Find and|a→|and|b→|, if and(a→+b→).(a→-b→)=8and|a→|=8|b→|. - Mathematics

Question

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

Sum

Solution

We have, $(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) = 8$

⇒ $\vec{a} \times \vec{a} - \vec{a} \times \vec{b} + \vec{b} \times \vec{a} - \vec{b} \times \vec{b} = 8$

but, $\vec{a} \times \vec{b} = \vec{b} \times \vec{a}$

∴ $\vec{a} \times \vec{a} - \vec{a} \times \vec{b} + \vec{a} \times \vec{b} - \vec{b} \times \vec{b} = 8$

= $\vec{a} \times \vec{a} . \vec{b} \times \vec{b} = 8$

= $64 {| \vec{b} |}^{2} - {| \vec{b} |}^{2} = 8$ $[∵ | \vec{a} | = 8 | \vec{b} |]$

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

∴ $| \vec{b} | = \sqrt{\frac{8}{63}} = \frac{2}{3} \sqrt{\frac{2}{7}}$

But $| \vec{a} | = 8 | \vec{b} |$

⇒ $| \vec{a} | = \frac{8 \sqrt{8}}{\sqrt{63}}$

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

Hence, $| \vec{a} | = \frac{16 \sqrt{2}}{3 \sqrt{7}}$

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

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Magnitude and Direction of a Vector

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Q 6 Q 5 Q 7

Chapter 10: Vector Algebra - Exercise 10.3 [Page 448]

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NCERT Mathematics [English] Class 12

Chapter 10 Vector Algebra
Exercise 10.3 | Q 6 | Page 448

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Find and|a→|and|b→|, if and(a→+b→).(a→-b→)=8and|a→|=8|b→|. - Mathematics

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