\(|A+B|=|A-B| \)
\(\sqrt { A^{ 2 }+B^{ 2 }+2ABcos\theta } =\sqrt { A^{ 2 }+B^{ 2 }-2ABcos\theta } \)
\(\Rightarrow 4ABcos\theta =0\Rightarrow cos\theta =0\)
Hence cos θ = 90⁰ or θ = π/2 

The sum of two vectors is maximum, when both the Vectors are in the same direction and is minimum when they act in opposite direction.
As, \(R =\sqrt { A^{ 2 }+B^{ 2 }+2ABcos\theta }\)
(i) For R to be maximum, cos θ = +1
\(R_{ max }\sqrt { A^{ 2 }+B^{ 2 }+2AB } =A+B\)
(ii) For R to be minimum
cos θ   = -1 or θ  = 180⁰
\(R_{ min }\sqrt { A^{ 2 }+B^{ 2 }+2AB\left( -1 \right) } =A-B\)

Area of parallelogram \(|A\times B|=AB\sin\theta =\frac { 1 }{ 2 } AB\)
\(\sin\theta =\frac { 1 }{ 2 } =\sin30^{ 0 }or\ \theta =30^{ 0 } \)

As we know, A x B = ABsin θ
A.B = AB cosθ
According to the question
AB sinθ = AB cosθ
\(\Rightarrow \frac { \sin\theta }{ \cos\theta } =1\Rightarrow \tan\vartheta \Rightarrow \theta =45^{ 0 }\)

As the vectors A + B and A - B are perpendicular to each other, therefore
 \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}(\vec { A } +\vec { B } ).(\vec { A } -\vec { B } )=0\)
\(\\ \vec { A } .\vec { A } -\vec { A } .\vec { B } .\vec { A } -\vec { B } .\vec { B } =0\)
or \(\vec { A } -\vec { B } =0\quad \ \quad [\therefore A.B=B.A]\ \)
\(\Rightarrow A=B\)

Given,A = 5 kg-wt, B = 10 kg-wt , \(\theta =120\)  then = \(\beta =?\)
 \(tan\beta =\frac { B\ sin\theta }{ A+B\cos\theta } =\frac { 10 \ sin\ 120^{ 0 } }{ 5+10cos120^{ 0 } } =\frac { 5sin \ { 60 }^{ 0 } }{ 10-5cos{ 60 }^{ 0 } }\)
\( =\frac { 5\times \sqrt { 3 } /2 }{ 10-5/2 } =\frac { 1 }{ \sqrt { 3 } } =tan\quad { 30 }^{ o }\)
\(\\ \therefore \beta =30^{ o }\) 

As we know that
\(\left| A+B \right| =\sqrt { A^{ 2 }+{ B }^{ 2 }+2AB\quad cos\quad \theta } \)
And    \(\left| A-B \right| =\sqrt { A^{ 2 }+{ B }^{ 2 }-2AB\quad cos\quad \theta } \)  
But as per question, we have
\(\sqrt { A^{ 2 }+{ B }^{ 2 }+AB\quad cos\quad \theta } =\sqrt { A^{ 2 }+{ B }^{ 2 }-2AB\quad cos\theta } \)
Squaring both sides, we have (4 AB cos ) = 0
\(\sqrt { A^{ 2 }+{ B }^{ 2 }+AB\quad cos\quad \theta } =\sqrt { A^{ 2 }+{ B }^{ 2 }-2AB\quad cos\theta } \)
Hence, the two vectors A and B are perpendicular to each other.

