Chapter 3, Problem 6Q

If $\text{<mover accent="true"> <mo> V </mo> <mo> → </mo> </mover>}={\text{<mover accent="true"> <mo> V </mo> <mo> → </mo> </mover>}}_1+{\text{<mover accent="true"> <mo> V </mo> <mo> → </mo> </mover>}}_2$ , is V necessarily greater than V₁, and/or V₂? Discuss.

Solution

To Determine:

Whether the magnitude of the sum of two vectors is necessarily greater than the magnitude of V₁ and/or V₂.

Solution:

No, the magnitude of the sum of two vectors is not necessarily greater than the magnitude of $V_1$ and/or $V_2$ .

Explanation:

The sum of the vector could result in a shorter vector than the original ones, therefore the magnitude would be less than the other vectors. The resultant magnitude depends on the direction in which the two vectors are inclined.

  $\overrightarrow{V}={\overrightarrow{V}}_1+{\overrightarrow{V}}_2$

The resultant magnitude of vector can be obtained by:

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

Here, the maximum value of $\cos \theta$ could be 1 and minimum value could be -1.

If $\cos \theta =1$

  $\begin{array}{l}V=\sqrt{V_1{}^2+V_2{}^2+2V_1{V}_2}\\ V=\sqrt{{\left(V_1+V_2\right)}^2}\\ V=\left(V_1+V_2\right)\end{array}$

This implies that the value of V is greater than the value of each $V_1$ and $V_2$ . If $\cos \theta =-1$

  $\begin{array}{l}V=\sqrt{V_1{}^2+V_2{}^2-2V_1{V}_2}\\ V=\sqrt{{\left(V_1-V_2\right)}^2}\\ V=\left(V_1-V_2\right)\end{array}$

This implies that the value of V is less than the value of each $V_1$ and $V_2$ .

Conclusions:

The magnitude of the sum of the vectors could be greater than the magnitude of any of the vectors, but not always.