The Question is from neet-physics motion-in-a-plane , and it was created by the author P.NARASIMHA

Question

If the magnitudes of scalar and vector products of two vectors and are 6 and $6\sqrt 3 $ respectively, then the angle  between two vectors is

P.NARASIMHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: C
## CONCEPT FRAMEWORK
	**Primary Concept**: Scalar and Vector products of vectors
	**Related Topics**: Dot product, Cross product, Trigonometry
	**Prerequisites**: Understanding of vector algebra, trigonometric functions
## SOLUTION PATHWAY
**Given Information**:
	* Magnitude of scalar product (dot product): $|\vec{a} \cdot \vec{b}| = 6$
	* Magnitude of vector product (cross product): $|\vec{a} 	imes \vec{b}| = 6\sqrt{3}$

**Method & Steps**:
	1.  Express the scalar and vector products in terms of magnitudes and the angle between the vectors.
		Reasoning: This allows us to relate the given magnitudes to the unknown angle.
		Formula/Rule:  $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos(	heta)$ and $|\vec{a} 	imes \vec{b}| = |\vec{a}||\vec{b}|\sin(	heta)$
	2.  Set up a ratio of the two equations.
		Working: Dividing the magnitude of the vector product by the scalar product: $\frac{|\vec{a} 	imes \vec{b}|}{|\vec{a} \cdot \vec{b}|} = \frac{|\vec{a}||\vec{b}|\sin(	heta)}{|\vec{a}||\vec{b}|\cos(	heta)} = \frac{\sin(	heta)}{\cos(	heta)} = 	an(	heta)$
		Key Point: The magnitudes $|\vec{a}|$ and $|\vec{b}|$ cancel out.
	3. Substitute the given values and solve for $	heta$.
		Result: $	an(	heta) = \frac{6\sqrt{3}}{6} = \sqrt{3} \Rightarrow 	heta = 	an^{-1}(\sqrt{3}) = 60^\circ$
		Verification: The angle whose tangent is $\sqrt{3}$ is 60 degrees.
 ## OPTION ANALYSIS
[A] :x: Incorrect angle.
[B] :x: Incorrect angle.
[C] ✓: Correct angle based on the ratio of vector and scalar products.
[D] :x: Incorrect angle.
 ## LEARNING AIDS
	**Common Mistakes**:
		1. [Error 1]: Incorrectly using sine or cosine for the dot and cross product.
		 [How to avoid]: Ensure you remember the definition of the scalar and vector product.
		2. [Error 2]: Forgetting to take the inverse tangent to find the angle
		 [How to avoid]: Make sure to find the angle by using tan-1
	**Quick Tips**:
		1. [Helpful shortcut]:  Recognizing the ratio $	an(	heta) = \frac{|\vec{a} 	imes \vec{b}|}{|\vec{a} \cdot \vec{b}|}$ helps to solve quickly.
		2. [Memory aid]: Remember that the dot product involves cosine and the cross product involves sine.

P.NARASIMHA · Answer

${15^0}$

P.NARASIMHA · Answer

${30^0}$

P.NARASIMHA · Answer

The correct Option is "${60^0}$"

P.NARASIMHA · Answer

${75^0}$

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