91Ó°ÊÓ

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

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:

$|\stackrel{→}{\mathrm{a}}â‹\dots \stackrel{→}{\mathrm{b}}|=\mathrm{²¹²ú³¦´Ç²õθ}$…… (1)

Vector product:

$|\stackrel{→}{\mathrm{a}}×\stackrel{→}{\mathrm{b}}|=\mathrm{ab}\mathrm{²õ¾\pm ²Ôθ}$……. (ii)

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

It can be written as,

$\stackrel{→}{\mathrm{v}}=\frac{\mathrm{b}}{\mathrm{d}}$.

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

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

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

It can be written as,

$\stackrel{→}{\mathrm{v}}=−\frac{\mathrm{b}}{\mathrm{d}}$

Multiplying a vector by negative scalar will reverse its direction.

As d is negative, the vector $\stackrel{→}{b}$/d is in $(−\text{y})$direction

The vector$\stackrel{→}{\mathrm{a}}$and $\stackrel{→}{\mathrm{b}}$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" $\stackrel{→}{\mathrm{a}}â‹\dots \stackrel{→}{\mathrm{b}}$role="math" localid="1660893523599" $.\stackrel{}{\stackrel{→}{\text{b}}}$is zero.

As the magnitude of $\stackrel{→}{\mathrm{a}}â‹\dots \stackrel{→}{\mathrm{b}}$is zero, the magnitude of $\stackrel{→}{\mathrm{a}}â‹\dots \stackrel{→}{\mathrm{b}}/\mathrm{d}$is also equal to zero.

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

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

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

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

$\begin{array}{c}|\stackrel{→}{\mathrm{a}}×\stackrel{→}{\mathrm{b}}|=\mathrm{absin}(90°)\\ =\mathrm{ab}\end{array}$

Therefore, the magnitude of vector $\stackrel{→}{a}×\stackrel{→}{b}$is$ab$

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

$\begin{array}{c}|\stackrel{→}{\mathrm{b}}×\stackrel{→}{\mathrm{a}}|=\mathrm{basin}(90°)\\ =\mathrm{ab}\end{array}$

Therefore, the magnitude of vector $\stackrel{→}{b}×\stackrel{→}{a}$is$ab$

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

Therefore, the magnitude of $\frac{\stackrel{→}{\mathrm{a}}×\stackrel{→}{\mathrm{b}}}{\mathrm{d}}$ is$\frac{\mathrm{ab}}{\mathrm{d}}$ .

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

Therefore, $\frac{\stackrel{→}{\mathrm{a}}×\stackrel{→}{\mathrm{b}}}{\mathrm{d}}$is along $+\mathrm{z}$direction.