Q79P - University Physics with Modern Physics

Question

Vectors $\stackrel{\rightharpoonup }{A}$and $\stackrel{\rightharpoonup }{B}$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 $\stackrel{\rightharpoonup }{A}$and $\stackrel{\rightharpoonup }{B}$is$\theta =-56.{31}^{\circ }$ .

Step 1: Identification of the given data

The given data can be expressed below as:

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

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:

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

Here,  $\stackrel{\rightharpoonup }{\mathrm{A}}\mathrm{and} \stackrel{\rightharpoonup }{\mathrm{B}}$ are the vectors having the scalar product which is:

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

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

$\stackrel{\rightharpoonup }{A}\times \overline{B}=\left|\stackrel{\rightharpoonup }{A}\right|\left|\stackrel{\rightharpoonup }{B}\right|\sin \theta$

Here,  and are the vectors having the vector product which is:

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

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

$$\begin{gathered} \frac{+9.00}{-6.00}=\frac{\left|\stackrel{\rightharpoonup }{A}\right|\left|\stackrel{\rightharpoonup }{B}\right|\sin \theta }{\left|\stackrel{\rightharpoonup }{A}\right|\left|\stackrel{\rightharpoonup }{B}\right|\cos \theta } \\ \tan \theta =\frac{9}{-6} \\ \theta ={\tan }^{-1}\left[\frac{9}{-6}\right] \\ \theta =-56.{31}^{°} \end{gathered}$$

Hence, the angle between two vectors $\stackrel{\rightharpoonup }{\mathrm{A}}\mathrm{and} \stackrel{\rightharpoonup }{\mathrm{B}}\mathrm{is} \theta =-56.{31}^{°}.$