The Question is from neet-physics motion-in-a-plane parallelogram-law-of-vectors, and it was created by the author NAGA SAI

Question

If the sum of the two unit vectors is also a unit vector, then magnitude of their difference is

NAGA SAI · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
	**Primary Concept**: Vector Operations and Magnitude
	**Related Topics**: Unit Vectors, Dot Product, Geometric Interpretation of Vectors
	**Prerequisites**: Understanding of vector addition, subtraction, dot product, and magnitude of a vector.
## SOLUTION PATHWAY
**Given Information**:
	Let the two unit vectors be  $\hat{a}$ and $\hat{b}$.
	|$\hat{a}$| = 1 and |$\hat{b}$| = 1
	The sum of the unit vectors is also a unit vector, so |$\hat{a} + \hat{b}$| = 1
**Method & Steps**:
	1. Square the magnitude of the sum vector: |$\hat{a} + \hat{b}$|^2 = 1^2 = 1
		Reasoning: Squaring the magnitude allows us to use the dot product.
		Formula/Rule: |$\vec{v}$|^2 = $\vec{v} \cdot \vec{v}$
	2. Expand the square using the dot product: ($\hat{a} + \hat{b}$) $\cdot$ ($\hat{a} + \hat{b}$) = $\hat{a} \cdot \hat{a}$ + 2($\hat{a} \cdot \hat{b}$) + $\hat{b} \cdot \hat{b}$ = 1
		Working: Using the distributive property of the dot product.
		Key Point:  $\hat{a} \cdot \hat{a}$ = |$\hat{a}$|^2 = 1 and  $\hat{b} \cdot \hat{b}$ = |$\hat{b}$|^2 = 1.
	3. Simplify the equation: 1 + 2($\hat{a} \cdot \hat{b}$) + 1 = 1, thus 2($\hat{a} \cdot \hat{b}$) = -1  and  $\hat{a} \cdot \hat{b}$ = -1/2
		Reasoning: Solving for the dot product of the two unit vectors.
	4. Find the magnitude of the difference vector: |$\hat{a} - \hat{b}$|^2 = ($\hat{a} - \hat{b}$) $\cdot$ ($\hat{a} - \hat{b}$)
		Reasoning: Similar to step 1, we use the dot product.
	5. Expand using the dot product: |$\hat{a} - \hat{b}$|^2 = $\hat{a} \cdot \hat{a}$ - 2($\hat{a} \cdot \hat{b}$) + $\hat{b} \cdot \hat{b}$ = 1 - 2(-1/2) + 1 = 1 + 1 + 1 = 3
		Working: Using the distributive property and substituting the value of $\hat{a} \cdot \hat{b}$.
	6. Find the magnitude: |$\hat{a} - \hat{b}$| = $\sqrt{3}$
		Result: The magnitude of the difference is $\sqrt{3}$
		Verification: The result matches the correct answer.
 ## OPTION ANALYSIS
[A] :x: Incorrect. $\sqrt{2}$ is not the magnitude of the difference in this scenario.
[B] ✓: Correct. The magnitude of the difference is $\sqrt{3}$.
[C] :x: Incorrect. $\sqrt{4}$ = 2, which is not the magnitude of the difference.
[D] :x: Incorrect. $\sqrt{7}$ is not the magnitude of the difference in this scenario.
 ## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Forgetting to use the dot product correctly when squaring the magnitude of a vector sum or difference. [How to avoid]: Remember that |$\vec{a}$ + $\vec{b}$|^2 = $\vec{a} \cdot \vec{a}$ + 2($\vec{a} \cdot \vec{b}$) + $\vec{b} \cdot \vec{b}$ and |$\vec{a}$ - $\vec{b}$|^2 = $\vec{a} \cdot \vec{a}$ - 2($\vec{a} \cdot \vec{b}$) + $\vec{b} \cdot \vec{b}$
		2.[Error 2]: Incorrectly calculating the dot product when the angle between the vectors is not 0 or 90 degrees. [How to avoid]: Remember the dot product definition: $\vec{a} \cdot \vec{b}$ = |$\vec{a}$||$\vec{b}$|cos($	heta$).
	**Quick Tips**:
		1.[Helpful shortcut]:  The fact that the sum of two unit vectors is a unit vector implies the angle between them is 120 degrees. Use this to directly find the dot product and then the difference magnitude.
		2.[Memory aid]: Remember that the dot product of a vector with itself is the square of its magnitude.  $\vec{v} \cdot \vec{v}$ = |$\vec{v}$|^2

NAGA SAI · Answer

$\sqrt 2 $

NAGA SAI · Answer

The correct Option is "$\sqrt 3 $"

NAGA SAI · Answer

$\sqrt 4 $

NAGA SAI · Answer

$\sqrt 7 $

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