Q79P - University Physics with Modern Physics

Question

Vectors $\overrightarrow{A}$  and $\overrightarrow{B}$  have scalar product $-6.00$, and their vector product has magnitude $+9.00$ . What is the angle between these two vectors?

Step-by-Step Solution

Answer

The angle between two vectors  and  is $\theta =56.31°$.

Step 1: Identification of the given data

The given data can be expressed below as:

• The scalar product of vectors $\overrightarrow{A } and \overrightarrow{B}$is $-6.00$.
• The vector product of vectors $\overrightarrow{A }and \overrightarrow{B}$is +9.00.

Step 2: Significance of the dot and cross product in identifying the angle

This scalar or dot product is the algebraic operation that takes “two equal-length” numbers and returns a single piece of number.

The vector or cross product is the two vector’s binary operation in the 3-D space.

Dividing the magnitude of the scalar and the vector product gives the angle between the two vectors.

Step 3: Determination of the angle between the two vectors

The magnitude of the dot product of the vector can be expressed as:

$\overrightarrow{A}.\overrightarrow{B}=\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\cos \theta$

Here, $\overrightarrow{A} and \overrightarrow{B}$are the vectors having the scalar product $-6.00$ which is:

$-6.00=\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\cos \theta .......................\left(1\right)$

The magnitude of the cross product of the vector can be expressed as:

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

Here, $\overrightarrow{A} and \overrightarrow{B}$are the vectors having the vector product +9.00  which is:

$+9.00=\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\sin \theta ......................\left(2\right)$

Now, divide equation (2) and (1), which results in,

$$\begin{gathered} \frac{+9.00}{-6.00}=\frac{\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\sin \theta }{\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\cos \theta } \\ \tan \theta =\frac{9}{-6} \\ \theta ={\tan }^{-1}\left(\frac{9}{-6}\right) \\ \theta =56.31° \end{gathered}$$

Hence, the angle between two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$  is  $\theta =-56.31°$.