91Ó°ÊÓ

The vectors are given below:

$$\begin{gathered} {\stackrel{→}{d}}_1=4.0\stackrel{Áåœ}{i}+5.01\stackrel{Áåœ}{j} \\ {\stackrel{→}{d}}_2=-3.0\stackrel{Áåœ}{i}+4.0\stackrel{Áåœ}{j} \end{gathered}$$

Usethe rules of vector product, dot product, and cross product. The dot product of two vectors is a scalar quantity and the cross product of two vectors is a vector quantity.

The angle between the vectors can be calculated as,

$\cos \mathrm{θ}=\frac{\left(\stackrel{→}{d_1}.{\stackrel{→}{d}}_2\right)}{\left(d_1.d_2\right)}$ (i)

The cross product is calculated as,

$\stackrel{→}{A}×\stackrel{→}{B}=\stackrel{Áåœ}{i}\left(A_y{B}_z-A_z{B}_y\right)-\stackrel{Áåœ}{j}\left(A_x{B}_z-A_z{B}_x\right)+\stackrel{Áåœ}{k}\left(A_x{B}_y-A_y{B}_x\right)$ (ii)

Given vectors do not have $\stackrel{Áåœ}{k}$components, so $\stackrel{Áåœ}{i}$and $\stackrel{Áåœ}{j}$components of the cross product would be zero. Hence, we have only $\stackrel{Áåœ}{k}$component in the cross product. The cross product can be found using equation (ii) as follows:

$$\begin{gathered} {\stackrel{→}{d}}_1×{\stackrel{→}{d}}_2=\left(4\stackrel{Áåœ}{i}+5.01\stackrel{Áåœ}{j}\right)×\left(-3\stackrel{Áåœ}{i}+04\stackrel{Áåœ}{j}\right) \\ =\stackrel{Áåœ}{i}\left[\left(5×0\right)-\left(0×4\right)\right]-\stackrel{Áåœ}{j}\left[\left(4×0\right)-\left(0×-3\right)\right]+\stackrel{Áåœ}{k}\left[\left(4×4\right)-\left(-3×5\right)\right] \\ =\stackrel{Áåœ}{i}\left[0\right]-\stackrel{Áåœ}{j}\left[0\right]+\stackrel{Áåœ}{k}\left[\left(16\right)-\left(-15\right)\right] \\ =31\stackrel{Áåœ}{k} \end{gathered}$$

Therefore, role="math" localid="1657946771878" ${\stackrel{→}{d}}_1×{\stackrel{→}{d}}_2$is $31\stackrel{Áåœ}{k}$.

Use the dot product formula to calculate the dot product${\stackrel{→}{d}}_1.{\stackrel{→}{d}}_2$.

role="math" localid="1657946958877" $$\begin{gathered} {\stackrel{→}{d}}_1.{\stackrel{→}{d}}_2=\left(4\stackrel{Áåœ}{i}+5.01\stackrel{Áåœ}{j}\right).\left(-3\stackrel{Áåœ}{i}+04\stackrel{Áåœ}{j}\right) \\ =8.0 \end{gathered}$$)

Therefore, dot product role="math" localid="1657946883238" ${\stackrel{→}{d}}_1.{\stackrel{→}{d}}_2$is 8.0 .

Now, to calculate the $\left({\stackrel{→}{d}}_1.{\stackrel{→}{d}}_2\right).{\stackrel{→}{d}}_2$, first simply the equation as follows.

$\left({\stackrel{→}{d}}_1.{\stackrel{→}{d}}_2\right).{\stackrel{→}{d}}_2={\stackrel{→}{d}}_1.{\stackrel{→}{d}}_2+d_2^2$

Now, calculate$d_2^2$ .

$$\begin{gathered} d_2^2={\left(-3\right)}^2+{\left(4\right)}^2 \\ =25 \end{gathered}$$

Use the value of ${\stackrel{→}{d}}_1.{\stackrel{→}{d}}_2$from step (4) to calculate $\left({\stackrel{→}{d}}_1.{\stackrel{→}{d}}_2\right){\stackrel{→}{.d}}_2$

$$\begin{gathered} \left({\stackrel{→}{d}}_1.{\stackrel{→}{d}}_2\right){\stackrel{→}{.d}}_2=8+25 \\ =33 \end{gathered}$$

The magnitude of ${\stackrel{→}{d}}_1$is as follows:

$$\begin{gathered} \left|d_1\right|=\sqrt{4^2+5^2} \\ =6.4\mathrm{m} \end{gathered}$$

Use equation (i) to find the direction.

$$\begin{gathered} \cos \mathrm{θ}=\frac{{\stackrel{→}{d}}_1.{\stackrel{→}{d}}_1}{d_1{d}_2} \\ \mathrm{θ}={\cos }^{-1}\left(\frac{\left(6.4\right).\left(5\right)}{8}\right) \\ =75.5° \end{gathered}$$

So horizontal component of${\stackrel{→}{d}}_1$ is as follows:

$$\begin{gathered} {\stackrel{→}{d}}_{1horizontal}=6.4.\cos \left(75.5\right) \\ =1.6 \end{gathered}$$

Therefore, the horizontal component of ${\stackrel{→}{d}}_1$is equal to 1.6 .