If the position vectors of the vertices of a triangle are $2\hat{i} - \hat{j} + \hat{k}$,$\hat{i} - 3\hat{j} - 5\hat{k}$,and $3\hat{i} - 4\hat{j} - 4\hat{k}$,then the triangle is:

Suppose $\overrightarrow{a}=\lambda \hat{i}-7 \hat{j}+3 \hat{k}$ and $\overrightarrow{b}=\lambda \hat{i}+\hat{j}+2 \lambda \hat{k}$. If the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is greater than $90^{\circ}$, then $\lambda$ satisfies the inequality:

Let $u$ and $v$ be two non-zero vectors. If $|u+v|=|u-v|$,then:

Let $(x, y) \in (R \times R)$ and $\vec{a} = x \hat{i} + 2 \hat{j} - \hat{k}$, $\vec{b} = 6 \hat{i} - y \hat{j} + 2 \hat{k}$ be two vectors. If $|\vec{a} \times \vec{b}|^2 + |\vec{a} \cdot \vec{b}|^2 = f(x) g(y)$, then $f(x) + g(y) - 46 = 0$ represents:

Let $\vec{u}=\hat{i}-\hat{j}-2\hat{k}$,$\vec{v}=2\hat{i}+\hat{j}-\hat{k}$,$\vec{v} \cdot \vec{w}=2$ and $\vec{v} \times \vec{w}=\vec{u}+\lambda\vec{v}$. Then $\vec{u} \cdot \vec{w}$ is equal to $......$

Find the angle between two vectors $\vec{a}$ and $\vec{b}$ with magnitudes $\sqrt{3}$ and $2$ respectively,having $\vec{a} \cdot \vec{b} = \sqrt{6}$.