Find the values of $x$ and $y$ so that the vectors $2 \hat{i} + 3 \hat{j}$ and $x \hat{i} + y \hat{j}$ are equal.

If $a, b, c$ are the position vectors of three collinear points,then the existence of scalars $x, y, z$ (not all zero) is such that:

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}$ be two vectors such that $\vec{a} \cdot \vec{b}=1$,$\cos(\theta) = \frac{1}{3}$ where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$,and the components of $\vec{b}$ with respect to $(\hat{i}, \hat{j}, \hat{k})$ are integers. Then the number of possible vectors that represent $\vec{b}$ is

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 $\vec{c}=\hat{i}-6 \hat{j}-7 \hat{k}$.

If $\overline{r} = -4 \hat{i} - 6 \hat{j} - 2 \hat{k}$ is a linear combination of the vectors $\overline{a} = -\hat{i} - 4 \hat{j} + 3 \hat{k}$ and $\overline{b} = -8 \hat{i} - \hat{j} + 3 \hat{k}$,then which of the following is true?

Let $ABC$ be a triangle. Let $u = \vec{AB}$ and $v = \vec{AC}$. If $D$ is the midpoint of $BC$,then $\vec{AD} =$