Vectors $\overrightarrow{\mathbf{A}}$ and $\overrightarrow{\mathit{B}}$ lie in an xy plane. $\overrightarrow{\mathbf{A}}$ has magnitude 8.00 and angle $\mathbf{130}\mathbf{°}$; has components localid="1657001111547" ${\mathit{B}}_{\mathbf{x}}\mathbf{=}\mathbf{-}\mathbf{7}\mathbf{.}\mathbf{72}$and ${\mathit{B}}_{\mathbf{y}}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{20}$. What are the angles between the negative direction of the y axis and (a) the direction of $\overrightarrow{\mathbf{A}}$, (b) the direction of the product $\overrightarrow{\mathbf{A}}\mathbf{\times }\overrightarrow{\mathbf{B}}$, and (c) the direction of localid="1657001453926" $\overrightarrow{\mathbf{A}}\mathbf{\times }\mathbf{(}\overrightarrow{\mathbf{B}}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{)}\stackrel{\mathbf{⏜}}{\mathbf{k}}$?

1) The magnitude of $\overrightarrow{A}=8.00$

2) The angle made by $\overrightarrow{A}$with x-axis is $130°$

3) The components of $\overrightarrow{\mathrm{B}}$are,

role="math" localid="1657001633950" $$\begin{gathered} B_x=-7.72 \\ {B}_y=-9.20 \end{gathered}$$

A scalar product is the product of the magnitude of one vector and the scalar component of the second vector along thedirection of the first vector.Using the formula for scalar product and vector product we can find the angle between two vectors.

$\overrightarrow{A}.\overrightarrow{B}=AB\cos \theta$ ...(i)

The vector product of two vectors $\overrightarrow{\mathbf{A}}$and $\overrightarrow{\mathbf{B}}$is written aslocalid="1657001852086" $\overrightarrow{\mathbf{A}}\mathbf{\times }\overrightarrow{\mathbf{B}}$and is a vector $\overrightarrow{\mathbf{C}}$whose magnitude is given by,

$C=AB\sin \theta$

In unit vector notation, the vector product is written as,

$\overrightarrow{A}\times \overrightarrow{B}=\stackrel{⏜}{i}\left(A_y{B}_z-A_z{B}_y\right)-\stackrel{⏜}{j}\left(A_x{B}_z-A_z{B}_x\right)+\stackrel{⏜}{k}\left(A_x{B}_y-A_y+B_x\right)$ …(ii)

The vector $\overrightarrow{A}$in unit vector notation is written as,

role="math" localid="1657002332987" $$\begin{gathered} \overrightarrow{A}=\left(8\cos \left(130°\right)\right)\stackrel{⏜}{i}+\left(8\sin \left(130°\right)\right)\stackrel{⏜}{j} \\ =\left(-5.14\right)\stackrel{⏜}{i}+\left(6.13\right)\stackrel{⏜}{j} \end{gathered}$$

Also, the vector $\overrightarrow{B}$in unit vector notation is written as,

$\overrightarrow{B}=\left(-7.22\right)\stackrel{⏜}{i}+\left(-9.20\right)\stackrel{⏜}{j}$

The direction of negative y axis is depicted by $-\stackrel{⏜}{j}$.

Therefore, the angle between –y axis and the direction of is,

$$\begin{gathered} \cos \theta =\frac{\left[\left(-\stackrel{⏜}{j}\right).\left(-5.14\right)\stackrel{⏜}{i}+\left(6.13\right)\stackrel{⏜}{j}\right]}{\sqrt{{\left(-5.14\right)}^2+6.{13}^2}} \\ \cos \theta =-\frac{6.13}{8} \\ \theta ={\cos }^{-1}\left(-\frac{6.13}{8}\right) \\ =140° \end{gathered}$$

Thus, the angle between –y axis and the direction of $\overrightarrow{A}$is$140°$

The vector product of $\overrightarrow{A}$and $\overrightarrow{B}$ is perpendicular to x and y-axis. So, $\overrightarrow{A}\times \overrightarrow{B}$is perpendicular to –y axis

Use equation (ii) to calculate the cross product and vector addition to add the vectors.

$$\begin{gathered} \overrightarrow{A}\times \left(\overrightarrow{B}+3.00\stackrel{⏜}{k}\right)=\left[\left(-5.14\right)\stackrel{⏜}{i}+\left(6.13\right)\stackrel{⏜}{j}\right]\times \left[\left(-7.22\right)\stackrel{⏜}{i}+\left(9.20\right)\stackrel{⏜}{j}+3\stackrel{⏜}{k}\right] \\ =\left[\left\{\left(6.13\times 3\right)-0\right\}\stackrel{⏜}{i}-\left\{3\times \left(-5.14\right)-0\right\}\stackrel{⏜}{j}+\left\{\left(-5.14\right)\times \left(9.2\right)-6.13\times \left(-7.22\right)\right\}\stackrel{⏜}{k}\right] \\ =18.39\stackrel{⏜}{i}+15.42\stackrel{⏜}{j}+91.54\stackrel{⏜}{k} \end{gathered}$$

Now, use equation (i) again to calculate the angle.

$$\begin{gathered} \cos \theta =\frac{\left[\left(-\stackrel{⏜}{j}\right).\left(18.39\stackrel{⏜}{i}+15.42\stackrel{⏜}{j}+91.54\stackrel{⏜}{k}\right)\right]}{\sqrt{{\left(18.39\right)}^2+{\left(-15.42\right)}^2+{\left(91.54\right)}^2}} \\ =-\frac{15.42}{94.63} \\ \theta ={\cos }^{-1}\left(\frac{15.42}{94.63}\right) \\ =99.37° \end{gathered}$$

Therefore, the angle between –y axis and the direction of $\overrightarrow{A}\times \left(\overrightarrow{B}+3.00\stackrel{⏜}{k}\right)is99.37°$.