Physics - Vectors Question with Solution | TestHub

Let $\vec{B}=B\hat{i}$, where $B$ is the magnitude of $\vec{B}$.

The resultant vector is $\vec{R}=\vec{A}+\vec{B}$.

Since $\vec{R}$ is along the y-axis and its magnitude is $3$ times that of $\vec{B}$, we can write $\vec{R}=3B\hat{j}$.

From the vector addition, we obtain $\vec{A}=\vec{R}-\vec{B}$.

Substituting the vectors gives $\vec{A}=3B\hat{j}-B\hat{i}$.

The magnitude of $\vec{A}$ is $|\vec{A}|=\sqrt{(-B{)}^2+(3B{)}^2}=\sqrt{B^2+9B^2}=B\sqrt{10}$.

Given that the magnitude of $\vec{A}$ is $6$, we equate them:

$B\sqrt{10}=6$

Squaring both sides gives:

$10B^2=36$

$B^2=3.6$

$B=\sqrt{3.6}$