The Question is from neet-chemistry atomic-structure scalar-product-or-dot-product, and it was created by the author Mr. S M HUSSAIN

Question

Two vectors $\vec{A}$ and $\vec{B}$ have magnitudes A and B respectively. If $\vec{A} \cdot \vec{B} = \frac{AB}{\sqrt{2}}$, then the magnitude of $\vec{A} - \vec{B}$ is:

Mr. S M HUSSAIN · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
 **Primary Concept**: Vector dot product and magnitude of vector difference.
 **Related Topics**: Vector algebra, geometry
 **Prerequisites**: Understanding of dot product, vector magnitudes, and trigonometric functions.
## SOLUTION PATHWAY
**Given Information**:
  $\vec{A} \cdot \vec{B} = \frac{AB}{\sqrt{2}}$
  Magnitudes of $\vec{A}$ and $\vec{B}$ are A and B respectively.
**Method & Steps**:
 1. Recall the definition of the dot product: $\vec{A} \cdot \vec{B} = ABcos	heta$, where $	heta$ is the angle between $\vec{A}$ and $\vec{B}$.
  Reasoning: We can use the given dot product to find the angle between the vectors.
  Formula/Rule: $\vec{A} \cdot \vec{B} = ABcos	heta$
 2. Solve for $	heta$: $\frac{AB}{\sqrt{2}} = ABcos	heta \implies cos	heta = \frac{1}{\sqrt{2}} \implies 	heta = \frac{\pi}{4}$ radians or 45 degrees.
  Working: The angle between the two vectors is 45 degrees.
  Key Point: This is a crucial step to determine the magnitude of the difference vector.
 3. Find the magnitude of $\vec{A} - \vec{B}$ using the law of cosines:
       $|\vec{A} - \vec{B}|^2 = A^2 + B^2 - 2ABcos(180-	heta) = A^2 + B^2 - 2ABcos(135) = A^2 + B^2 - 2AB(-\frac{1}{\sqrt{2}}) = A^2 + B^2 + \sqrt{2}AB$
       Therefore, $|\vec{A} - \vec{B}| = \sqrt{A^2 + B^2 + \sqrt{2}AB}$
  Result: The magnitude of $\vec{A} - \vec{B}$ is $\sqrt{A^2 + B^2 + \sqrt{2}AB}$. However, there is a mistake in the calculation.  The correct calculation is shown below.
       $|\vec{A} - \vec{B}|^2 = A^2 + B^2 - 2ABcos	heta = A^2 + B^2 - 2AB(\frac{1}{\sqrt{2}}) = A^2 + B^2 - \sqrt{2}AB$
       Therefore, $|\vec{A} - \vec{B}| = \sqrt{A^2 + B^2 - \sqrt{2}AB}$
  Verification: The result should be consistent with the geometric interpretation of vector subtraction.
 ## OPTION ANALYSIS
A ✓: $\sqrt{A^2 + B^2 - \sqrt{2}AB}$: This is the correct answer, derived from the law of cosines.
B :x: $\sqrt{A^2 + B^2 + \sqrt{2}AB}$: This would be correct if the angle between the vectors were 135 degrees.
C :x: $\sqrt{A^2 - B^2 + \sqrt{2}AB}$: This formula is incorrect.
D :x: $\sqrt{2AB}$: This formula is incorrect and doesn't account for the magnitudes A and B.
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Incorrectly applying the dot product formula or the law of cosines: Carefully review the formulas and their applications in vector algebra.
  2.Making errors in trigonometric calculations: Double-check trigonometric values and ensure consistent units (radians or degrees).
 **Quick Tips**:
  1.Visualize the vectors and their subtraction geometrically. This can help in understanding the application of the law of cosines.
  2.Remember the relationship between the dot product and the angle between two vectors.

Mr. S M HUSSAIN · Answer

The correct Option is "$\sqrt{A^2 + B^2 - \sqrt{2}AB}$"

Mr. S M HUSSAIN · Answer

$\sqrt{A^2 + B^2 + \sqrt{2}AB}$

Mr. S M HUSSAIN · Answer

$\sqrt{A^2 - B^2 + \sqrt{2}AB}$

Mr. S M HUSSAIN · Answer

$\sqrt{2AB}$

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