Question

The resultant of two vectors $\vec{A}$ and $\vec{B}$ is $\vec{C}$. If the magnitude of $\vec{B}$ is doubled, the new resultant vector becomes perpendicular to $\vec{A}$, then the magnitude of $\vec{C}$ is

• A. 4 B
• B. 3 B
• C. B
• D. 2 B

Hint:

Use the dot product property $A \cdot B = 0$ whenever you see "perpendicular" in vector problems.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We need to find the magnitude of the initial resultant vector $\vec{C}$.
The problem states that when vector $\vec{B}$ is doubled to $2\vec{B}$, the new resultant vector is perpendicular to vector $\vec{A}$.
Two vectors are perpendicular if their dot product is zero.
Step 2: Key Formula or Approach:
The resultant vector is $\vec{C} = \vec{A} + \vec{B}$.
The magnitude squared is $C^2 = A^2 + B^2 + 2\vec{A} \cdot \vec{B}$.
The condition for perpendicularity between $\vec{X}$ and $\vec{Y}$ is $\vec{X} \cdot \vec{Y} = 0$.
Step 3: Detailed Explanation:
The new resultant vector when $\vec{B}$ is doubled is $\vec{R}' = \vec{A} + 2\vec{B}$.
Since $\vec{R}'$ is perpendicular to $\vec{A}$, their dot product is zero: \[ (\vec{A} + 2\vec{B}) \cdot \vec{A} = 0 \] Expanding the dot product: \[ \vec{A} \cdot \vec{A} + 2(\vec{B} \cdot \vec{A}) = 0 \] Since $\vec{A} \cdot \vec{A} = A^2$, we have: \[ A^2 + 2\vec{A} \cdot \vec{B} = 0 \] Rearranging to find the dot product term: \[ 2\vec{A} \cdot \vec{B} = -A^2 \] Now, let's find the magnitude of the original resultant vector $\vec{C}$: \[ C^2 = |\vec{A} + \vec{B}|^2 = A^2 + B^2 + 2\vec{A} \cdot \vec{B} \] Substitute the expression we derived for $2\vec{A} \cdot \vec{B}$: \[ C^2 = A^2 + B^2 + (-A^2) \] \[ C^2 = B^2 \] Taking the square root: \[ C = B \] Step 4: Final Answer:
The magnitude of $\vec{C}$ is B.