If theta is an obtuse angle between vectors o | vedclass

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(B) Given that $|\overline{a}|=5$ and $|\overline{b}|=3$.We know that $|\overline{a} 	imes \overline{b}| = |\overline{a}| |\overline{b}| \sin 	heta$

Vedclass · Accepted Answer

$-10$

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(B) Given that $|\overline{a}|=5$ and $|\overline{b}|=3$.We know that $|\overline{a} 	imes \overline{b}| = |\overline{a}| |\overline{b}| \sin 	heta$.Substituting the given values: $5 \sqrt{5} = 5 	imes 3 	imes \sin 	heta$.$5 \sqrt{5} = 15 \sin 	heta$.$\sin 	heta = \frac{5 \sqrt{5}}{15} = \frac{\sqrt{5}}{3}$.Since $\sin^2 	heta + \cos^2 	heta = 1$,we have $\cos^2 	heta = 1 - \sin^2 	heta = 1 - (\frac{\sqrt{5}}{3})^2 = 1 - \frac{5}{9} = \frac{4}{9}$.Thus,$\cos 	heta = \pm \frac{2}{3}$.Since $	heta$ is an obtuse angle,$\cos 	heta$ must be negative,so $\cos 	heta = -\frac{2}{3}$.Now,$\overline{a} \cdot \overline{b} = |\overline{a}| |\overline{b}| \cos 	heta$.$\overline{a} \cdot \overline{b} = 5 	imes 3 	imes (-\frac{2}{3}) = 15 	imes (-\frac{2}{3}) = -10$.

If $\theta$ is an obtuse angle between vectors $\overline{a}$ and $\overline{b}$ such that $|\overline{a}|=5$,$|\overline{b}|=3$ and $|\overline{a} \times \overline{b}|=5 \sqrt{5}$,then $\overline{a} \cdot \overline{b}=$

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