Let \(\overrightarrow{A},\overrightarrow{B}\) and \(\overrightarrow{C}\) be represented in component form,
\(\overrightarrow{A}=\overrightarrow{A}_z\hat{i}+\hat{j}\overrightarrow{A}_y\hat{j}+\overrightarrow{A}_z\hat{k}\)
\(\overrightarrow{B}=\overrightarrow{B}_z\hat{i}+\overrightarrow{B}_y\hat{j}+\overrightarrow{B}_z\hat{k}\)
\(\overrightarrow{C}=\overrightarrow{C}_x\hat{i}+\overrightarrow{C}_y\hat{j}+\overrightarrow{C}_z\hat{k}\)
Let \(\overrightarrow{D}\) be their summation vector,
\(\overrightarrow{D}=(\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C})\)
\(=\left( (\overrightarrow{A} _x\hat{i}+\overrightarrow{A}_y\hat{j}+\overrightarrow{A}_z\hat{k})+(\overrightarrow{B}_x\hat{i}+\overrightarrow{B}_y\hat{j}+\overrightarrow{B}_z\overrightarrow{k} +\overrightarrow{C}_x\hat{i}+\overrightarrow{C}_y\hat{j}+\overrightarrow{C}_z\hat{k}\right)\)
Addition of vectors obey the commutative as well as associative laws.
\(A_y+B_y+C_yD=(A_x+B_x+C_x){2 u \sin \theta\over g}+(A_y+B_y+C_y)\hat{j}+(A_z+B_z+C_z)\hat{k}+(A_x+B_x+C_x)\hat{i}+(A_yB_y+C_y){2\ u\ \sin \theta\over g}(A_z+B_z+C_z)\hat{k}+(A_x+B_x+C_x)\hat{i}+(A_y+B_y+C_y)\hat{j}+(A_z+B_z+C_z){2\ u \sin \theta \over g}\)D_x = A_x + B_x+ C_x
D_y = A_y + B_y+ C_y
D_z = A_z + B_z + C_z

Magnitude of vectors \(\hat { i } +\hat { j } \)
\(=\left| \hat { i } +\hat { j } \right| \)
\(=\sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 } } =\sqrt { 2 } \)
\(tan \ \theta =\frac { |\hat { j } | }{ |\hat { i } | } =\frac { 1 }{ 1 } =1\)
\(\theta ={ tan }^{ -1 }1=45° \ with \ x \ axis\)
Magnitude of vectors \(\hat { i } -\hat { j } \)
=|\(\hat { i } -\hat { j } \)|
\(=\sqrt { { 1 }^{ 2 }+1-{ 1 }^{ 2 } } =\sqrt { 2 } \)

\(tan \ \theta =\frac { |-\hat { j } | }{ |\hat { i } | } =\frac { -1 }{ 1 } =-1\)
\(\theta ={ tan }^{ -1 }-1\)
= - 45°with x axis.

