91Ó°ÊÓ

The magnitude of first vector is A.

The magnitude of second vector is B.

The displacement is resolved along the horizontal and vertical directions; then the sum of all the horizontal components provides the horizontal component of net displacement. The sum of the vertical components provides the vertical component of net displacement.

The resultant of two vectors is given by

$\stackrel{→}{R}=\stackrel{→}{A}+\stackrel{→}{B}$

The magnitude of resultant is given by

$$\begin{gathered} {\left|\stackrel{→}{R}\right|}^2={\left|\stackrel{→}{A}\right|}^2+{\left|\stackrel{→}{B}\right|}^2+2\left|\stackrel{→}{A}\right|\left|\stackrel{→}{B}\right|\cos \mathrm{θ} \\ {R}^2=A^2+B^2+2AB\cos \mathrm{θ} \\ R=\sqrt{A^2+B^2+2AB\cos \mathrm{θ}}......\left(1\right) \end{gathered}$$

Here, $\mathrm{θ}$ is the angle between vector A and vector B. The magnitude of the resultant will be maximum if the value of angle $\mathrm{θ}$ is $0°$.

Substitute $\mathrm{θ}=0°$ values in the equation (1).

$$\begin{gathered} R_{\max }=\sqrt{A^2+B^2+2AB\left(\cos 0°\right)} \\ {R}_{\max }=\sqrt{A^2+B^2+2AB} \\ {R}_{\max }=\sqrt{{\left(A+B\right)}^2} \\ {R}_{\max }=A+B \end{gathered}$$

The relative direction of vectors A and B should be role="math" $0°$ for the maximum magnitude of the resultant.

Substitute $\mathrm{θ}=180°$ values in the equation (1) for the minimum magnitude of the resultant.

$$\begin{gathered} R_{\min }=\sqrt{A^2+B^2+2AB\left(\cos 180°\right)} \\ {R}_{\min }=\sqrt{A^2+B^2-2AB} \\ {R}_{\min }=\sqrt{{\left(A-B\right)}^2} \\ {R}_{\min }=A-B \end{gathered}$$

The relative direction of vectors A and B should be opposite to each other for the minimum magnitude of the resultant.

Therefore, the relative direction for maximum magnitude $\left(A+B\right)$should be in the same direction and the relative direction for minimum magnitude $\left(A-B\right)$ should be in the opposite direction.