Two vectors have length$V_1=3.5\mathrm{km}$and$V_2=4.0\mathrm{km}$. What are maximum and minimum magnitude of their vector sum?

Vector summation, or vector addition, is an operation performed to determine the resultant vector of two or more vectors.

The magnitude of the first vector,$V_1=3.5\mathrm{km}$

The magnitude of the second vector,$V_2=4.0\mathrm{km}$

Let us assume that the angle between the first and second vectors is $\theta$and the magnitude of the resultant vector is V.

When two vectors $\overrightarrow{A}$and $\overrightarrow{B}$ having magnitude A and B, respectively, are aligned at an angle $\theta$with respect to each other, the magnitude of the vector sum of these two vectors is written as$R=\sqrt{A^2+B^2+2AB\cos \theta }$

Here, R is the magnitude of the resultant vector.

The length of a vector signifies the magnitude of that vector. Thus, using the formula mentioned above, the magnitude of the resultant vector can be written as$V=\sqrt{V_1^2+V_2^2+2V_1{V}_2\cos \theta }$

From the above equation of V, you can see that the value of V is dependent on $\cos \theta$. When $\cos \theta$ will be the maximum, V will also be the maximum. $\cos \theta$is the maximum at $\theta =0°$. ($\cos 0°=1$)

Thus, the maximum value of the vector sum will be

$V=\sqrt{V_1^2+V_2^2+2V_1{V}_2\cos 0°}$

Substituting the values in the above equation, you get:

$\begin{array}{c}{V}_{\max }=\sqrt{{3.5}^2+{4.0}^2+\left(2\times 3.5\times 4.0\right)}\\ =7.5\mathrm{km}\end{array}$

Thus, the maximum value of the vector sum will be 7.5 km.

From the equation of V, you can see that when $\cos \theta$will be the minimum, V will also be the minimum. $\cos \theta$is the minimum at $\theta =180°$. ($\cos 180°=-1$)

Thus, the minimum value of the vector sum will be

$V=\sqrt{V_1^2+V_2^2+2V_1{V}_2\cos 180°}$

Substituting the values in the above equation, you get:

$\begin{array}{c}{V}_{\min }=\sqrt{{3.5}^2+{4.0}^2-\left(2\times 3.5\times 4.0\right)}\\ =0.5\mathrm{km}\end{array}$

Thus, the minimum value of the vector sum will be 0.5 km.