The position vectors of points P and Q are $\vec{p}$ and $\vec{q}$ respectively. The point $R$ divides line segment $P Q$ in the ratio $3: 1$ and S is the mid-point of line segment PR. The position vector of $S$ is

Assertion (A): The vectors

$$\begin{aligned} & \vec{a}=6 \hat{i}+2 \hat{j}-8 \hat{k} \\ & \vec{b}=10 \hat{i}-2 \hat{j}-6 \hat{k} \\ & \vec{c}=4 \hat{i}-4 \hat{j}+2 \hat{k} \end{aligned}$$

represent the sides of a right angled triangle.

Reason (R): Three non-zero vectors of which none of two are collinear forms a triangle if their resultant is zero vector or sum of any two vectors is equal to the third.

Two vector $$\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k} \quad$$ and $$\vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}$$ are collinear if:

The magnitude of the vector $$6 \hat{i}-2 \hat{j}+3 \hat{k}$$ is :

Find a vector of magnitude 4 units perpendicular to each of the vectors $2 \hat{i}-\hat{j}+\hat{k}$ and $\hat{i}+\hat{j}-\hat{k}$ and hence verify your answer.

(a) If the vectors $$\vec{a}$$ and $$\vec{b}$$ are such that $$|\vec{a}|=3,|\vec{b}|=\frac{2}{3}$$ and $$\vec{a} \times \vec{b}$$ is a unit vector, then find the angle between $$\vec{a}$$ and $$\vec{b}$$.

OR

(b) Find the area of a parallelogram whose adjacent side are determined by the vectors $$\vec{a}=\hat{i}-\hat{j}+3 \hat{k}$$ and $$\vec{b}=2 \hat{i}-7 \hat{j}+\hat{k}$$

Find the distance between the lines:

$$\begin{aligned} & \vec{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k}) ; \\ & \vec{r}=(3 \hat{i}+3 \hat{j}-5 \hat{k})+\mu(4 \hat{i}+6 \hat{j}+12 \hat{k}) \end{aligned}$$