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 $\vec{a}=\hat{i}+2\hat{j}+\hat{k}$,$\vec{b}=\hat{i}-\hat{j}+4\hat{k}$,and $\vec{c}=\hat{i}+\hat{j}+\hat{k}$ are such that $\vec{a}+\lambda\vec{b}$ is perpendicular to $\vec{c}$,then the value of $\lambda$ is

Let $\vec{a}$ and $\vec{b}$ be two unit vectors and $\theta$ is the angle between them. Then $\vec{a}+\vec{b}$ is a unit vector if $\theta =$

If $|\vec{a}|=3, |\vec{b}|=4, |\vec{c}|=5$ and each one of $\vec{a}, \vec{b}, \vec{c}$ is perpendicular to the sum of the remaining two,then find $|\vec{a}+\vec{b}+\vec{c}|$.

If $\vec{a} = 4 \hat{i} + 5 \hat{j} - 3 \hat{k}$ and $\vec{b} = 6 \hat{i} - 2 \hat{j} - 2 \hat{k}$ are two vectors, then the magnitude of the component of $\vec{b}$ parallel to $\vec{a}$ is: (in $\sqrt{2}$)

$A$ particle is acted upon by constant forces $4\hat{i} + \hat{j} - 3\hat{k}$ and $3\hat{i} + \hat{j} - \hat{k}$. The displacement of the particle from the point $\hat{i} + 2\hat{j} + 3\hat{k}$ to the point $5\hat{i} + 4\hat{j} + \hat{k}$ is given. Find the total work done by the forces in units.