The vector in the direction of the sum of the vectors $\vec{a}=2 \hat{i}-2 \hat{j}+5 \hat{k}$ and $\vec{b}=-2 \hat{i}+5 \hat{j}-3 \hat{k}$ is

If $\overline{AB} \parallel \overline{CD}$ and the position vectors of the four points $A, B, C,$ and $D$ are $2i + j, i - 3j, 3i + 2j,$ and $i + \lambda j$ respectively,then the value of $\lambda$ is .....

If the points with position vectors $(\alpha \hat{i}+10 \hat{j}+13 \hat{k})$,$(6 \hat{i}+11 \hat{j}+11 \hat{k})$,and $(\frac{9}{2} \hat{i}+\beta \hat{j}-8 \hat{k})$ are collinear,then $(19 \alpha-6 \beta)^2=$

If $a$ and $b$ are two non-collinear vectors and the vector $a+b$ bisects the angle between $a$ and $b$,then

If $O$ is the origin and the position vector of $A$ is $4\,i + 5\,j$,then a unit vector parallel to $\overrightarrow{OA}$ is:

The line joining the points $6 \overrightarrow{a}-4 \overrightarrow{b}+4 \overrightarrow{c}$ and $-4 \overrightarrow{c}$ and the line joining the points $-\overrightarrow{a}-2 \overrightarrow{b}-3 \overrightarrow{c}$ and $\overrightarrow{a}+2 \overrightarrow{b}-5 \overrightarrow{c}$ intersect at: