A vector vec{A} having magnitude 6 units is a | vedclass

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(C) Let $\vec{B} = B\hat{i}$ and $\vec{A} = A_x\hat{i} + A_y\hat{j}$. Given $|\vec{A}| = 6$,so $A_x^2 + A_y^2 = 36$. Since the resultant $\vec{R} = \v

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$\sqrt{3.6}$

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(C) Let $\vec{B} = B\hat{i}$ and $\vec{A} = A_x\hat{i} + A_y\hat{j}$. Given $|\vec{A}| = 6$,so $A_x^2 + A_y^2 = 36$. Since the resultant $\vec{R} = \vec{A} + \vec{B} = (A_x + B)\hat{i} + A_y\hat{j}$ is along the $y$-axis,the $x$-component must be zero: $A_x + B = 0$,so $A_x = -B$. Substituting $A_x = -B$ into the magnitude equation: $(-B)^2 + A_y^2 = 36$,which gives $A_y^2 = 36 - B^2$. The resultant vector is $\vec{R} = A_y\hat{j}$,so its magnitude is $|\vec{R}| = |A_y| = \sqrt{36 - B^2}$. Given $|\vec{R}| = 3B$,we have $\sqrt{36 - B^2} = 3B$. Squaring both sides: $36 - B^2 = 9B^2$,which simplifies to $10B^2 = 36$. Thus,$B^2 = 3.6$,so $B = \sqrt{3.6}$ units.

$A$ vector $\vec{A}$ having magnitude $6$ units is added to vector $\vec{B}$,which is along the $x$-axis. The resultant of $\vec{A}$ and $\vec{B}$ is along the $y$-axis. If the magnitude of the resultant of $\vec{A}$ and $\vec{B}$ is three times that of $\vec{B}$,then the magnitude of $\vec{B}$ is:

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