The Question is from neet-physics motion-in-a-plane vector-product-or-cross-product, and it was created by the author VENKATESWARARAO

Question

The component of $\overrightarrow A $ along $\overrightarrow B $  is 2 times that of the component of $\overrightarrow B $  along $\overrightarrow A $. Then A : B is

VENKATESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
	**Primary Concept**: Vector Projection
	**Related Topics**: Dot Product, Vector Magnitude
	**Prerequisites**: Understanding of vector projection, dot product, and magnitude of a vector.
## SOLUTION PATHWAY
**Given Information**:
	* The component of $\overrightarrow{A}$ along $\overrightarrow{B}$ is 2 times the component of $\overrightarrow{B}$ along $\overrightarrow{A}$.
**Method & Steps**:
	1.  Express the component of $\overrightarrow{A}$ along $\overrightarrow{B}$:
		Reasoning: The component of $\overrightarrow{A}$ along $\overrightarrow{B}$ is given by $\frac{\overrightarrow{A} \cdot \overrightarrow{B}}{|\overrightarrow{B}|}$.
		Formula/Rule:  $A_B = \frac{\overrightarrow{A} \cdot \overrightarrow{B}}{|\overrightarrow{B}|}$
	2.  Express the component of $\overrightarrow{B}$ along $\overrightarrow{A}$:
		Reasoning: The component of $\overrightarrow{B}$ along $\overrightarrow{A}$ is given by $\frac{\overrightarrow{B} \cdot \overrightarrow{A}}{|\overrightarrow{A}|}$.
		Formula/Rule:  $B_A = \frac{\overrightarrow{B} \cdot \overrightarrow{A}}{|\overrightarrow{A}|}$
	3.  Set up the equation according to the given condition:
		Working: We are given that $A_B = 2 B_A$, so $\frac{\overrightarrow{A} \cdot \overrightarrow{B}}{|\overrightarrow{B}|} = 2 \frac{\overrightarrow{B} \cdot \overrightarrow{A}}{|\overrightarrow{A}|}$. Since $\overrightarrow{A} \cdot \overrightarrow{B} = \overrightarrow{B} \cdot \overrightarrow{A}$, we have $\frac{\overrightarrow{A} \cdot \overrightarrow{B}}{|\overrightarrow{B}|} = 2 \frac{\overrightarrow{A} \cdot \overrightarrow{B}}{|\overrightarrow{A}|}$.
		Key Point:  The dot product is commutative, so $\overrightarrow{A} \cdot \overrightarrow{B} = \overrightarrow{B} \cdot \overrightarrow{A}$.
	4.  Simplify the equation and solve for the ratio of magnitudes:
		Working: Dividing both sides by $\overrightarrow{A} \cdot \overrightarrow{B}$ (assuming it's non-zero), we get $\frac{1}{|\overrightarrow{B}|} = \frac{2}{|\overrightarrow{A}|}$. Rearranging, we find $|\overrightarrow{A}| = 2|\overrightarrow{B}|$. Thus, $A : B = 2:1$.
		Result:  $A:B = 2:1$
		Verification: The ratio found matches the correct answer.
 ## OPTION ANALYSIS
[A] :x:  Incorrect ratio.  It does not satisfy the given condition.
[B] ✓: Correct ratio.  It satisfies the given condition, $A_B = 2B_A$.
[C] :x:  Incorrect ratio.  It does not satisfy the given condition.
[D] :x: Incorrect ratio. It does not satisfy the given condition.
 ## LEARNING AIDS
	**Common Mistakes**:
		1. **Confusing component with projection**: The component is a scalar value, while the projection is a vector. Be careful not to confuse them.
		2. **Incorrectly setting up the equation**: Ensure you correctly relate the component of A along B to the component of B along A as specified in the problem.
	**Quick Tips**:
		1.  Remember the formula for the component of one vector along another. The component of $\overrightarrow{A}$ along $\overrightarrow{B}$ is $\frac{\overrightarrow{A} \cdot \overrightarrow{B}}{|\overrightarrow{B}|}$.
		2.  Use the commutative property of the dot product. This simplifies the equation and makes it easier to solve.

VENKATESWARARAO · Answer

$1{\rm{ }}:{\rm{ }}4$

VENKATESWARARAO · Answer

The correct Option is "$2{\rm{ }}:{\rm{ }}1$"

VENKATESWARARAO · Answer

$\sqrt 3 :2$

VENKATESWARARAO · Answer

$6{\rm{ }}:{\rm{ }}1$

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