Given direction ratios are 1, 2, 3
Let x = 1, y = 2, z = 3
r = \(\sqrt{x^2+y^2+z^2}=\sqrt{1+4+9}=\sqrt{14}\)
The direction cosines are \({x\over r},{y\over r},{z\over r}\)
Thus, the direction cosines are \({1\over \sqrt{14}},{2\over \sqrt{14}},{3\over \sqrt{14}}\)

\( \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1 \)
\(Here \cos \alpha=\frac{1}{5} \)
\(\cos \beta=\frac{3}{5} \)
\( \cos \gamma =\frac{4}{5} \)
\(\text { LHS } =\left(\frac{1}{5}\right)^2+\left(\frac{3}{5}\right)^2+\left(\frac{4}{5}\right)^2 \)
\( =\frac{1}{25}+\frac{9}{25}+\frac{16}{25} \)
\( =\frac{26}{25} \) 
\( \neq 1\)
They are not direction cosines.

\(\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\hat{i}-4\hat{j}+\hat{k}\)
Direction cosines are \({1\over \sqrt{18}},{-4\over \sqrt{18}},{1\over \sqrt{18}}\).

The direction cosines are \({x\over \sqrt{x^2+y^2+z^2}},{y \over \sqrt{x^2+y^2+z^2}},{z\over \sqrt{x^2+y^2+z^2}}\)
That is, \({2\over 7},{3\over 7},{-6\over7}.\)

We know that a unit vector along the direction of the vector \(\overrightarrow{a}\) is given by \({\overrightarrow{a}\over|\overrightarrow{a}|}\). 
So a unit vector along the direction of 5 \(\hat{i}\) - 3 \(\hat{j}\) + 4 \(\hat{k}\) is given by \({5\hat{i}-3\hat{j}+4\hat{k}\over |5\hat{i}-3\hat{j}+4\hat{k}|}={5\hat{i}-3\hat{j}+4\hat{k}\over\sqrt{5^2+3^2+4^2}}={5\hat{i}-3\hat{j}+4\hat{k}\over \sqrt{50}}\).

45cm 30°north of east.

The vector \(\overrightarrow{OP}\) represents a displacement of 45 cm, 30° north of east.

(i)

(ii) 

Let \(\overrightarrow{a},\overrightarrow{b} , \overrightarrow{c}\in V\), where V is the set of all vectors.
Let R be the relation defined by\(\overrightarrow{a}=\overrightarrow{b} \)
(i) Reflexive : \(\overrightarrow{a}=\overrightarrow{a} \Rightarrow aRa \Rightarrow R\) is Reflexive.
(ii) Symmetric: \(\overrightarrow{a}=\overrightarrow{b} \Rightarrow \overrightarrow{b} =\overrightarrow{a}\)
\(\therefore aRb \overrightarrow{a} \Rightarrow bRa \Rightarrow R\) is Symmetric.
(iii) Transitive : \(\overrightarrow{a}=\overrightarrow{b},\overrightarrow{b}=\overrightarrow{c} \Rightarrow\overrightarrow{a}=\overrightarrow{c}\)
\(\therefore aRb, bRc\Rightarrow aRc \)
\(\therefore \) R is transitive.
\(\therefore \) The relation R is an equivalence relation on V.
Hence Proved.

Let the position vector of the vertices of the \(\triangle\) ABC be \(\overrightarrow{a},\overrightarrow{b}\) and \(\overrightarrow{c}\) respectively.
\(\therefore \overrightarrow{OA}=\overrightarrow{a},\overrightarrow{OB}=\overrightarrow{b},\overrightarrow{OC}=\overrightarrow{c}.\)
Since G is the centroid of \(\triangle\) ABC, we have
\(\Rightarrow \overrightarrow{OG}={\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\over3}\) \(\Rightarrow 3\overrightarrow{OG}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\)
Now,LHS \(=\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}\)
\(=\overrightarrow{OA}-\overrightarrow{OG}+\overrightarrow{OB}-\overrightarrow{OG}+\overrightarrow{OC}-\overrightarrow{OG}=(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC})\)\(-3\overrightarrow{OG}\)
\(=(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC})-(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{0}=RHS\)
Hence proved.