Question

Vector ${\overrightarrow{\mathbf{d}}}_{\mathbf{1}}$is in the negative direction of a y axis, and vector $\overrightarrow{{\mathbf{d}}_{\mathbf{2}}}$is in the positive direction of an x axis. What are the directions of (a) $\overrightarrow{{\mathbf{d}}_{\mathbf{2}}}\mathbf{/}\mathbf{4}$and (b) ${\mathit{d}}_{\mathbf{1}}\mathbf{/}\left(-4\right)$? What are the magnitudes of products (c) ${\overrightarrow{\mathbf{d}}}_{\mathbf{1}}\mathbf{-}{\overrightarrow{\mathbf{d}}}_{\mathbf{2}}$and (d) ${\overrightarrow{\mathbf{d}}}_{\mathbf{1}}.\left({\overrightarrow{d}}_2/4\right)$? What is the direction of the vector resulting from (e) ${\overrightarrow{\mathbf{d}}}_{\mathbf{1}}\mathbf{\times }{\overrightarrow{\mathbf{d}}}_{\mathbf{2}}$and (f)${\overrightarrow{\mathbf{d}}}_{\mathbf{1}}\mathbf{\times }{\overrightarrow{\mathbf{d}}}_{\mathbf{2}}$? What is the magnitude of the vector product in (g) part (e) and (h) part (f)? What are the (i) magnitude and (j) direction of${\overrightarrow{\mathbf{d}}}_{\mathbf{1}}\times \left({\overrightarrow{d}}_2/4\right)$?

Short answer

a) The direction of ${\overrightarrow{d}}_2/4$is along x axis.

b) The direction of ${\overrightarrow{d}}_1/\left(-4\right)$is along y axis.

c) The magnitude of the product ${\overrightarrow{d}}_1.{\overrightarrow{d}}_2$is 0

d) The magnitude of the product${\overrightarrow{d}}_1.\left({\overrightarrow{d}}_2/4\right)$ is 0

e) The direction of ${\overrightarrow{d}}_1\times \overrightarrow{d_2}$is along +z axis.

f) The direction of $\overrightarrow{d_2}\times \overrightarrow{d_1}$is along –z direction

g) The magnitude of vector product${\overrightarrow{d}}_1\times {\overrightarrow{d}}_2$ is$d_1{d}_2$

h) The magnitude of vector product${\overrightarrow{d}}_2\times {\overrightarrow{d}}_1$ is$d_1{d}_2$

i) The magnitude of$\overrightarrow{d_1}\times \left(\overrightarrow{d_1}/4\right)$ is$d=\frac{d_1{d}_2}{4}$

j) The direction of $\overrightarrow{d_1}\times \left(\overrightarrow{d_2}/4\right)$is along +z axis.

Step-by-step solution

Given data

Two vectors, $\overrightarrow{d_1}$and $\overrightarrow{d_2}$are given as,

$$\begin{gathered} \overrightarrow{d_1}=-d_1\stackrel{⏜}{j} \\ \overrightarrow{d_2}=d_2\stackrel{⏜}{i} \end{gathered}$$

Understanding the concept

Use the laws of dot product and cross product to find the direction and magnitude of different operations described in problem. Magnitude of dot product is scalar quantity and cross product of vectors is vector quantity. The direction of the cross product can be found using the right-hand rule.

The dot product can be found using the formula,

$\overrightarrow{a}.\overrightarrow{b}=ab\cos \theta$……..(i)

The cross product can be found using the formula,

$(\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}})=({\mathrm{a}}_{\mathrm{y}}{\mathrm{b}}_{\mathrm{z}}-{\mathrm{b}}_{\mathrm{y}}{\mathrm{a}}_{\mathrm{z}})\stackrel{⏜}{\mathrm{i}}+({\mathrm{a}}_{\mathrm{z}}{\mathrm{b}}_{\mathrm{x}}-{\mathrm{b}}_{\mathrm{z}}{\mathrm{a}}_{\mathrm{x}})\stackrel{⏜}{\mathrm{j}}+({\mathrm{a}}_{\mathrm{x}}{\mathrm{b}}_{\mathrm{y}}-{\mathrm{b}}_{\mathrm{y}}{\mathrm{a}}_{\mathrm{x}})\stackrel{⏜}{\mathrm{k}}$……….(ii)

Step 3: (a) Calculate the directions of d→2/4

Multiplying a vector by a positive scalar quantity does not affect the direction of the vector. Therefore, the direction of${\overrightarrow{d}}_2/4$ is same as vector${\overrightarrow{d}}_2$ which is in positive x direction.

Step 4: (b) Calculate the directions of d→1/(-4) 

Dividing a vector by a negative scalar quantity reverses the direction of the vector. Therefore, the direction of$\overrightarrow{d_1}/\left(-4\right)$ is in opposite to$\overrightarrow{d_1}$ i.e., in positive y direction.

