Find $|\vec{a}-\vec{b}|$,if two vectors $\vec{a}$ and $\vec{b}$ are such that $|\vec{a}|=2, |\vec{b}|=3$ and $\vec{a} \cdot \vec{b}=4$.

If $\vec{a}$ and $\vec{b}$ are non-zero vectors which are linearly dependent such that $\frac{|\vec{a} + \vec{b}|}{|\vec{a} - \vec{b}|} = 2$ and $|\vec{b}| > |\vec{a}|$,then:

Let $ABC$ be a triangle and $\bar{a}, \bar{b}, \bar{c}$ be the position vectors of $A, B, C$ respectively. If $D$ divides $BC$ in the ratio $2:3$ internally and $E$ divides $CA$ in the ratio $2:1$ internally,then the position vector of the point $P$ which divides $DE$ in the ratio $3:5$ internally is

If $\vec{a} = 4\hat{i} + 3\hat{j}$ and $\vec{b} = 2\hat{i} + \lambda\hat{j}$ are parallel vectors,then the value of $\lambda$ is:

If $\vec{a}$ and $\vec{b}$ are two collinear vectors,then which of the following is incorrect?

Find the sum of the vectors $\vec{a}=\hat{i}-2 \hat{j}+\hat{k}$,$\vec{b}=-2 \hat{i}+4 \hat{j}+5 \hat{k}$,and $\vec{c}=\hat{i}-6 \hat{j}-7 \hat{k}$.