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Question:

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

Solution:

$(\vec{a} \cdot \vec{b}) \cdot(\vec{a}-\vec{b})=8$

$\Rightarrow \vec{a} \cdot \vec{a}-\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}-\vec{b} \cdot \vec{b}=8$

$\Rightarrow|\vec{a}|^{2}-|\vec{b}|^{2}=8$

$\Rightarrow(8|\vec{b}|)^{2}-|\vec{b}|^{2}=8 \quad[|\vec{a}|=8|\vec{b}|]$

$\Rightarrow 64|\vec{b}|^{2}-|\vec{b}|^{2}=8$

$\Rightarrow 63|\vec{b}|^{2}=8$

$\Rightarrow|\vec{b}|^{2}=\frac{8}{63}$

$\Rightarrow|\vec{b}|=\sqrt{\frac{8}{63}}$                          [Magnitude of a vector is non-negative]

$\Rightarrow|\vec{b}|=\frac{2 \sqrt{2}}{3 \sqrt{7}}$

$|\vec{a}|=8|\vec{b}|=\frac{8 \times 2 \sqrt{2}}{3 \sqrt{7}}=\frac{16 \sqrt{2}}{3 \sqrt{7}}$

 

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