Question

Which of the following expressions give(s) unit magnitude?

• A. $\frac{\overrightarrow{a}+\overrightarrow{b}}{|\overrightarrow{a}+\overrightarrow{b}|}$
• B. $|\hat{a}+\hat{b}|$ when angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is $120^{\circ}$
• C. $\frac{\hat{a}-\hat{b}}{|\hat{a}-\hat{b}|}$
• D. $|\hat{a}-\hat{b}|$ when angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is $60^{\circ}$

Answer 1

Answer: (A), (B), (C), (D)

(A), (B), (C), (D)

Answer 2

Let's analyze each option:

(A) $\frac{\overrightarrow{a}+\overrightarrow{b}}{|\overrightarrow{a}+\overrightarrow{b}|}$

This expression represents a vector. Let $\overrightarrow{v} = \overrightarrow{a}+\overrightarrow{b}$. The expression is $\frac{\overrightarrow{v}}{|\overrightarrow{v}|}$. This is the definition of a unit vector in the direction of $\overrightarrow{v}$, provided $\overrightarrow{v} \neq \overrightarrow{0}$. The magnitude of a unit vector is always 1. So, if $\overrightarrow{a}+\overrightarrow{b} \neq \overrightarrow{0}$, the magnitude of this expression is 1. If $\overrightarrow{a}+\overrightarrow{b} = \overrightarrow{0}$, the expression is undefined. Assuming the expression is well-defined, it gives a unit magnitude.

(B) $|\hat{a}+\hat{b}|$ when angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is $120^{\circ}$.

Here, $\hat{a}$ and $\hat{b}$ are unit vectors, so $|\hat{a}| = 1$ and $|\hat{b}| = 1$. The angle between $\hat{a}$ and $\hat{b}$ is given as $\theta = 120^{\circ}$.

The magnitude of the sum of two vectors is given by $|\vec{u}+\vec{v}| = \sqrt{|\vec{u}|^2 + |\vec{v}|^2 + 2|\vec{u}||\vec{v}|\cos\theta}$.

$|\hat{a}+\hat{b}| = \sqrt{|\hat{a}|^2 + |\hat{b}|^2 + 2|\hat{a}||\hat{b}|\cos(120^{\circ})}$

$|\hat{a}+\hat{b}| = \sqrt{1^2 + 1^2 + 2(1)(1)\cos(120^{\circ})}$

$|\hat{a}+\hat{b}| = \sqrt{1 + 1 + 2(-\frac{1}{2})}$

$|\hat{a}+\hat{b}| = \sqrt{2 - 1} = \sqrt{1} = 1$.

The expression gives the scalar value 1, whose magnitude is 1.

(C) $\frac{\hat{a}-\hat{b}}{|\hat{a}-\hat{b}|}$

This expression represents a vector. Let $\overrightarrow{w} = \hat{a}-\hat{b}$. The expression is $\frac{\overrightarrow{w}}{|\overrightarrow{w}|}$. This is the definition of a unit vector in the direction of $\overrightarrow{w}$, provided $\overrightarrow{w} \neq \overrightarrow{0}$. The magnitude of a unit vector is always 1. So, if $\hat{a}-\hat{b} \neq \overrightarrow{0}$ (i.e., $\hat{a} \neq \hat{b}$), the magnitude of this expression is 1. If $\hat{a} = \hat{b}$, the expression is undefined. Assuming the expression is well-defined, it gives a unit magnitude.

(D) $|\hat{a}-\hat{b}|$ when angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is $60^{\circ}$.

Here, $\hat{a}$ and $\hat{b}$ are unit vectors, so $|\hat{a}| = 1$ and $|\hat{b}| = 1$. The angle between $\hat{a}$ and $\hat{b}$ is given as $\theta = 60^{\circ}$.

The magnitude of the difference of two vectors is given by $|\vec{u}-\vec{v}| = \sqrt{|\vec{u}|^2 + |\vec{v}|^2 - 2|\vec{u}||\vec{v}|\cos\theta}$.

$|\hat{a}-\hat{b}| = \sqrt{|\hat{a}|^2 + |\hat{b}|^2 - 2|\hat{a}||\hat{b}|\cos(60^{\circ})}$

$|\hat{a}-\hat{b}| = \sqrt{1^2 + 1^2 - 2(1)(1)\cos(60^{\circ})}$

$|\hat{a}-\hat{b}| = \sqrt{1 + 1 - 2(\frac{1}{2})}$

$|\hat{a}-\hat{b}| = \sqrt{2 - 1} = \sqrt{1} = 1$.

The expression gives the scalar value 1, whose magnitude is 1.

All four expressions, when well-defined or under the given conditions, result in a quantity with a magnitude of 1. The question asks which expressions give unit magnitude, implying potentially multiple correct options.