Vectors vec{A} and vec{B} make angles of 20^c | vedclass

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(B) The angle between vector $\vec{A}$ and vector $\vec{B}$ is $	heta = 110^\circ - 20^\circ = 90^\circ$.The magnitude of the resultant vector $R$ is

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(B) The angle between vector $\vec{A}$ and vector $\vec{B}$ is $	heta = 110^\circ - 20^\circ = 90^\circ$.The magnitude of the resultant vector $R$ is given by the formula:$R = \sqrt{A^2 + B^2 + 2AB \cos 	heta}$Substituting the given values $A = 5 \, m$,$B = 12 \, m$,and $	heta = 90^\circ$:$R = \sqrt{5^2 + 12^2 + 2(5)(12) \cos 90^\circ}$Since $\cos 90^\circ = 0$:$R = \sqrt{25 + 144 + 0} = \sqrt{169} = 13 \, m$.Thus,the magnitude of the resultant vector is $13 \, m$.

Vectors $\vec{A}$ and $\vec{B}$ make angles of $20^\circ$ and $110^\circ$ with the $X$-axis,respectively. The magnitudes of these vectors are $5 \, m$ and $12 \, m$,respectively. The magnitude of the resultant vector of these two vectors is ....... $m$.

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