Question

If the component of the vector $\vec{A}$ along the vector $\vec{B}$ is twice the component of $\vec{B}$ along $\vec{A}$, then the ratio of magnitudes of vectors $\vec{A}$ and $\vec{B}$ is

• A. $1:2$
• B. $3:2$
• C. $2:1$
• D. $3:1$

Hint:

The projection of $\vec{P}$ on $\vec{Q}$ is $\vec{P} \cdot \hat{Q}$. This depends inversely on the magnitude of the vector being projected *on*.

Answer 1

The Correct Option is C

Step 1: Write Formulae for Components:
Component of $\vec{A}$ along $\vec{B}$ = $A \cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|}$. Component of $\vec{B}$ along $\vec{A}$ = $B \cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}|}$.
Step 2: Apply Condition:
Given: $(\text{Comp of } \vec{A} \text{ on } \vec{B}) = 2 \times (\text{Comp of } \vec{B} \text{ on } \vec{A})$. \[ \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|} = 2 \left( \frac{\vec{A} \cdot \vec{B}}{|\vec{A}|} \right) \]
Step 3: Solve for Ratio:
Assuming $\vec{A} \cdot \vec{B} \neq 0$, cancel the dot product term: \[ \frac{1}{|\vec{B}|} = \frac{2}{|\vec{A}|} \] \[ |\vec{A}| = 2 |\vec{B}| \] \[ \frac{|\vec{A}|}{|\vec{B}|} = \frac{2}{1} \]