If $|a|=3, |b|=4$ and the angle between $a$ and $b$ is $120^{\circ}$,then $|4a+3b|$ is equal to

Let $\vec{a}=4 \hat{i}+3 \hat{j}$ and $\vec{b}=3 \hat{i}-4 \hat{j}+5 \hat{k}$ and $\vec{c}$ is a vector such that $\vec{c} \cdot(\vec{a} \times \vec{b})+25=0, \vec{c} \cdot(\hat{i}+\hat{j}+\hat{k})=4$ and the projection of $\vec{c}$ on $\vec{a}$ is $1$. Then,the projection of $\vec{c}$ on $\vec{b}$ equals:

The orthogonal projection vector of $\vec{a} = 2\hat{i} + 3\hat{j} + 3\hat{k}$ on $\vec{b} = \hat{i} - 2\hat{j} + \hat{k}$ is

The points represented by $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are coplanar and $(\sin A)\vec{a} + (2\sin 2B)\vec{b} + (3\sin 3C)\vec{c} - 4\vec{d} = \vec{0}$. Then the least value of $\frac{21}{8}(\sin^2 A + \sin^2 2B + \sin^2 3C)$ is:

If $|\vec{a}|=3, |\vec{b}|=5$ and $|\vec{c}|=7$ and $\vec{a}+\vec{b}+\vec{c}=\vec{0}$,then the angle between $\vec{a}$ and $\vec{b}$ is

If $\vec{a}$ is a unit vector and $(\vec{x}-\vec{a}) \cdot (\vec{x}+\vec{a}) = 8$,then $|\vec{x}| = $ . . . . . . .