Question

the resultant of vector A and vector B is perpendicular to vector B.what is the angle between a and b

Answer 1

The angle made by the resultant of the two vectors $\vec{\sf{A}}$ and $\vec{\sf{B}}$ with $\vec{\sf{B}}$ is given by,

$\longrightarrow\alpha=\tan^{-1}\left(\dfrac{\mathsf{A}\sin\theta}{\mathsf{B+A\cos\theta}}\right)\quad\quad\dots\sf{(1)}$

where $\theta$ is the angle between them.

Here the resultant is perpendicular to $\vec{\sf{B}}$ so it makes 90° with it.

$\sf{\longrightarrow\alpha=90^o}$

From (1),

$\longrightarrow\tan^{-1}\left(\dfrac{\mathsf{A}\sin\theta}{\mathsf{B+A\cos\theta}}\right)=\sf{90^o}$

$\longrightarrow\dfrac{\mathsf{A}\sin\theta}{\mathsf{B+A\cos\theta}}=\tan\sf{90^o}$

$\longrightarrow\dfrac{\mathsf{A}\sin\theta}{\mathsf{B+A\cos\theta}}=\mathsf{\dfrac{1}{0}}$

Equating the denominators,

$\sf{\longrightarrow B+A\cos\theta=0}$

$\sf{\longrightarrow \cos\theta=-\dfrac{B}{A}}$

$\sf{\longrightarrow\underline{\underline{\theta=\cos^{-1}\left(-\dfrac{B}{A}\right)}}}$