Question
$\text{Find}\ |\vec{a}|\ \text{and}\ \big|\vec{b}\big|,\text{if}\ (\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b})=8\ \text{and}\ |\vec{a}|=8|\vec{b}|.$
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Answer
$\text{Given:}\ (\vec{a}+\vec{b}).(\vec{a}-\vec{b})=8\ \text{and}\ |\vec{a}|=8|\vec{b}|\ \ \ \ \ ....(\text{i})$
$\Rightarrow\ \ {\vec{a}.\vec{a}-\vec{a}.\vec{b}+\vec{b}.\vec{a}-\vec{b}.\vec{b}=8}{}$ $ \Rightarrow\ \ \big|\vec{a}\big|^2-\vec{a}.\vec{b}+\vec{a}.\vec{b}-\big|\vec{b}\big|^2=8$
$\Rightarrow\ \ \big|\vec{a}\big|^2-\big|\vec{b}\big|^2=8\ \ \ .....\text{(ii)}$
$\text{Putting}\ \big|\vec{a}\big|=8\big|\vec{b}\big|\ \text{in eq. (ii),}$
$ 64\bigg|\vec{b}\bigg|^2-\bigg|\vec{b}\bigg|^2=8 \ \ \Rightarrow\ \ \ (64-1)\bigg|\vec{b}\bigg|^2=8$
$\Rightarrow\ \ 63\bigg|\vec{b}\bigg|^2=8\ \ \Rightarrow\ \ \bigg|\vec{b}\bigg|^2=\frac{8}{63}$
$\Rightarrow\ \ \Big|\vec{b}\Big|=\sqrt{\frac{8}{63}}=\frac{2\sqrt{2}}{3\sqrt{7}}$
$\text{Putting}\ \Big|\vec{b}\Big|=\frac{2\sqrt{2}}{3\sqrt{7}}\ \text{in eq (i),}$
$\big|\vec{a}\big|=8\Big(\frac{2\sqrt{2}}{3\sqrt{7}}\Big)=\frac{16}{3}\sqrt{\frac{2}{7}}$
$\Rightarrow\ \ {\vec{a}.\vec{a}-\vec{a}.\vec{b}+\vec{b}.\vec{a}-\vec{b}.\vec{b}=8}{}$ $ \Rightarrow\ \ \big|\vec{a}\big|^2-\vec{a}.\vec{b}+\vec{a}.\vec{b}-\big|\vec{b}\big|^2=8$
$\Rightarrow\ \ \big|\vec{a}\big|^2-\big|\vec{b}\big|^2=8\ \ \ .....\text{(ii)}$
$\text{Putting}\ \big|\vec{a}\big|=8\big|\vec{b}\big|\ \text{in eq. (ii),}$
$ 64\bigg|\vec{b}\bigg|^2-\bigg|\vec{b}\bigg|^2=8 \ \ \Rightarrow\ \ \ (64-1)\bigg|\vec{b}\bigg|^2=8$
$\Rightarrow\ \ 63\bigg|\vec{b}\bigg|^2=8\ \ \Rightarrow\ \ \bigg|\vec{b}\bigg|^2=\frac{8}{63}$
$\Rightarrow\ \ \Big|\vec{b}\Big|=\sqrt{\frac{8}{63}}=\frac{2\sqrt{2}}{3\sqrt{7}}$
$\text{Putting}\ \Big|\vec{b}\Big|=\frac{2\sqrt{2}}{3\sqrt{7}}\ \text{in eq (i),}$
$\big|\vec{a}\big|=8\Big(\frac{2\sqrt{2}}{3\sqrt{7}}\Big)=\frac{16}{3}\sqrt{\frac{2}{7}}$
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