Find |vec{a}| and |vec{b}|, if (vec{a}+vec{b} | vedclass

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(A) Given: $(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=8$ and $|\vec{a}|=8|\vec{b}|.$Expanding the dot product:$\vec{a} \cdot \vec{a} - \vec{a} \cdot \v

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$\frac{16\sqrt{2}}{3\sqrt{7}}, \frac{2\sqrt{2}}{3\sqrt{7}}$

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(A) Given: $(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=8$ and $|\vec{a}|=8|\vec{b}|.$Expanding the dot product:$\vec{a} \cdot \vec{a} - \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} - \vec{b} \cdot \vec{b} = 8$Since $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$,we have:$|\vec{a}|^2 - |\vec{b}|^2 = 8$Substitute $|\vec{a}|=8|\vec{b}|$ into the equation:$(8|\vec{b}|)^2 - |\vec{b}|^2 = 8$$64|\vec{b}|^2 - |\vec{b}|^2 = 8$$63|\vec{b}|^2 = 8$$|\vec{b}|^2 = \frac{8}{63}$$|\vec{b}| = \sqrt{\frac{8}{63}} = \frac{2\sqrt{2}}{3\sqrt{7}}$Now,find $|\vec{a}|$: $|\vec{a}| = 8|\vec{b}| = 8 	imes \frac{2\sqrt{2}}{3\sqrt{7}} = \frac{16\sqrt{2}}{3\sqrt{7}}$

Find $|\vec{a}|$ and $|\vec{b}|,$ if $(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=8$ and $|\vec{a}|=8|\vec{b}|.$

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