Two forces vec F_1 = 5hat i + 10hat j - 20hat | vedclass

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(B) The angle $	heta$ between two vectors is found using the dot product formula: $\cos 	heta = \frac{\vec F_1 \cdot \vec F_2}{|\vec F_1| |\vec F_2|

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(B) The angle $	heta$ between two vectors is found using the dot product formula: $\cos 	heta = \frac{\vec F_1 \cdot \vec F_2}{|\vec F_1| |\vec F_2|}$.First,calculate the dot product: $\vec F_1 \cdot \vec F_2 = (5 	imes 10) + (10 	imes -5) + (-20 	imes -15) = 50 - 50 + 300 = 300$.Next,calculate the magnitudes: $|\vec F_1| = \sqrt{5^2 + 10^2 + (-20)^2} = \sqrt{25 + 100 + 400} = \sqrt{525} \approx 22.91$.$|\vec F_2| = \sqrt{10^2 + (-5)^2 + (-15)^2} = \sqrt{100 + 25 + 225} = \sqrt{350} \approx 18.71$.Now,substitute these into the formula: $\cos 	heta = \frac{300}{\sqrt{525} 	imes \sqrt{350}} = \frac{300}{\sqrt{183750}} \approx \frac{300}{428.66} \approx 0.70$. Since $\cos 45^\circ \approx 0.707$,the angle $	heta$ is approximately $45^\circ$.

Column $-I$	Column $-II$
$(a) \vec A \cdot \vec B = 0$	$(i) \theta = 0^\circ$
$(b) \vec A \cdot \vec B = +8$	$(ii) \theta = 90^\circ$
$(c) \vec A \cdot \vec B = 4$	$(iii) \theta = 180^\circ$
$(d) \vec A \cdot \vec B = -8$	$(iv) \theta = 60^\circ$

Two forces $\vec F_1 = 5\hat i + 10\hat j - 20\hat k$ and $\vec F_2 = 10\hat i - 5\hat j - 15\hat k$ act on a single point. The angle between $\vec F_1$ and $\vec F_2$ is nearly . . . . . . $^\circ$.

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