(i) (c): Position vector of \(\overrightarrow{A B} \) 
\(=(2-1) \hat{i} \dot{+}(1-1) \hat{j}+(3-1) \hat{k}=\hat{i}+2 \hat{k}\)
(ii) (b): Position vector of  \(\overrightarrow{A C} \)
\(=(3-1) \hat{i}+(2-1) \hat{j}+(2-1) \hat{k}=2 \hat{i}+\hat{j}+\hat{k}\)
(iii) (d): Position vector of \(\overrightarrow{AD} \)
\(=(3-1) \hat{i}+(3-1) \hat{j}+(4-1) \hat{k}=2 \hat{i}+2 \hat{j}+3 \hat{k}\)
(iv) (b): Area of \(\Delta A B C\) = \(\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{A C}|\)
\(\overrightarrow{A B} \times \overrightarrow{A C}=\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 2 \\ 2 & 1 & 1 \end{array}\right|=\hat{i}(0-2)-\hat{j}(1-4)+\hat{k}(1-0)\)
\(=-2 \hat{i}+3 \hat{j}+\hat{k}\)
\( \Rightarrow |\overrightarrow{A B} \times \overrightarrow{A C}| =\sqrt{(-2)^{2}+3^{2}+1^{2}} \)
\(=\sqrt{4+9+1}=\sqrt{14}\)
Area of \(\Delta A B C\) \(=\frac{1}{2} \sqrt{14} \text { sq. units }\)
(v) (a): Unit vector along \(\overrightarrow{A D}=\frac{\overrightarrow{A D}}{|\overrightarrow{A D}|}\)
\(=\frac{2 \hat{i}+2 \hat{j}+3 k}{\sqrt{2^{2}+2^{2}+3^{2}}}=\frac{2 \hat{i}+2 \hat{j}+3 \hat{k}}{\sqrt{4+4+9}}=\frac{1}{\sqrt{17}}(2 \hat{i}+2 \hat{j}+3 \hat{k})\)

Here, \(\left|\vec{F}_{1}\right|=\sqrt{(4)^{2}+0^{2}}=4 \mathrm{KN}\)
\( \left|\vec{F}_{2}\right|=\sqrt{(-2)^{2}+4^{2}}=\sqrt{20} \mathrm{KN} \)
\(\left|\vec{F}_{3}\right|=\sqrt{(-3)^{2}+(-3)^{2}}=\sqrt{18} \mathrm{KN}\)
(i) (a): Since, \(\sqrt 20\) is larger. So, team B will win the game.
(ii) (b): Let F be the combined force
\(\therefore \vec{F}=\vec{F}_{1}+\vec{F}_{2}+\vec{F}_{3}=4 \hat{i}+0 \hat{j}-3 \hat{i}-3 \hat{j}-2 \hat{i}+4 \hat{j} \)
\(=-\hat{i}+\hat{j} \)
\(\therefore |\vec{F}|=\sqrt{(-1)^{2}+1^{2}}=\sqrt{2}=1.4 \mathrm{KN}\)
(iii) (c) : We have,  \(\vec{F}=-\hat{i}+\hat{j}\)
\(\therefore \theta=\tan ^{-1}\left(\frac{F_{y}}{F_{x}}\right)=\tan ^{-1}\left(\frac{1}{-1}\right)=\frac{3 \pi}{4} \text { radian }\)
= 0.75 x 3.14 radian = 2.3555 radian ≈ 2.4 radian 
(iv) (a): Magnitude of force of Team B = \(\sqrt 20\) KN
= 2\(\sqrt 5\) KN
(v) (b): 4 KN force is applied by team A.

(i) (b): Here (5, 3) are the coordinates of B.
\(\therefore\) P.V. of B = \(5 \hat{i}+3 \hat{j}\)
(ii) (d) : Here (9, 8) are the coordinates of D.
\(\therefore\) P.V. of D = \({9} \hat{i}+8 \hat{j}\)
(iii) (b) : P.V. of B =  \(5 \hat{i}+3 \hat{j}\)     and ஃ P.V. of C = \(6 \hat{i}+5 \hat{j}\)
\(\therefore \ \overrightarrow{B C}=(6-5) \hat{i}+(5-3) \hat{j}=\hat{i}+2 \hat{j}\)
(iv) (b): Since P.V. of A = \(2 \hat{i}+2 \hat{j}\) , P.V. of D = \({9} \hat{i}+8 \hat{j}\)
\( \therefore \overrightarrow{A D}=(9-2) \hat{i}+(8-2) \hat{j}=7 \hat{i}+6 \hat{j} \)
\(|\overrightarrow{A D}|^{2}=7^{2}+6^{2}=49+36=85 \)
\(\Rightarrow |\overrightarrow{A D}|=\sqrt{85} \text { units }\)
(v) (a): We have, \(\vec{M}=4 \hat{j}+3 \hat{k}\) ,
\( \therefore |\vec{M}|=\sqrt{4^{2}+3^{2}}=\sqrt{16+9}=\sqrt{25}=5 \)
\(\therefore \hat{M}=\frac{\vec{M}}{|\vec{M}|}=\frac{4 \hat{j}+3 \hat{k}}{5}=\frac{4}{5} \hat{j}+\frac{3}{5} \hat{k}\)

