The Question is from neet-chemistry , and it was created by the author THIRUPATHI REDDY

Question

If ${\rm{\vec A}}\,and\,\overrightarrow B $ denote the sides of a parallelogram and its area is $\frac{1}{2}AB$(A and B are the magnitude of ${\rm{\vec A}}\,and\,\overrightarrow B $ respectively), the angle between ${\rm{\vec A}}\,and\,\overrightarrow B $ is

THIRUPATHI REDDY · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
	**Primary Concept**: Area of a parallelogram using vectors
	**Related Topics**: Vector cross product, Sine function, Trigonometry
	**Prerequisites**: Definition of cross product, Area of parallelogram using vectors, basic trigonometry
## SOLUTION PATHWAY
**Given Information**:
	Area of parallelogram = $\frac{1}{2}AB$
	Sides of parallelogram are given by vectors $\vec{A}$ and $\vec{B}$
**Method & Steps**:
	1. The area of a parallelogram formed by two vectors $\vec{A}$ and $\vec{B}$ is given by the magnitude of their cross product: Area = $|\vec{A} 	imes \vec{B}|$
		Reasoning: This is a standard formula for the area of a parallelogram using vectors.
		Formula/Rule: Area = $|\vec{A} 	imes \vec{B}|$
	2. The magnitude of the cross product can be written as $|\vec{A} 	imes \vec{B}| = AB \sin{	heta}$, where $	heta$ is the angle between $\vec{A}$ and $\vec{B}$.
		Working: Substituting the given area $\frac{1}{2}AB$, we have $\frac{1}{2}AB = AB \sin{	heta}$. 
		Key Point: This relates the area to the angle between the vectors.
	3.  Solving for $\sin{	heta}$, we get $\sin{	heta} = \frac{1}{2}$. Therefore, $	heta = 30^0$.
		Result: The angle between the vectors is 30 degrees.
		Verification:  The $\sin{30^0} = \frac{1}{2}$, which is consistent with the result.
 ## OPTION ANALYSIS
[A] ✓: The angle between the vectors is 30 degrees, so this is the correct option.
[B] :x: If the angle was 60 degrees, the area would be  $AB \sin{60^0} = \frac{\sqrt{3}}{2}AB$ not $\frac{1}{2}AB$.
[C] :x: If the angle was 45 degrees, the area would be  $AB \sin{45^0} = \frac{1}{\sqrt{2}}AB$ not $\frac{1}{2}AB$.
[D] :x: If the angle was 120 degrees, the area would be  $AB \sin{120^0} = \frac{\sqrt{3}}{2}AB$ not $\frac{1}{2}AB$.
 ## LEARNING AIDS
	**Common Mistakes**:
		1. Using cosine instead of sine. Area formula uses sine.
		2. Forgetting to take the magnitude of the cross product.
	**Quick Tips**:
		1.  Remember the area of a parallelogram is given by magnitude of the cross product of the vectors.
		2. Recall the sine values for common angles like 30, 45, 60, and 90 degrees.

THIRUPATHI REDDY · Answer

The correct Option is "${30^0}$"

THIRUPATHI REDDY · Answer

${60^0}$

THIRUPATHI REDDY · Answer

${45^0}$

THIRUPATHI REDDY · Answer

${120^0}$

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