(c) Calculate the magnitude of d1→.d2→

The vector $\overrightarrow{{\mathrm{d}}_1}$and $\overrightarrow{{\mathit{d}}_{\mathit{2}}}$are perpendicular to each other. Therefore, the angle between them is $90°$. From equation (i)

$$\begin{gathered} \overrightarrow{{\mathrm{d}}_1}.\overrightarrow{{\mathrm{d}}_2}=d_1{d}_{2}cos\theta \\ =d_1{d}_{2}cos\left(90\right) \\ =0 \end{gathered}$$

Therefore, the dot product of the vectors, $\overrightarrow{{\mathit{d}}_{\mathit{1}}}$and $\overrightarrow{{\mathit{d}}_{\mathit{2}}}$, is equal to 0.

(d) Calculate the magnitude of d1→.(d2→/4)

The vector$\overrightarrow{{\mathit{d}}_{\mathit{1}}}$and$\overrightarrow{{\mathit{d}}_{\mathit{2}}}$are perpendicular to each other. Therefore, the angle between them is $90°$. From equation (i)

$$\begin{gathered} \overrightarrow{{\mathrm{d}}_1}.\left(\frac{\overrightarrow{{\mathrm{d}}_2}}{4}\right)=\frac{1}{4}\left(d_1{d}_{2}cos\theta \right) \\ =\frac{1}{4}({\mathrm{d}}_1{\mathrm{d}}_2\cos \left(90\right)) \\ =0 \end{gathered}$$

Therefore, the dot product$\overrightarrow{{\mathrm{d}}_1}.\left(\frac{\overrightarrow{{\mathrm{d}}_2}}{4}\right)$ is equal to 0.

(e) Calculate the direction of the vector d→1×d2→

The cross product of vector${\overrightarrow{\mathit{d}}}_{\mathit{1}}$and${\overrightarrow{d}}_{\mathit{2}}$ can be calculated from equation (ii). Substituting the components in equation (ii) and simplifying them,

$$\begin{gathered} {\overrightarrow{d}}_{\mathit{1}}\times {\overrightarrow{d}}_{\mathit{2}}=-d_{\mathbf{1}}{d}_2\left(\stackrel{⏜}{j}\times \stackrel{⏜}{i}\right) \\ =d_{\mathbf{1}}{d}_2\stackrel{⏜}{k} \end{gathered}$$

Since$\stackrel{⏜}{k}$ is along the z-axis, direction of the cross product of vector $\overrightarrow{d_1}$and$\overrightarrow{d_2}$ is along the z-axis.

(f) Calculate direction of the vector d→2×d→1

$$\begin{gathered} {\overrightarrow{d}}_{\mathbf{2}}\times {\overrightarrow{d}}_1=-d_1{d}_2\left(\stackrel{⏜}{i}\times \stackrel{⏜}{j}\right) \\ =-d_1{d}_2\stackrel{⏜}{k} \end{gathered}$$

Since$-\stackrel{⏜}{k}$ is along the negative z-axis, direction of the cross product of vector ${\overrightarrow{d}}_2$and${\overrightarrow{d}}_1$ is along the z-axis.

(g) Calculate the magnitude of d→1×d→2

From part e, we can write,

${\overrightarrow{d}}_{\mathbf{1}}\times {\overrightarrow{d}}_2=d_1{d}_2\stackrel{⏜}{k}$

Therefore, the magnitude of${\overrightarrow{d}}_{\mathbf{1}}\times {\overrightarrow{d}}_2$ is equal to$d_1{d}_2$

(h) Calculate the magnitude of the vector d→2×d→1/4

Again, from part (f), we can write,

$$\begin{gathered} {\overrightarrow{d}}_{\mathbf{2}}\times {\overrightarrow{d}}_1=d_1{d}_2 \\ \left|{\overrightarrow{d}}_{\mathbf{2}}\times {\overrightarrow{d}}_1\right|=d_1{d}_2 \end{gathered}$$

Therefore, the magnitude of ${\overrightarrow{d}}_{\mathbf{2}}\times {\overrightarrow{d}}_1$is$d_1{d}_2$

(i) Calculate the magnitude of d→1×d→2/4

From part e, we can write,

$$\begin{gathered} {\overrightarrow{d}}_{\mathbf{1}}\times \frac{{\overrightarrow{d}}_2}{4}=\frac{1}{4}\left({\overrightarrow{d}}_1\times {\overrightarrow{d}}_2\right) \\ =\frac{d_1{d}_2}{4}\stackrel{⏜}{k} \end{gathered}$$

Therefore, the magnitude is$\frac{d_1{d}_2}{4}$

(j) Calculate the direction of d1→×d2→/4

From part e, we can write,

$$\begin{gathered} \overrightarrow{d_1}\times \frac{\overrightarrow{d_2}}{4}=\frac{1}{4}\left(\overrightarrow{d_1}\times \overrightarrow{d_2}\right) \\ =\frac{d_1{d}_2}{4}\stackrel{⏜}{k} \end{gathered}$$

Since the $\stackrel{⏜}{k}$is along the positive z-axis, the direction of the cross product$\overrightarrow{d_1}\times \frac{\overrightarrow{d_2}}{4}$is also along the positive z-axis.