Chapter MA, Problem 1.1MA

(i)

To determine

To obtain: The value of $\theta$ from the inequality $\left|A+B\right|<\left|A-B\right|$.

Answer to Problem 1.1MA

The angle between the two vectors $A\text{ and }B$  is $\theta >90°$.

Explanation of Solution

The angle between the two vectors $A\text{ and }B$ is $\theta$.

Suppose the inequality $\left|A+B\right|<\left|A-B\right|$ exists.

This implies the magnitude of $A-B$ is greater than the magnitude of $A+B$.

It further implies that the vector $B$ of higher magnitude when subtracted from the vector $A$ of lower magnitude, the resultant vector with greater magnitude is obtained.

However, it’s not just the magnitude but the signs of the components of the vectors.

When the vector $B$ of opposite sign to that of the vector $A$, is subtracted from $A$; it adds to the value of $A$, hence magnitude of the resultant vector increases

When the vector $B$ of opposite sign to that of the vector $A$, is added to $A$; it reduces the value of $A$, hence the magnitude of the resultant vector decreases.

Due to change in signs of the components of two vectors, the angle between the two vectors is always greater than $90°$.

Example: $A=3a_x\text{+2}{a}_y\text{ and }B=-a_x-3a_y$

Both the vectors lie in different quadrant, hence the angle between them is greater than $90°$.

$\begin{array}{c}A+B=3a_x\text{+2}{a}_y-a_x-3a_y\\ =2a_x-a_y\\ \left|A+B\right|=\sqrt{2^2+{\left(-1\right)}^2}\\ =\sqrt{4+1}=\sqrt{5}\end{array}$

$\begin{array}{c}A-B=3a_x\text{+2}{a}_y-\left(-a_x-3a_y\right)\\ =4a_x+5a_y\\ \left|A-B\right|=\sqrt{4^2+5^2}\\ =\sqrt{16+25}=\sqrt{41}\end{array}$

Thus, the inequality $\left|A+B\right|<\left|A-B\right|$ is verified.

Therefore, when $\left|A+B\right|<\left|A-B\right|$, the angle between the two vectors $A\text{ and }B$  is $\theta >90°$.

(ii)

To determine

To obtain: The value of $\theta$ from the inequality $\left|A+B\right|=\left|A-B\right|$.

Answer to Problem 1.1MA

The angle between the two vectors $A\text{ and }B$  is $\theta =90°$.

Explanation of Solution

The angle between the two vectors $A\text{ and }B$ is $\theta$.

Suppose the inequality $\left|A+B\right|=\left|A-B\right|$ exists.

Example: $A=a_x\text{ and }B=a_y$

Both the vectors are on the axis, hence the angle between them is $90°$.

$\begin{array}{c}A+B=a_x+a_y\\ \left|A+B\right|=\sqrt{1^2+1^2}\\ =\sqrt{1+1}=\sqrt{2}\end{array}$

$\begin{array}{c}A-B=a_x-a_y\\ \left|A-B\right|=\sqrt{1^2+{\left(-1\right)}^2}\\ =\sqrt{1+1}=\sqrt{2}\end{array}$

Thus, the inequality $\left|A+B\right|=\left|A-B\right|$ is verified.

Therefore, when $\left|A+B\right|=\left|A-B\right|$, the angle between the two vectors $A\text{ and }B$  is $90°$.

(i)

To determine

To obtain: The value of $\theta$ from the inequality $\left|A+B\right|<\left|A-B\right|$.

Answer to Problem 1.1MA

The angle between the two vectors $A\text{ and }B$  is $\theta <90°$.

Explanation of Solution

The angle between the two vectors $A\text{ and }B$ is $\theta$.

Suppose the inequality $\left|A+B\right|>\left|A-B\right|$ exists.

This implies the magnitude of $A+B$ is greater than the magnitude of $A-B$.

It further implies that the vector $B$ when subtracted from the vector $A$, the resultant vector with greater magnitude is obtained.

However, it’s not just the magnitude but the signs of the components of the vectors.

When the vector $B$ of same sign to that of the vector $A$, is added to $A$; it adds to the value of $A$, hence magnitude of the resultant vector increases

When the vector $B$ of same sign to that of the vector $A$, is subtracted from $A$; it reduces the value of $A$, hence the magnitude of the resultant vector decreases.

As the sign of the components of two vectors are same, the angle between the two vectors is always less than $90°$.

Example: $A=-3a_x\text{+2}{a}_y\text{ and }B=-a_x+3a_y$

Both the vectors lie in same quadrant, hence the angle between them is less than $90°$.

$\begin{array}{c}A+B=-3a_x\text{+2}{a}_y-a_x+3a_y\\ =-4a_x+5a_y\\ \left|A+B\right|=\sqrt{{\left(-4\right)}^2+5^2}\\ =\sqrt{16+25}=\sqrt{41}\end{array}$

$\begin{array}{c}A-B=-3a_x\text{+2}{a}_y+a_x-3a_y\\ =-2a_x-a_y\\ \left|A-B\right|=\sqrt{{\left(-2\right)}^2+{\left(-1\right)}^2}\\ =\sqrt{4+1}=\sqrt{5}\end{array}$

Thus, the inequality $\left|A+B\right|>\left|A-B\right|$ is verified.

Therefore, when $\left|A+B\right|>\left|A-B\right|$, the angle between the two vectors $A\text{ and }B$  is $\theta <90°$.