Which of the following relations is true for two unit vectors $\hat{ A }$ and $\hat{ B }$ making an angle $\theta$ to each other$?$
$|\hat{ A }+\hat{ B }|=|\hat{ A }-\hat{ B }| \tan \frac{\theta}{2}$
$|\hat{ A }-\hat{ B }|=|\hat{ A }+\hat{ B }| \tan \frac{\theta}{2}$
$|\hat{ A }+\hat{ B }|=|\hat{ A }-\hat{ B }| \cos \frac{\theta}{2}$
$|\overrightarrow{ A }-\hat{ B }|=|\overrightarrow{ A }+\hat{ B }| \cos \frac{\theta}{2}$
Two forces of magnitude $P$ & $Q$ acting at a point have resultant $R$. The resolved part of $R$ in the direction of $P$ is of magnitude $Q$. Angle between the forces is :
A particle is situated at the origin of a coordinate system. The following forces begin to act on the particle simultaneously (Assuming particle is initially at rest)
${\vec F_1} = 5\hat i - 5\hat j + 5\hat k$ ${\vec F_2} = 2\hat i + 8\hat j + 6\hat k$
${\vec F_3} = - 6\hat i + 4\hat j - 7\hat k$ ${\vec F_4} = - \hat i - 3\hat j - 2\hat k$
Then the particle will move
The sum of two forces acting at a point is $16\, N.$ If the resultant force is $8\, N$ and its direction is perpendicular to minimum force then the forces are
The position vector of a particle is determined by the expression $\vec r = 3{t^2}\hat i + 4{t^2}\hat j + 7\hat k$ The distance traversed in first $10 \,sec$ is........ $m$
If $\vec{P}+\vec{Q}=\vec{P}-\vec{Q}$, then