If $a = x^2 \hat{i} + x \hat{j} + 3 \hat{k}$ and $b = x \hat{i} - 4 \hat{j} + 2 \hat{k}$ and $a \cdot b > 6$,then:

Three vectors $\vec{a}, \vec{b}, \vec{c}$ satisfy the condition $\vec{a}+\vec{b}+\vec{c}=\vec{0}$. If $|\vec{a}|=1, |\vec{b}|=3, |\vec{c}|=4$,then find the value of $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$.

Let $\vec{b}$ and $\vec{c}$ be non-collinear vectors satisfying $\vec{a} \times (\vec{b} \times \vec{c}) + (\vec{a} \cdot \vec{b})\vec{b} = (4 - 2x - \sin y)\vec{b} + (x^2 - 1)\vec{c}$ and $(\vec{c} \cdot \vec{c})\vec{a} = \vec{c}$,then $x$ is equal to

If $\vec{a}=5 \hat{i}-\hat{j}-3 \hat{k}$ and $\vec{b}=\hat{i}+3 \hat{j}-5 \hat{k},$ then show that the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are perpendicular.

If $a, b, c$ are unit vectors satisfying the relation $a+b+\sqrt{3} c=0$, then the angle between $a$ and $b$ is

If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}|=|\vec{b}|=\sqrt{2}$ and $\vec{a} \cdot \vec{b}=-1$,then the angle between $\vec{a}$ and $\vec{b}$ is