The sum of the magnitudes of two vectors vec{ | vedclass

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(A) Let the magnitudes of the vectors be $A$ and $B$. Given $A + B = 8$,so $B = 8 - A$. Let the resultant vector be $\vec{R} = \vec{A} + \vec{B}$,with

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$3, 5$

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(A) Let the magnitudes of the vectors be $A$ and $B$. Given $A + B = 8$,so $B = 8 - A$. Let the resultant vector be $\vec{R} = \vec{A} + \vec{B}$,with magnitude $R = 4$. Since $\vec{R}$ is perpendicular to $\vec{A}$,the angle between $\vec{A}$ and $\vec{R}$ is $90^{\circ}$. Using the triangle law of vector addition,we have the relation $B^2 = A^2 + R^2$. Substituting $B = 8 - A$ and $R = 4$: $(8 - A)^2 = A^2 + 4^2$ $64 - 16A + A^2 = A^2 + 16$ $64 - 16 = 16A$ $48 = 16A$ $A = 3$. Then $B = 8 - 3 = 5$. Thus,the magnitudes are $3$ and $5$.

The sum of the magnitudes of two vectors $\vec{A}$ and $\vec{B}$ is $8$,and the magnitude of their resultant is $4$. If the resultant vector is perpendicular to one of the vectors,then the magnitudes of the two vectors $\vec{A}$ and $\vec{B}$ are:

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