91影视

The given data can be expressed below as:

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

This scalar or dot product is the algebraic operation that takes 鈥渢wo equal-length鈥� numbers and returns a single piece of number.

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

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

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

$\stackrel{鈫�}{A}.\stackrel{鈫�}{B}=\left|\stackrel{鈫�}{A}\right|\left|\stackrel{鈫�}{B}\right|\cos 胃$

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

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

$\stackrel{鈫�}{A}脳\stackrel{鈫�}{B}=\left|\stackrel{鈫�}{A}\right|\left|\stackrel{鈫�}{B}\right|\sin 胃$

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

$+9.00=\left|\stackrel{鈫�}{A}\right|\left|\stackrel{鈫�}{B}\right|\sin 胃......................\left(2\right)$

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

$$\begin{gathered} \frac{+9.00}{-6.00}=\frac{\left|\stackrel{鈫�}{A}\right|\left|\stackrel{鈫�}{B}\right|\sin 胃}{\left|\stackrel{鈫�}{A}\right|\left|\stackrel{鈫�}{B}\right|\cos 胃} \\ \tan 胃=\frac{9}{-6} \\ 胃={\tan }^{-1}\left(\frac{9}{-6}\right) \\ 胃=56.31掳 \end{gathered}$$

Hence, the angle between two vectors $\stackrel{鈫�}{A}$and $\stackrel{鈫�}{B}$ is$胃=-56.31掳$.