{ 0 }^{ o }=AB\ (maximum)\)

(i) \(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -2 & 3 & -4 \\ 3 & -4 & 5 \end{array}\right|\)
\(=\hat{i}\left|\begin{array}{rr} 3 & -4 \\ -4 & 5 \end{array}\right|-\hat{j}\left|\begin{array}{rr} -2 & -4 \\ 3 & 5 \end{array}\right|+\hat{k}\left|\begin{array}{rr} -2 & 3 \\ 3 & -4 \end{array}\right|\)
\(=\hat{i}(15-16)-\hat{j}(-10+12)+\hat{k}(8-9)\)
\(=-\hat{i}-2 \hat{j}-\hat{k}\)
(ii) \(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=(-2 \hat{i}+3 \hat{j}-4 \hat{k}) \cdot(3 \hat{i}-4 \hat{j}+5 \hat{k})\)
= - 6 - 12 - 20 = -38

\(\overrightarrow{\mathrm{A}}=\hat{i}+2 \hat{j}+3 \hat{k}\)
\(\overrightarrow{\mathrm{B}}=2 \hat{i}-\hat{j}\)
\(\overrightarrow{\mathrm{A}}\) is perpendicular to \(\overrightarrow{\mathrm{B}}\)
\(\text { if } \quad \overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=0\)
\(\therefore(\hat{i}+2 \hat{j}+3 \hat{k}) \cdot(2 \hat{i}-\hat{j})=2-2=0\)
Thus \(\overrightarrow{\mathrm{A}}\) is perpendicular to \(\overrightarrow{\mathrm{B}}\).

Area of the parallelogram = \(|\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}|\)
\(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}=\left|\begin{array}{rrr} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & -3 & 1 \end{array}\right|=11 \hat{i}+5 \hat{j}-7 \hat{k}\)
\(\therefore \quad|\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}|=\sqrt{121+25+49}=\sqrt{195} \text { sq. units }\) 

Let the velocity of rain be \(\overrightarrow{v_{A}}=(10 \sqrt{3}) \mathrm{ms}^{-1},\) vertically downwards
Velocity of the man on the bicycle \(\overrightarrow{v_{B}}=10 \mathrm{~ms}^{-1}\) north to south
Relative velocity of rain W.r.t. man
\(\vec{v}=\overrightarrow{v_{A}}-\overrightarrow{v_{B}}\)
From ΔOLM \(\overrightarrow{\mathrm{ML}}=\overrightarrow{v_{\mathrm{A}}}=(10 \sqrt{3}) \mathrm{ms}^{-1}\)
\(\overrightarrow{\mathrm{LO}}=-\overrightarrow{v_{B}}=-10 \mathrm{~ms}^{-1}\)
\(\vec{v}=\overrightarrow{v_{A}}+\left(-\overrightarrow{v_{B}}\right)=\overrightarrow{\mathrm{MO}}\)

Using trigonometric considerations,
\(\tan \theta=\frac{10 \sqrt{3}}{10}=\sqrt{3}=\tan 60^{\circ}\)
ஃ θ = 60°
Thus, in order to guard himself, the man should keep his umbrella along OM., i.e., at 60° to the horizontal.

Two balls of different masses m₁ and m₂ (m₁ > m₂ ) are thrown vertically upward with the same initial speed v₀ simultaneously.
(a) Both balls will rise to the same height
\(\because v^{2}-v_{0}^{2}=-2 g h\) here at highest point, v = 0
\(\therefore h=\frac{v_{0}^{2}}{2 g}\)
(b) Both balls will come back with the same speed V₀ to the point of projection.
(c) Both balls will come back at the same time to the point of projection.
\(\because t=2 v_{0} / g\) 

Let \(3 \hat{i}+7 \hat{j}+4 \hat{k}=\vec{a}\)
\(|\vec{a}|=\sqrt{3^{2}+7^{2}+4^{2}}=\sqrt{9+49+16}=\sqrt{74}\)
Unit vector in the direction of
\(\vec{a}=\frac{\vec{a}}{|\vec{a}|}=\frac{3 \hat{i}+7 \hat{j}+4 \hat{k}}{\sqrt{74}}\)

(i) Consider an object moving in a straight line with uniform acceleration 'a', let at any instant of time 't', dx be the displacement of the objects.
ஃ instantaneous velocity, v \(=\frac{d x}{d t} \text { or } d x=v \mathrm{dt}\)
dx = (u + at) dt (∵v = u + at)
let x₀ and x be the displacements of the object at time 'zero' and 't'
\(\int_{x_{0}}^{x} d x=\int_{0}^{t}(u+a t) d t=u \int_{0}^{t} d t+a \int_{0}^{t} t d t\)
\((x)_{x_{0}}^{x}=u(t)_{0}^{t}+a\left(\frac{t^{2}}{2}\right)_{0}^{t}\)
\(x-x_{0}=u t+\frac{1}{2} a t^{2}\)
If x = x₀ = s (distance) covered by the object.
\(s=u t+\frac{1}{2} a t^{2}\)
(ii) Consider a particle moving in a straight line with initial velocity 'v' and acceleration 'a'.
Then \(a=\frac{d v}{d t}=\frac{d v}{d x} \times \frac{d x}{d t}=\frac{d v}{d x} \times v\)
adx = vdv
Integrating,\(\int_{x_{0}}^{x} a d x \quad=\int_{u}^{V} v d v\)
\(\Rightarrow \quad a|x| x_{0}^{x}=\left.\frac{v^{2}}{2_{}}\right|_{u} ^{V}\)
\(a\left(x-x_{0}\right)=\frac{v^{2}}{2}-\frac{u^{2}}{2}\)
v ² - u ² = 2as

