Displacement $\overrightarrow{{\mathbf{d}}_{\mathbf{1}}}$is in the yz plane $\mathbf{63}\mathbf{.}\mathbf{0}\mathbf{°}$from the positive direction of the y axis, has a positive z component, and has a magnitude of 4.50 m. Displacement is in the xz plane $\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{°}$from the positive direction of the x axis, has a positive z component, and has magnitude 1.40 m. What are (a)${\overrightarrow{\mathbf{d}}}_{\mathbf{1}}$.${\overrightarrow{\mathbf{d}}}_{\mathbf{2}}$, (b) role="math" localid="1656999023128" ${\overrightarrow{\mathbf{d}}}_{\mathbf{1}}\mathbf{\times }{\overrightarrow{\mathbf{d}}}_{\mathbf{2}}$, and (c) the angle between ${\overrightarrow{\mathbf{d}}}_{\mathbf{1}}$and ${\overrightarrow{\mathbf{d}}}_{\mathbf{2}}$?

$$\begin{gathered} d_1=4.50\left(\cos \left(63\right)\stackrel{⏜}{j}+\sin \left(63\right)\stackrel{⏜}{k}\right) \\ =2.04\stackrel{⏜}{j}+4.01\stackrel{⏜}{k} \\ {d}_2=1.40\left(\cos \left(30\right)\stackrel{⏜}{i}+\sin \left(30\right)\stackrel{⏜}{k}\right) \\ =1.21\stackrel{⏜}{i}+0.7\stackrel{⏜}{k} \end{gathered}$$

Here, use rule of vector product, dot product and cross product. Dot product of two vectors is scalar quantity and cross product of two vectors is vector quantity.

Formula is as follows:

$\cos \theta =\frac{\left(\overrightarrow{d_1}.\overrightarrow{d_2}\right)}{\overrightarrow{d_1}\overrightarrow{d_2}}$

Where,

role="math" localid="1656999888380" $\overrightarrow{d_1},\overrightarrow{d_2}$Are displacement vectors.

Using the formula for the dot product,

$$\begin{gathered} \overrightarrow{d_1}.\overrightarrow{d_2}=\left(2.04\stackrel{⏜}{j}+4.01\stackrel{⏜}{k}\right).{\left(1.21\stackrel{⏜}{i}+0.7\stackrel{⏜}{k}\right)}_{)} \\ =2.81 \end{gathered}$$

Thus,the dot product is 2.81.

Using the formula for the cross product,

$$\begin{gathered} \overrightarrow{d_1}\times \overrightarrow{d_2}=\left(2.04\stackrel{⏜}{j}+4.0\stackrel{⏜}{k}\right)\times \left(1.21\stackrel{⏜}{i}+0.7\stackrel{⏜}{k}\right) \\ =1.43\stackrel{⏜}{i}+4.86\stackrel{⏜}{j}-2.48\stackrel{⏜}{k} \end{gathered}$$

Thus, cross product is $1.43\stackrel{⏜}{i}+4.86\stackrel{⏜}{j}-2.48\stackrel{⏜}{k}$.

Magnitude of$d_1$ is as follow:

$$\begin{gathered} d_1=\sqrt{2.{04}^2+4.{01}^2} \\ =4.50 \end{gathered}$$

Magnitude of is as follow:

$$\begin{gathered} d_2=\sqrt{1.{21}^2+0.7^2} \\ =1.40 \end{gathered}$$

Therefore, the direction is given as,

$$\begin{gathered} \cos \theta =\frac{\overrightarrow{d_1}.\overrightarrow{d_2}}{\overrightarrow{d_1}\overrightarrow{d_2}} \\ =\frac{2.81}{4.5\times 1.40} \\ \theta =63.5° \end{gathered}$$

Hence, thedirection is $63.5°$.

Therefore, from the given information, vectors in the unit vector notation form can be written. Using the formula for the dot product and cross product, the answers can be found. It is also possible to find the angle between them using the property of dot product.