Chapter 10, Problem 2P

To determine

To Choose: The correct option.

Answer to Problem 2P

Option (b)

Explanation of Solution

Given:

Two nonzero vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ .

Formula used:

The formula of vector product:

  $\overrightarrow{A}\times \overrightarrow{B}=\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\sin \theta \widehat{n}$

$·\left|\overrightarrow{A}\right|=$ Magnitude of vector B

$·\left|\overrightarrow{A}\right|=$ Magnitude of vector A

$·\theta =$ Angle between the vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ .

$·\widehat{n}=$ Unit vector perpendicular to the plane $\overrightarrow{A}\times \overrightarrow{B}$ .

Calculation:

When $\overrightarrow{A}$ and $\overrightarrow{B}$ are parallel:

  $\theta =0^o$

  $\begin{array}{c}\overrightarrow{A}\times \overrightarrow{B}=\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\sin \theta \widehat{n}\\ =\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\sin 0^o\widehat{n}\\ =\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\times 0\widehat{n}\because \left(\sin 0^o=0\right)\\ =0\widehat{n}\end{array}$

The magnitude of the vector product:

  $\left|\overrightarrow{A}\times \overrightarrow{B}\right|=\sqrt{0^2}=0$

When $\overrightarrow{A}$ and $\overrightarrow{B}$ are perpendicular:

  $\theta ={90}^o$

  $\begin{array}{c}\overrightarrow{A}\times \overrightarrow{B}=\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\sin \theta \widehat{n}\\ =\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\sin {90}^o\widehat{n}\\ =\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\times 1\widehat{n}\because \left(\sin {90}^o=1\right)\\ =\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\widehat{n}\end{array}$

The magnitude of the vector product:

  $\begin{array}{c}\left|\overrightarrow{A}\times \overrightarrow{B}\right|=\sqrt{{\left(\text{coefficient of vector}\right)}^2}\\ =\sqrt{{\left(\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\right)}^2}\\ =\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\end{array}$

When $\overrightarrow{A}$ and $\overrightarrow{B}$ are antiparallel:

  $\theta ={180}^o$

  $\begin{array}{c}\overrightarrow{A}\times \overrightarrow{B}=\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\sin \theta \widehat{n}\\ =\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\sin {180}^o\widehat{n}\\ =\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\times 0\widehat{n}\because \left(\sin {180}^o=0\right)\\ =0\widehat{n}\end{array}$

The magnitude of the vector product:

  $\left|\overrightarrow{A}\times \overrightarrow{B}\right|=\sqrt{0^2}=0$

When $\overrightarrow{A}$ and $\overrightarrow{B}$ are at ${45}^o$ :

  $\theta ={45}^o$

  $\begin{array}{c}\overrightarrow{A}\times \overrightarrow{B}=\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\sin \theta \widehat{n}\\ =\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\sin {45}^o\widehat{n}\\ =\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\times \frac{1}{\sqrt{2}}\widehat{n}\because \left(\sin {90}^o=1\right)\\ =\frac{\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|}{\sqrt{2}}\widehat{n}\end{array}$

The magnitude of the vector product:

  $\begin{array}{c}\left|\overrightarrow{A}\times \overrightarrow{B}\right|=\sqrt{{\left(\text{coefficient of vector}\right)}^2}\\ =\sqrt{{\left(\frac{\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|}{\sqrt{2}}\right)}^2}\\ =\frac{\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|}{\sqrt{2}}\end{array}$

The greatest magnitude of the vector product is $\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|$ .

Conclusion:

Hence, when the vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ are perpendicular to each other, the magnitude of the vector productwould be greatest.

Option (b) is correct.