Question

Consider $\overrightarrow{\mathbf{a}}$in the positive direction of x, $\overrightarrow{\mathbf{b}}$in the positive direction of y, and a scalar d. What is the direction of $\stackrel{}{\overrightarrow{\mathbf{b}}\mathbf{/}\mathbf{d}}$if d is

(a) positive and

(b) negative? What is the magnitude of

(c)$\overrightarrow{\mathbf{a}}\mathbf{\cdot }\overrightarrow{\mathbf{b}}$and (d)$\overrightarrow{\mathbf{a}}\mathbf{\cdot }\overrightarrow{\mathbf{b}}\mathbf{/}\mathbf{d}$?

What is the direction of the vector resulting from (e)$\overrightarrow{\mathbf{a}}\mathbf{\times }\overrightarrow{\mathbf{b}}$and (f)$\overrightarrow{\mathbf{b}}\mathbf{\times }\overrightarrow{\mathbf{a}}$?

(g) What is the magnitude of the vector product in (e)?

(h) What is the magnitude of the vector product in (f)? What are

(i) the magnitude and

(j) the direction of $\overrightarrow{\mathbf{a}}\mathbf{\times }\overrightarrow{\mathbf{b}}\mathbf{/}\mathbf{d}$if d is positive?

Short answer

1. The direction of $\overrightarrow{\mathrm{b}}/\mathrm{d}$if d is positive is in$(+\mathrm{y})$direction.
2. The direction of$\overrightarrow{\mathrm{b}}/\mathrm{d}$if d is negative is in$(-\mathrm{y})$direction.
3. The magnitude of$\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}$is zero.
4. The magnitude of $\frac{\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}}{\mathrm{d}}$is zero.
5. The direction of the vector $\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}$is in$(+\mathrm{z})$ direction.
6. The direction of the vector $\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{a}}$ is along$(–\mathrm{z})$axis.
7. The magnitude of vector $\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}$is$\mathrm{ab}$ .
8. The magnitude of the vector$\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{a}}$ is$\mathrm{ab}$ .
9. The magnitude of $\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}/\mathrm{d}$, if d is positive is $\mathrm{ab}/\mathrm{d}$.
10. The direction of $\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}/\mathrm{d}$, if d is positive is along $(+\mathrm{z})$direction.

Step-by-step solution

Given data

1. $\overrightarrow{\mathrm{a}}$is in the $\mathrm{x}$direction,$\overrightarrow{\mathrm{b}}$ is in the$y$direction and$\mathrm{d}$ is a scalar.
2. The vector role="math" localid="1660892974496" $\overrightarrow{a}$can be written as,$\overrightarrow{\mathrm{a}}=\mathrm{a}\widehat{\text{i}}$
3. The vector$\overrightarrow{\mathrm{b}}$ can be written as,$\overrightarrow{\mathrm{b}}=\mathrm{b}\widehat{\text{j}}$

Understanding the concept

Using the formula for scalar product and vector product we can find the magnitude and direction of different vectors which are resultant of the two vectors and scalars.

Formulae:

Scalar product:

$|\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}|=\mathrm{abcos\theta }$…… (1)

Vector product:

$|\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}|=\mathrm{ab}\mathrm{sin\theta }$……. (ii)

(a) Calculate the direction of  b→/d if d is positive

Let’s assume that the vector$\overrightarrow{\mathrm{b}}/\mathrm{d}$ is$\overrightarrow{\mathrm{v}}$ .

It can be written as,

$\overrightarrow{\mathrm{v}}=\frac{\mathrm{b}}{\mathrm{d}}$.

Multiplying a vector by positive scalar does not affect its direction.

Therefore,as d is positive, the vector$\overrightarrow{\mathrm{b}}/\mathrm{d}$ is in $(+\mathrm{y})$direction.

(b) Calculate the direction of  b→/d if d is negative

Let’s assume that the vector $\overrightarrow{b}/d$is vector $\overrightarrow{v}$.

It can be written as,

$\overrightarrow{\mathrm{v}}=-\frac{\mathrm{b}}{\mathrm{d}}$

Multiplying a vector by negative scalar will reverse its direction.

As d is negative, the vector $\overrightarrow{b}$/d is in $(-\text{y})$direction

(c) Calculate the magnitude of a→⋅b→

The vector$\overrightarrow{\mathrm{a}}$and $\overrightarrow{\mathrm{b}}$are perpendicular to each other. Therefor, the angle between them is$90°$ and hence,$\cos (90°)=0$.

Using equation (i), it can be said that the magnitude of role="math" localid="1660893498520" $\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}$role="math" localid="1660893523599" $.\stackrel{}{\overrightarrow{\text{b}}}$is zero.

(d) Calculate the magnitude of a→⋅b→/d

As the magnitude of $\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}$is zero, the magnitude of $\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}/\mathrm{d}$is also equal to zero.

(e) Calculate the direction of the vector resulting from a→×b→

$\overrightarrow{a}$and $\overrightarrow{b}$are perpendicular to each other and$\sin (90°)=1$ . Using the right-hand rule, we can conclude that the direction of $\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}$is along the $+\text{z}$direction.

(f) Calculate the direction of the vector resulting from b→×a→

From the right-hand rule, we can conclude that $\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{a}}$ is opposite to $\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}$.

Therefore,role="math" localid="1660893867310" $\overrightarrow{b}\times \overrightarrow{a}$ is along$-z$ axis.

(g) Calculate the magnitude of the vector resulting from a→×b→

$\overrightarrow{\mathrm{a}}$and $\overrightarrow{b}$are perpendicular to each other and$\sin (90°)=1$ . Using equation (ii), we can conclude that,

$\begin{array}{c}|\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}|=\mathrm{absin}(90°)\\ =\mathrm{ab}\end{array}$

Therefore, the magnitude of vector $\overrightarrow{a}\times \overrightarrow{b}$is$ab$

(h) Calculate the magnitude of the vector resulting from b→×a→

Since $\overrightarrow{a}$and$\overrightarrow{b}$ are perpendicular to each other and,$\sin (90°)=1$. Using equation (ii), we can conclude that,

$\begin{array}{c}|\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{a}}|=\mathrm{basin}(90°)\\ =\mathrm{ab}\end{array}$

Therefore, the magnitude of vector $\overrightarrow{b}\times \overrightarrow{a}$is$ab$

(i) Calculate the magnitude of the a→×b→d  , if d is positive

From part (g), the magnitude of vector$\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}$is $ab$.

Therefore, the magnitude of $\frac{\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}}{\mathrm{d}}$ is$\frac{\mathrm{ab}}{\mathrm{d}}$ .

(j) Calculate the direction of  a→×b→d if d is positive 

From part (e), the direction of$\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}$ is along +z direction. As d is a positive scalar quantity, it will not affect the direction of $\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}$.

Therefore, $\frac{\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}}{\mathrm{d}}$is along $+\mathrm{z}$direction.