Angle between velocity vector and acceleration vector in uniform circular motion is 90°.

The identity is possible only if \(​​​​\overset\rightarrow{B}\)= \(​​​​\overset\rightarrow{0}\)i.e.,\(​​​​\overset\rightarrow{B}\) is a null vector.

No,adding any two vectors is not meaningful because only vectors of the same dimensions i.e. having same unit can be added.

Yes, Multiplying any two scalars is meaningful. Density p and volume V both the scalar quantities.When density is multiplied by volume,then we get \(p\times VB=m\)   , mass of the body,which is scalar quantity

\(\therefore \)   Ratio is 
\(\frac { A.B }{ \left| A.B \right| } =\frac { AB\quad cos\quad \theta }{ AB\quad sin\quad \theta } =cot\quad \theta \)
\(\\ =cot\quad { 60 }^{ o }=\frac { 1 }{ \sqrt { 3 } } \)
\(\\ As,\quad\theta ={ 60 }^{ o },cot{ 60 }^{ o }=\frac { 1 }{ \sqrt { 3 } } \) 

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,   \(A.B=ABcos{ 0 }^{ o }=AB\ (maximum)\)

Given \({ \nu }_{ A }=6\overset { \wedge }{ i } ;{ \nu }_{ B }=6\cos { 30^{ \circ } } \overset { \wedge }{ i } +6sin30^{ \circ }\overset { \wedge }{ j } \)
\(=\frac { 6\sqrt { 3 } }{ 2 } \overset { \wedge }{ i } +\frac { 6 }{ 2 } \overset { \wedge }{ j } =3\sqrt { 3 } \overset { \wedge }{ i } +3\overset { \wedge }{ j } \)
\(\\ \therefore \quad { \nu }_{ BA }={ \nu }_{ B }-{ \nu }_{ A }=(3\sqrt { 3 } \overset { \wedge }{ i } +3\overset { \wedge }{ j } )-6\overset { \wedge }{ i } \)
\( { \nu }_{ BA }=(3\sqrt { 3 } -6)\overset { \wedge }{ i } +3\overset { \wedge }{ j }\)
\( { \nu }_{ BA }=\sqrt { (3\sqrt { 3 } -6)^{ 2 }+({ 3) }^{ 2 } } =3.11m/s\)
\(\\ \tan { \alpha =\frac { 3 }{ 3\sqrt { 3 } -6 } =\frac { 1 }{ \sqrt { 3 } -2 } \Rightarrow \alpha =105.13^{ \circ } } \)

The position of the particle is given by
\(r(t)={ \nu }_{ 0 }t+\frac { 1 }{ 2 } at^{ 2 }=15\overset { \wedge }{ i } t+\frac { 1 }{ 2 } (15\overset { \wedge }{ i } +10\overset { \wedge }{ j } ){ t }^{ 2 }\)
\(=(15t+7.5{ t }^{ 2 })\overset { \wedge }{ i } +5\overset { \wedge }{ j } { t }^{ 2 }\)
\(\\ \therefore \quad x(t)=15t+7.5{ t }^{ 2 }\Rightarrow y(t)=5t^{ 2 }\)
\(\\ If\quad x(t)=125m,t=?\)
\(\\ 125=15t+7.5{ t }^{ 2 }\Rightarrow 1.5{ t }^{ 2 }+3{ t }-25=0\)
\(\\ t=3.2s\) 
\(\\ \therefore \quad y(t)=5\times (3.2)^{ 2 }=51.2m\) 

The two bodies will collide, if they reach at
a point acquiring the same vertical distance in the same time. Therefore
y = u¹ sin 60⁰\(\times\) t-\(\frac{1}{2}\)gt²   = u² t - \(\frac{1}{2}\)gt²
\(\Rightarrow\)u¹ sin 60⁰\(\times\)t = u₂t 
or \(\frac { { u }_{ 1 } }{ { u }_{ 2 } } =\frac { 1 }{ sin\quad { 60 }^{ 0 } } =\frac { 2 }{ \sqrt { 3 } } \)

Acceleration needed for a particle to undergo uniform circular motion is called 'centripetal acceleration'. It is directed along the radius of circular path towards its centre. Two common examples are:
(i) An electron revolving around the nucleus of an atom in a uniform circular motion experiences a centripetal acceleration on account of Coulombian electrostatic force on electron due to nucleus.
(ii) A satellite revolving around the earth in a circular orbit experiences a centripetal acceleration on account of gravitational force due to the earth.

Let the body be projected horizontally with a velocity u, from the top of a tower of height h.
Time taken to reach the ground, t = \(\sqrt{2h/g}\)
Since the initial vertical velocity is zero and there is no acceleration in the horizontal.
Thus, x = ut
x = u\(\sqrt{2h/g}\)
i.e., h = x²  h=x²\(\frac{g}{2u^2}\)
As h \(\alpha\)x², the path is a parabola.