The Question is from neet-chemistry , and it was created by the author P SRAVAN KUMAR

Question

Two vectors $\vec{A}$ and $\vec{B}$ have equal magnitudes. Under what condition will their difference have the same magnitude as either of the vectors?

P SRAVAN KUMAR · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
	**Primary Concept**: Vector Subtraction and Magnitude
	**Related Topics**: Vector Addition, Dot Product, Law of Cosines
	**Prerequisites**: Understanding of vectors, magnitude, and angle between vectors.
## SOLUTION PATHWAY
**Given Information**:
	$\|\vec{A}\| = \|\vec{B}\| = A$
	$\|\vec{A} - \vec{B}\| = A$
**Method & Steps**:
	1. **Express the magnitude of the difference using the Law of Cosines**:
		Reasoning: The magnitude of the vector difference $\vec{A} - \vec{B}$ can be found using the Law of Cosines, treating the vectors as sides of a triangle.
		Formula/Rule: $\|\vec{A} - \vec{B}\|^2 = \|\vec{A}\|^2 + \|\vec{B}\|^2 - 2\|\vec{A}\|\|\vec{B}\|\cos(	heta)$, where $	heta$ is the angle between $\vec{A}$ and $\vec{B}$.
	2. **Substitute the given information and solve for** $	heta$:
		Working: $A^2 = A^2 + A^2 - 2A^2\cos(	heta)$
		Key Point: Simplify the equation to isolate $\cos(	heta)$.
	3. **Solve for** $	heta$:
		Result: $A^2 = 2A^2 - 2A^2\cos(	heta) \implies 2A^2\cos(	heta) = A^2 \implies \cos(	heta) = \frac{1}{2} \implies 	heta = 60^\circ$
		Verification: If $	heta = 60^\circ$, then $\|\vec{A} - \vec{B}\| = \sqrt{A^2 + A^2 - 2A^2\cos(60^\circ)} = \sqrt{2A^2 - A^2} = A$.
## OPTION ANALYSIS
[A] ✓: The angle between the two vectors is 60 degrees. This is the correct answer, as derived above.
[B] :x: The angle between the two vectors is 90 degrees. If $	heta = 90^\circ$, then $\|\vec{A} - \vec{B}\| = \sqrt{A^2 + A^2} = \sqrt{2}A$, which is not equal to $A$.
[C] :x: The angle between the two vectors is 120 degrees. If $	heta = 120^\circ$, then $\|\vec{A} - \vec{B}\| = \sqrt{A^2 + A^2 - 2A^2\cos(120^\circ)} = \sqrt{3}A$, which is not equal to $A$.
[D] :x: The angle between the two vectors is 45 degrees. If $	heta = 45^\circ$, then $\|\vec{A} - \vec{B}\| = \sqrt{A^2 + A^2 - 2A^2\cos(45^\circ)} = \sqrt{2A^2 - \sqrt{2}A^2} = A\sqrt{2-\sqrt{2}}$, which is not equal to $A$.
## LEARNING AIDS
	**Common Mistakes**:
		1. Incorrectly applying the Law of Cosines: Ensure the correct signs and terms are used in the formula.
		2. Confusing vector subtraction with addition: Remember that vector subtraction involves changing the direction of the subtracted vector.
	**Quick Tips**:
		1. Visualize the vectors as sides of a triangle to understand the relationship between their magnitudes and the angle between them.
		2. Remember the special case where $	heta = 60^\circ$ results in an equilateral triangle formed by $\vec{A}$, $\vec{B}$, and $\vec{A} - \vec{B}$.

P SRAVAN KUMAR · Answer

The correct Option is "The angle between the two vectors is 60 degrees."

P SRAVAN KUMAR · Answer

The angle between the two vectors is 90 degrees.

P SRAVAN KUMAR · Answer

The angle between the two vectors is 120 degrees.

P SRAVAN KUMAR · Answer

The angle between the two vectors is 45 degrees.

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