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?

Short answer

The angle between two vectors $\stackrel{\rightharpoonup }{A}$and $\stackrel{\rightharpoonup }{B}$is$\theta =-56.{31}^{\circ }$ .

Step-by-step solution

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 $\stackrel{\rightharpoonup }{\mathrm{A}}\mathrm{and}\stackrel{\rightharpoonup }{\mathrm{B}}\mathrm{is}+9.00.$

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.

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:

role="math" localid="1655791391574" $-6.00=\left|\stackrel{\rightharpoonup }{A}\right|\left|\stackrel{\rightharpoonup }{B}\right|\cos \theta ........\dots \left(1\right)$

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

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

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