91Ó°ÊÓ

This problem is based on the product rule in which the vector product and scalar product are the two ways of multiplying vectors.Using the given components, the magnitude of the vector can be found. Also, the magnitude angle notation can be used to find out the direction of the vector.

Given,

Magnitude of vector $\stackrel{→}{A}$is 8.00 and angle is $130°$

$\stackrel{→}{B}$has components ${\stackrel{→}{B}}_x=-7.72$ and${\stackrel{→}{B}}_y=9.20$

Using the above quantities, vector $\stackrel{→}{A}$and $\stackrel{→}{B}$vector can be written as

$$\begin{gathered} \stackrel{→}{\mathrm{A}}=8.00\left(\mathrm{COS}\left(130\right)\stackrel{Áåœ}{\mathrm{i}}+\sin \left(130\right)\stackrel{Áåœ}{\mathrm{j}}\right) \\ \stackrel{→}{\mathrm{A}}=-5.14\stackrel{Áåœ}{\mathrm{i}}+6.13\stackrel{Áåœ}{\mathrm{j}} \\ \stackrel{→}{\mathrm{B}}={\mathrm{B}}_{\mathrm{x}}\stackrel{Áåœ}{\mathrm{i}}+{\mathrm{B}}_{\mathrm{y}}\stackrel{Áåœ}{\mathrm{j}} \\ \stackrel{→}{\mathrm{B}}=-7.72\stackrel{Áåœ}{\mathrm{i}}+9.20\stackrel{Áåœ}{\mathrm{j}} \end{gathered}$$ (i)

Using equation (i) and (ii),$5\stackrel{→}{A}.\stackrel{→}{B}$ can be written as

$$\begin{gathered} 5\stackrel{→}{A}.\stackrel{→}{B}=5\left[\left(-5.14\stackrel{Áåœ}{\mathrm{i}}+6.13\stackrel{Áåœ}{\mathrm{j}}\right).\left(-7.72\stackrel{Áåœ}{\mathrm{i}}+9.20\stackrel{Áåœ}{\mathrm{j}}\right)\right] \\ 5\stackrel{→}{A}.\stackrel{→}{B}=5\left[\left(-5.14\left)\right(-7.72\right)+\left(6.13\left)\right(-9.20\right)\right] \\ 5\stackrel{→}{A}.\stackrel{→}{B}=-83.4 \end{gathered}$$

In unit vector notation,

$$\begin{gathered} 4\stackrel{→}{A}×3\stackrel{→}{B}=12\stackrel{→}{A}×\stackrel{→}{B} \\ 4\stackrel{→}{A}×3\stackrel{→}{B}=12\left[\left(-5.14\stackrel{Áåœ}{\mathrm{i}}+6.13\stackrel{Áåœ}{\mathrm{j}}\right)×\left(-7.72\stackrel{Áåœ}{\mathrm{i}}+9.20\stackrel{Áåœ}{\mathrm{j}}\right)\right] \end{gathered}$$

With the standard vector product operation, it gives

$$\begin{gathered} 4\stackrel{→}{A}×3\stackrel{→}{B}=12\left(94.6\stackrel{Áåœ}{k}\right) \\ 4\stackrel{→}{A}×3\stackrel{→}{B}=1.14×{10}^3\stackrel{Áåœ}{k} \end{gathered}$$

In spherical coordinate system $\left(\mathrm{r},\mathrm{θ},\mathrm{ϕ}\right)$gives radial distance, polar angle and azimuthal angle. In this case $\mathrm{θ}$ is not defined. Azimuthal angle along z axis is also not defined. Therefore the radial distance is$1.14×{10}^3$

Here $\stackrel{→}{A}$is in xy plane and $\stackrel{→}{A}×\stackrel{→}{B}$is perpendicular to the plane, the angle between directions of $\stackrel{→}{A}$ and$4\stackrel{→}{A}×3\stackrel{→}{B}$ is$90°$

$\stackrel{→}{\mathrm{A}}=-5.14\stackrel{Áåœ}{\mathrm{i}}+6.13\stackrel{Áåœ}{\mathrm{j}}+3.00\stackrel{Áåœ}{\mathrm{k}}$

Using the Pythagorean theorem, the magnitude of vector $\stackrel{→}{A}$is given by

$$\begin{gathered} A=\sqrt{\stackrel{→}{A}+3.00\stackrel{Áåœ}{k}} \\ A=\sqrt{{\left(5.14\right)}^2+{\left(6.13\right)}^2+{\left(3.00\right)}^2} \\ A=8.54 \\ \mathrm{The}\mathrm{angle}\mathrm{θ}=130° \\ \mathrm{Azimuthal}\mathrm{angle}\mathrm{is}\mathrm{given}\mathrm{by} \\ \mathrm{Ï•}={\cos }^{-1}\left(\frac{3.00}{8.54}\right) \\ \mathrm{Ï•}=69.4° \end{gathered}$$