The Question is from neet-chemistry atomic-structure different-types-of-vectors, 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.

NEET Chemistry Structure Of Atom Different Types Of Vectors MCQs

Select Topic

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

NEET Subjects

Our Bots

Company