(i) Complete the given figure to explain their entire movement plan along the respective vectors. (ii) Find vectors $\vec{AC}$ and $\vec{BC}$. (iii) (a) If $\vec{a} \cdot \vec{b} = 1$, distance of O to A is 1 km and that from O to B is 2 km, then find the angle between $\overrightarrow{OA}$ and $\overrightarrow{OB}$. Also, find $| \vec{a} \times \vec{b}|$. OR (iii) (b) If $\vec{a} = 2\hat{i} - \hat{j} + 4\hat{k}$ and $\vec{b} = \hat{j} - \hat{k}$, then find a unit vector perpendicular to $(\vec{a}+\vec{b})$ and $(\vec{a}-\vec{b})$.

Find the position vector of point C which divides the line segment joining points A and B having position vectors $\hat{i}+2\hat{j}-\hat{k}$ and $-\hat{i}+\hat{j}+\hat{k}$ respectively in the ratio $4:1$ externally. Further, find $|\vec{AB}|:|\vec{BC}|$ .

If $\vec{AB}=\hat{j}+\hat{k}$ and $\vec{AC}=3\hat{i}-\hat{j}+4\hat{k}$ represent the two vectors along the sides AB and AC of $\Delta ABC$, prove that the median $\vec{AD}=\frac{\vec{AB}+\vec{AC}}{2}$ where D is midpoint of BC. Hence, find the length of median AD.

Let $\vec{a}$, $\vec{b}$, $\vec{c}$ be three vectors such that $\vec{a}\cdot\vec{b}=\vec{a}\cdot\vec{c}$ and $\vec{a}\times\vec{b}=\vec{a}\times\vec{c}$, $\vec{a}\ne\vec{0}$. Show that $\vec{b}=\vec{c}$.

Let three toys A, B and C be placed in the same straight line. If the position vectors of A, B and C are $5\hat{i}-2\hat{j}$, $5\hat{i}+8\hat{j}$ and $a\hat{i}-52\hat{j}$ respectively, find the value of 'a'.