Let \(\overrightarrow{A}\) and \(\overrightarrow{B}\) be the two vectors
\(\overrightarrow{A}=\overrightarrow{A}_x\hat{i}+{\overrightarrow{A}}_{y}\hat{j}+{\overrightarrow{A}}_{z}\hat{k}\)
\(\overrightarrow{B}=\overrightarrow{B}_x\hat{i}+\overrightarrow{B}_y\hat{j}+\overrightarrow{B}_z\hat{k}\)
Cross product of \(\overrightarrow{A}\) and \(\overrightarrow{B}\) ,
\(\overrightarrow{A}\times\overrightarrow{B}=(\overrightarrow{A}_x\hat{i}+\overrightarrow{A}_y\hat{j}+\overrightarrow{A}_z\hat{k})\times({\overrightarrow{B}}_{x}\hat{i}+\overrightarrow{B}_y\hat{j}+\overrightarrow{B}_z\hat{k})\)
\(=\overrightarrow{A}_x\overrightarrow{B}_x\hat{i}\times\hat{i}+\overrightarrow{A}_x\overrightarrow{B}_x\hat{i}\times\hat{i}+{\overrightarrow{A}}_{x}{\overrightarrow{B}}_{x}\overrightarrow{i}\times\overrightarrow{j}+\overrightarrow{A}_x\overrightarrow{B}_z\hat{i}\times\hat{j}\)
\(=\overrightarrow{A}_z\overrightarrow{B}_x\hat{k}\times\hat{i}+\overrightarrow{A}_z\overrightarrow{B}_y\hat{k}\times\hat{j}+\overrightarrow{A}_z\overrightarrow{B}_z\hat{k}+\hat{k}\)
\(\overrightarrow{A}\times\overrightarrow{B}=\overrightarrow{A}_z\overrightarrow{B}_z(0)+\overrightarrow{A}_x\overrightarrow{B}_y(\hat{k})+A_xA_z(\hat{j})+\overrightarrow{A}_y{\overrightarrow{B}}_{x}(-\hat{k})+\overrightarrow{A}_y\overrightarrow{B}_y(0)+\overrightarrow{A}_y\overrightarrow{B}{}_{z}(\hat{i})+9\overrightarrow{A}_z\overrightarrow{B}_z(\hat{i})+\overrightarrow{A}_z\overrightarrow{B}_y(-\hat{i})+\overrightarrow{A}_z\overrightarrow{B}_z(0)\)
\((\overrightarrow{A}\times\overrightarrow{B})=(\overrightarrow{A}_z\overrightarrow{B}_x-\overrightarrow{A}_z\overrightarrow{B}_y)\hat{i}+(A_yB_y-A_yB_x)\hat{k}+(A_zB_x-A_x-B_x)\hat{j}\)
\([\because \hat{i}\times\hat{i}=\hat{j}\times\hat{j}=\hat{k}\times\hat{k}=0 \hat{i}\times\hat{j}=k,\hat{i}\times\hat{k}]=-\hat{j},\hat{j}\times\hat{i}=-k,\hat{j}\times\hat{k}=\hat{i} \hat{k}\times\hat{i}=-\hat{j},\hat{k}\times\hat{j}=-\hat{i}\)
It can be in determinant form as
\(\overrightarrow{A\times\overrightarrow{B}}=\begin{vmatrix} \hat{i}&\hat{j}&\hat{k}\\A_x&A_y&A_z\\B_x&B_y&B_z \end{vmatrix}\)
\(=\hat{i}(A_yB_z-B_y-A_z)-\hat{j}(A_xB_z-B_xA_z)+\hat{k}(A_xB_y-B_x-A_Y)\)
Dot product of \(\overrightarrow{A}\) and \(\overrightarrow{B}\),
\(\overrightarrow{A}={\overrightarrow{A}}_{z},\hat{i}+\overrightarrow{A}_y\hat{j}+\overrightarrow{A}_z\hat{k}\)
\(\overrightarrow{B}=\overrightarrow{B}_x\hat{i}+\overrightarrow{B}_g\hat{j}+\overrightarrow{B}_z\hat{k}\)
\(\overrightarrow{A}.\overrightarrow{B}=(\overrightarrow{A}_x\hat{i}+\overrightarrow{A}_y\hat{j}+\overrightarrow{A}_z\hat{k}).(\overrightarrow{B}_x\hat{i}+\overrightarrow{B}_y\hat{j}+\overrightarrow{B}_z\hat{k})\)
\(=(\overrightarrow{A}_x\overrightarrow{B}_x(\hat{i}.\hat{j})+\overrightarrow{A}_x\overrightarrow{B}_y(\hat{i}.\hat{j})+\overrightarrow{A}_x\overrightarrow{B}_z(\hat{i}\hat{k})+\overrightarrow{A}_y\overrightarrow{B}_z(\hat{j}.\hat{i})+\overrightarrow{A}_y\overrightarrow{B}_y(\hat{i}.\hat{j})+\overrightarrow{A}_z\overrightarrow {B}_x(\hat{k}.\hat{i})+\overrightarrow{A}_z\overrightarrow{B}_y(\hat{k}.\hat{j})+A_zB_z(\hat{k}.\hat{k})\)
\(\overrightarrow{A}.\overrightarrow{B}=\overrightarrow{A}_x\overrightarrow{B}_x(1)+\overrightarrow{A}_x\overrightarrow{B}_y(0)+A_x\overrightarrow{B}_z(0)+\overrightarrow{A}_y\overrightarrow{B}_x(0)+\overrightarrow{A}_y\overrightarrow{B}-Y(1)+\overrightarrow{A}_y\overrightarrow{B}_z()+\overrightarrow{A}_z\overrightarrow{B}_z(0)+A_zB_Z(1)\)
\(\overrightarrow{A}.\overrightarrow{B}=\overrightarrow{A}_x\overrightarrow{B}_x(X)+\overrightarrow{A}_y\overrightarrow{B}_y+\overrightarrow{A}_z\overrightarrow{B}_z\)
\([\because \hat{i}.\hat{j}=\hat{j}.\hat{j}=\hat{k}.\hat{k}=1\hat{i}.\hat{j}=\hat{i}.\hat{j}=\hat{i}.\hat{k}=0\hat{k}.\hat{i}=\hat{k}.\hat{j}=0]\)

Mass M₁=3g, \(\overrightarrow { { r }_{ 1 } } =2\hat { i } +\hat { j } +3u\)
Mass M₂=6g, \(\overrightarrow { { r }_{ 2 } } =\hat { j } +3\hat { j } +2u\)
Position vector of centre of mass
\(=\frac { { M }_{ 1 }{ r }_{ 1 }+{ M }_{ 2 }{ r }_{ 2 } }{ { M }_{ 1 }+{ M }_{ 2 } } \)
\(=3\frac { (2\hat { i } +\hat { j } +3u)+6(\hat { j } +3\hat { j } +2u) }{ 3+6 } \)
\(=\frac { 6\hat { i } +3\hat { j } +6u+6i+18\hat { j } +12u }{ 9 } \)
\(=\frac { 12\hat { i } +21\hat { j } +18u }{ 9 } =\frac { 4 }{ 3 } \hat { i } +\frac { 7 }{ 3 } \hat { j } +\frac { 6 }{ 3 } u\)
\(\because\) Position vectors of centre of mass
\(=\frac { 4 }{ 3 } \hat { i } +\frac { 7 }{ 3 } \hat { j } +\frac { 6 }{ 3 } u\)

AB=|A||B| cos \(\theta\)
\(cos\theta =\frac { A.B }{ |A||B| } \)
A.B=\((\hat { i } +2\hat { j } -u).(-4\hat { i } +\hat { j } -2u)\)
-4+2+2u = -4+4u
= -4 + 4 =0
cos \(\theta\) =0
\(\theta\) = cos^-1
\(\theta\) = 90°
Hence, the angle between two vectors of A and B is 90°