If $|\vec{a}|=2, |\vec{b}|=5$ and $|\vec{a} \times \vec{b}|=8$,then $|\vec{a} \cdot \vec{b}|$ is equal to :

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:

If $a(\vec{\alpha} \times \vec{\beta}) + b(\vec{\beta} \times \vec{\gamma}) + c(\vec{\gamma} \times \vec{\alpha}) = \overrightarrow{0}$, where $a, b, c$ are non-zero scalars, then the vectors $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$ are

If $a=\hat{i}+\hat{j}+t \hat{k}$ and $b=\hat{i}+2 \hat{j}+3 \hat{k}$, then the values of $t$ for which $(a+b)$ and $(a-b)$ are perpendicular are:

If $a$ and $b$ are unlike vectors,then $a \cdot b =$

Let $\vec{a}, \vec{b}, \vec{c}$ be unit vectors such that $2 \vec{a}+3 \vec{b}+4 \vec{c}=\vec{0}$. Then $|\vec{b} \times \vec{c}|=$