\(\vec{F}=(3 \vec{i}+\overrightarrow{4 j}-5 \vec{k})\)
\(\vec{d}=(5 \vec{i}+4 \vec{j}+3 \vec{k})\)
\(F=\sqrt{9+16+25}=\sqrt{50}\)
\(d=\sqrt{25+16+9}=\sqrt{50}\)
\(\vec{F} \cdot \vec{d}=(3 \hat{i}+4 \hat{j}-5 \hat{k}) \cdot(5 \hat{i}+4 \hat{j}+3 \hat{k})\)
= 15 + 16 - 15 = 16
Let Q be the angle between \(\vec{F} \text { on } \vec{d}\)
\(W=\vec{F} \cdot \vec{d}=F d \cos \theta\)
\(16=\sqrt{50} \sqrt{50} \cos \theta\)
\(\frac{16}{50}=\cos \theta \Rightarrow \theta=\cos ^{-1}(0.32)=71.3^{\circ}\)
Unit vector along d is
\(\hat{d}=\frac{5 \hat{i}+4 \hat{j}+3 \hat{k}}{\sqrt{5^{2}+4^{2}+3^{2}}}=\frac{5 \hat{i}+4 \hat{j}+3 \hat{k}}{\sqrt{50}}\)
\(\vec{F} \cdot \vec{d}=\frac{16}{\sqrt{50}}\)
component vector of \(\vec{F}\) along \(\vec{d}\)is
\((\vec{F} \cdot \vec{d}) \hat{d}=\frac{16}{\sqrt{50}}\left(\frac{5 \hat{i}+4 \hat{j}+3 \hat{k}}{\sqrt{50}}\right)\)
\(=0.32(5 \hat{i}+4 \hat{j}+3 \hat{k})\) 

Width of the river, d= 1km
Velocity of swimmer, v _s = 5 km/h
Velocity of water flowing in river v_r = 3 km/h along OQ

(i) The swimmer wants to cross the river straight if the resultant of the velocity of the river flow and swimmer acts perpendicular to the direction of the river flow i.e. along OP. This will be so if the swimmer moves making an angle \(\alpha\) with upstream i.e. along OR.
\(\text { But, } \alpha+\theta=90^{\circ} \text { or } \theta=90^{\circ}-\alpha\)
From  \(\triangle\)OPR, we have
\(\sin \theta=\sin \left(90^{\circ}-\alpha\right)=\cos \alpha=\frac{R P}{R O}=\frac{3}{5}=0.6\)
\(\therefore \cos \alpha=\cos 53^{\circ} 8^{\prime} \Rightarrow \alpha=53^{\circ} 8^{\prime}\) upstream
(ii)The resultant velocity along oris given by
\(v=\sqrt{v_{s}^{2}-v_{r}^{2}}=\sqrt{5^{2}-3^{2}}=4 \mathrm{~km} / \mathrm{h}\)
(iii) Time taken by swimmer to cross the river
\(t=\frac{d}{v}=\frac{1}{4}=0.25 \mathrm{~h}=15 \mathrm{~min}\) 

The given vectors are
\(\mathbf{A}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-\mathbf{k}\)
and \(\mathbf{B}=-6 \hat{\mathbf{i}}+9 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\)
Then, the vectors are parallel, if Ax B = 0
\(\therefore \quad \mathbf{A} \times \mathbf{B}=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & -3 & -1 \\ -6 & 9 & 3 \end{array}\right|\)
\(=\hat{\mathbf{i}}(-9+9)-\hat{\mathbf{j}}(6-6)+\hat{\mathbf{k}}(18-18)=0\)
\(\text { But }|\mathbf{A} \times \mathbf{B}|=0\)
or AB sin \(\theta\) = 0       \([\because \mathbf{A} \neq 0 \text { and } \mathbf{B} \neq 0]\)
\(\therefore\) sin \(\theta\) = 0 or \(\theta\) = 0
Hence, the vectors A and B are parallel.

(i) A vector in general has no definite location in space because a vector remains unaffected whenever it is displaced anywhere in space provided its magnitude and direction do not change. However, a position vector has a definite location in space.
(ii) A vector can vary with time e.g. the velocity vector of an accelerated particle varies with time.
(iii) Two equal vectors at different locations in space do not necessarily have same physical effects. e.g. two equal forces acting at two different points on a body which can cause the rotation of a body about an axis will not produce equal turning effect.

(i) The magnitude of resultant R of two vectors A and B is
 Given by, 
 \(R=\sqrt { { A }^{ 2 }+B^{ 2 }+2AB\quad cos\quad \theta } \)
\(=\sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 }+2\times 3\times 4\quad cos\quad \theta } \)
R is 7m,if \(\theta ={ 0 }^{ o }\)
(ii) The magnitude of resultant R of two vectors A and B is
 Given by,
 \(R=\sqrt { { A }^{ 2 }+B^{ 2 }+2AB\quad cos\quad \theta } \)
\(=\sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 }+2\times 3\times 4\quad cos\quad \theta } \)
R is m,if   \(\theta ={ 180 }^{ o }\)
(iii) The magnitude of resultant R of two vectors A and B is
 Given by,
 \(R=\sqrt { { A }^{ 2 }+B^{ 2 }+2AB\quad cos\quad \theta }\)
\(=\sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 }+2\times 3\times 4\quad cos\quad \theta } \)
R is m ,if   \(\theta ={ 90 }^{ o }\)