(i) (c) : Let OAB be a triangle such that

\(\overrightarrow{A O}=-\vec{p}, \overrightarrow{A B}=\vec{q}, \)\(\overrightarrow{B O}=\vec{r}\)
Now, \(\vec{q}+\vec{r} =\overrightarrow{A B}+\overrightarrow{B O}\)
\(=\overrightarrow{A O} =-\vec{p}\)
(ii) (c) : From triangle law of vector addition,
\(\overrightarrow{A C}+\overrightarrow{B D}=\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{B C}+\overrightarrow{C D}\)

\( =\overrightarrow{A B}+2 \overrightarrow{B C}+\overrightarrow{C D}\)
\(=\overrightarrow{A B}+2 \overrightarrow{B C}-\overrightarrow{A B}=2 \overrightarrow{B C} \)
 (iii) (b) : \(\operatorname{In} \Delta A B C, \overrightarrow{A C}=2 \vec{a}+2 \vec{b}\)

and in \(\Delta A B D, 2 \vec{b}=2 \vec{a}+\overrightarrow{B D}\)..(ii)
[By triangle law of addition]
Adding (i) and (ii), we have
\( \overrightarrow{A C}+2 \vec{b}=4 \vec{a}+\overrightarrow{B D}+2 \vec{b} \)
⇒\( \overrightarrow{A C}-\overrightarrow{B D}=4 \vec{a} \)
(iv) (d) :  \(\text { In } \Delta A B C, \overrightarrow{B A}+\overrightarrow{A C}=\overrightarrow{B C}\)... (i)
[By triangle law]
In \(\Delta B C D, \overrightarrow{B C}+\overrightarrow{C D}=\overrightarrow{B D}\) ..(i)
From (i) and (ii), \(\overrightarrow{B A}+\overrightarrow{A C}=\overrightarrow{B D}-\overrightarrow{C D}\)
\(\Rightarrow \overrightarrow{B A}+\overrightarrow{C D}=\overrightarrow{B D}-\overrightarrow{A C}=\overrightarrow{B D}+\overrightarrow{C A}\)
(v) (b): Since T is the mid point of YZ.
So,\(\overrightarrow{Y T}=\overrightarrow{T Z}\)
Now,\(\overrightarrow{X Y}+\overrightarrow{X Z}=(\overrightarrow{X T}+\overrightarrow{T Y})+(\overrightarrow{X T}+\overrightarrow{T Z})\)
[By triangle law]
\(=2 \overrightarrow{X T}+\overrightarrow{T Y}+\overrightarrow{T Z}=2 \overrightarrow{X T} \quad[\because \overrightarrow{T Y}=-\overrightarrow{Y T}]\)

(i) (b): Here, (4, 10) are the coordinates of A.
\(\therefore\) P.V. of A = \(4 \hat{i}+10 \hat{j}\)
(ii) (c): Here, (9, 7) are the coordinates of B.
\(\therefore\) P.V. of B \(=9 \hat{i}+7 \hat{j}\)
(iii) (b): Here, P.V. of A  = \(4 \hat{i}+10 \hat{j}\) and P.V. of \(C=4 \hat{i}+2 \hat{j}\)
\(\therefore \ \overrightarrow{A C}=(4-4) \hat{i}+(2-10) \hat{j}=-8 \hat{j}\)
(iv) (a): Here \(\vec{A}=\hat{i}+2 \hat{j}+3 \hat{k}\)
\( \therefore \ |\vec{A}|=\sqrt{1^{2}+2^{2}+3^{2}}=\sqrt{1+4+9}=\sqrt{14} \)
\(\therefore \ \hat{A}=\frac{\vec{A}}{|\vec{A}|}=\frac{\hat{i}+2 \hat{j}+3 \hat{k}}{\sqrt{14}}=\frac{1}{\sqrt{14}} \hat{i}+\frac{2}{\sqrt{14}} \hat{j}+\frac{3}{\sqrt{14}} \hat{k}\)
(v) (d): We have, \(\vec{A}=4 \hat{i}+3 \hat{j}\) and \( \vec{B}=3 \hat{i}+4 \hat{j}\)
\(\therefore|\vec{A}|=\sqrt{4^{2}+3^{2}}=\sqrt{16+9}=\sqrt{25}=5\)
and \(|\vec{B}|=\sqrt{3^{2}+4^{2}}=\sqrt{9+16}=\sqrt{25}=5\)
Thus, \(|\vec{A}|+|\vec{B}|=5+5=10\)