class 12 maths vector algebra

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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📘 Vector Algebra NCERT,Page 448,Ex.10.3,Q.No 6 SA

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

Official Solution

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We have, $(\vec a + \vec b) \cdot (\vec a - \vec b) = 8 \Rightarrow \vec a \cdot (\vec a - \vec b) + \vec b \cdot (\vec a - \vec b) = 8$

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

$\Rightarrow 64|\vec b{|^2} - |\vec b{|^2} = 8$
$\Rightarrow 63|\vec b{|^2} = 8 \Rightarrow |\vec b{|^2} = \cfrac{8}{{63}} \Rightarrow |\vec b| = \sqrt {\cfrac{8}{{63}}} = \cfrac{{2\sqrt 2 }}{{3\sqrt 7 }}$

But $|\vec a| = 8|\vec b| \Rightarrow |\vec a| = \cfrac{{8\sqrt 8 }}{{\sqrt {63} }} = \cfrac{{16\sqrt 2 }}{{3\sqrt 7 }}$

Hence, $\vec a = \cfrac{{16\sqrt 2 }}{{3\sqrt 7 }}$ and $|\vec b| = \cfrac{{2\sqrt 2 }}{{3\sqrt 7 }}$

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