\(\theta ={ 45 }^{ o }\)

We're asked to find the necessary launch angle (with respect to the vertical) of the anti-aircraft shell to hit an incoming plane flying overhead.
I'll assume the shell launcher will fire as soon as the plane is directly overhead.
The position component equations for the plane are given by
x_plane=(200lm/s)t
y_plane=2000lm
And for the anti-aircraft shell:
x_shell=v₀cosα₀t
y_shell=v₀sinα₀t−1/2gt
Ideally, the x-velocity for the plane and shell must be equal, because the x-component does not change; if they were different, it would never hit the target.
With that being said, the two quantities
(200lm/s)t
and v0cosα0t
should be the same; i.e.
v₀cosα₀=200lm/s:
The shell's initial velocity is 600 m/s:
(600lm/s)cosα₀=200lm/s
\(\alpha_0=\arccos \left[\frac{200 \mathrm{~m} / \mathrm{s}}{600 \mathrm{~m} / \mathrm{s}}\right]={70.5^{\circ}}\)
However, this is the angle with respect to the horizontal. The angle measured from the vertical is
\(\text { angle }=90^{\circ}-70.5^{\circ}={19.5^{\circ}}\)
We can even use this value to find the time it takes the projectile to hit the target, knowing that the height of the plane (2000m) should be set equal to the y-position equation for the shell:
\(2000 \mathrm{~m}=(600 \mathrm{~m} / \mathrm{s}) \sin \left[70.5^{\circ}\right] t-\frac{1}{2} g t^2 \)
\(2000 \mathrm{~m}=593 t-5 t^2 \)
\(t=3.48 \mathrm{~s}\)
\(\theta \) = 19.5⁰,H = 16km

For any vector 
\(a\Rightarrow \left| a \right| ^{ 2 }=a.a\)
\(\\ \therefore \left| \hat { a } -\hat { b } \right| ^{ 2 }=(\hat { a } -\hat { b } ).(\hat { a } -\hat { b } )\)
\(\\ =\hat { a } .\hat { a } -\hat { a } .\hat { b } -\hat { b } .\hat { a } +\hat { b } .\hat { b } \)
\(\\ =1-2\hat { a } .\hat { b } +1\quad \quad \quad [\therefore \hat { a } .\hat { a } =1\times 1\times cos0^{ o }=1]\)
\(\\ =2-2\times 1\times cos\quad \theta \)
\(\\ -2(1-cos\quad \theta )\)
\(\\ =2.2sin^{ 2 }\frac { \theta }{ 2 } =4sin^{ 2 }\frac { \theta }{ 2 } \quad \quad [\therefore 1-cos2\theta =2sin^{ 2 }\theta ]\)
Hence, \(\\ \left| \hat { a } -\hat { b } \right| =2sin\frac { \theta }{ 2 } \) 

Velocity of plane, v_p=720\(\frac{5}{18}\)ms^-1= 200 ms^-1
Velocity of shell = 600 ms^-1;
sin \(\theta \) = \(\frac{200}{600}=\frac{1}{3}\)
or \(\theta \) = sin^-1\(\left( \frac { 1 }{ 3 } \right) \) = 19.47⁰
This angle is with the vertical.
Let h be the required minimum height.
Using equation
 v²- u² = 2 as,we get
(0)²-(600 cos \(\theta \))²= -2x10xh
or,    h=\(\frac { 600\times 600(1-sin^{ 2 }\theta ) }{ 20 } \)
= 30\(\times\)600\(\left( 1-\frac { 1 }{ 9 } \right) =\frac { 8 }{ 9 } \)\(\times\)30\(\times\)600m
= 16km.

The vector v_b representing the velocity of the motorboat and the vector v_c representing the water current are shown in directions specified by the problem. Using the parallelogram method of addition, the resultant R is obtained in the direction shown in the figure.

We can obtain the magnitude of R using the Law of cosine
\(R=\sqrt{v_{\mathrm{b}}^{2}+v_{\mathrm{c}}^{2}+2 v_{\mathrm{b}} v_{\mathrm{c}} \cos 120^{\circ}}\)
\(=\sqrt{25^{2}+10^{2}+2 \times 25 \times 10(-1 / 2)} \cong 22 \mathrm{~km} / \mathrm{h}\)
To obtain the direction, we apply the Law of sines
\(\frac{R}{\sin \theta}=\frac{v_{c}}{\sin \phi} \text { or, } \sin \phi=\frac{v_{c}}{R} \sin \theta\)
\(=\frac{10 \times \sin 120^{\circ}}{21.8}=\frac{10 \sqrt{3}}{2 \times 21.8} \cong 0.397\)
\(\phi \cong 23.4^{\circ}\)