Suppose $\vec{a}+\vec{b}+\vec{c}=0$,$|\vec{a}|=3$,$|\vec{b}|=5$,$|\vec{c}|=7$. Then the angle between $\vec{a}$ and $\vec{b}$ is:

If $a, b, c$ are unit vectors such that $a + b + c = 0,$ then $a \cdot b + b \cdot c + c \cdot a = $

Show that the vectors $2 \hat{i}-\hat{j}+\hat{k}, \hat{i}-3 \hat{j}-5 \hat{k}$ and $3 \hat{i}-4 \hat{j}-4 \hat{k}$ form the vertices of a right-angled triangle.

If $a, b$ and $c$ are unit vectors such that $a+b+c=0$,then $a \cdot b+b \cdot c+c \cdot a$ is equal to

Let $a = \sin^2 x \hat{i} + \cos^2 x \hat{j} + \hat{k}$,where $x \in R$. If the pairs of vectors $(a, \hat{i})$,$(a, \hat{j})$,and $(a, \hat{k})$ are adjacent sides of $3$ distinct parallelograms and $A$ is the sum of the squares of the areas of these parallelograms,then $A$ lies in the interval

If $|\vec{a}| = \sqrt{27}$,$|\vec{b}| = 7$ and $|\vec{a} \times \vec{b}| = 35$,then $\vec{a} \cdot \vec{b}$ is equal to