Question

A vector \(\vec{A}\) has magnitude \(6\) units is added to vector \(\vec{B}\) which is along \(x\)-axis. The resultant of \(\vec{A}\) and \(\vec{B}\) is along \(y\)-axis. If the magnitude of the resultant of \(\vec{A}\) and \(\vec{B}\) is \(3\) times that of \(\vec{B}\) then the magnitude of \(\vec{B}\) is

• A. \(3\)
• B. \(\sqrt{3.6}\)
• C. \(6\)
• D. \(4\)

Hint:

The resultant lies along the y-axis while B lies along the x-axis, so the three vectors form a right-angled triangle with A as the hypotenuse.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Since \(\vec{R}=\vec{A}+\vec{B}\) with \(\vec{B}\) along the x-axis and \(\vec{R}\) along the y-axis, the vectors \(\vec{B}\) and \(\vec{R}\) are perpendicular.
So \(\vec{A}=\vec{R}-\vec{B}\) is the hypotenuse of a right-angled triangle.

Step 2: Write the Relation
\(A^{2}=R^{2}+B^{2}\)
Given \(A=6\) and \(R=3B\).

Step 3: Calculation
\(36=9B^{2}+B^{2}=10B^{2}\)
\(B^{2}=3.6\)
\(B=\sqrt{3.6}\)

Step 4: Compare the options
The value \(\sqrt{3.6}\) matches option (B).

Final Answer:
The magnitude of B is \(\sqrt{3.6}\). \[ \boxed{\text{(B) }\sqrt{3.6}} \]