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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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Find $| \overrightarrow a |$ and $| \overrightarrow b |$ , if $(\overrightarrow a + \overrightarrow b) ⋅ (\overrightarrow a − \overrightarrow b) = 8$ and $| \overrightarrow a |= 8 | \overrightarrow b |.$

$\begin{array}{1 1}|\overrightarrow a|=\large\frac{\sqrt2}{\sqrt{63}},\:|\overrightarrow b|=\large\frac{2\sqrt 2}{\sqrt{63}} \\|\overrightarrow a|=\large\frac{16\sqrt2}{\sqrt{63}},\:|\overrightarrow b|=\large\frac{2\sqrt 2}{\sqrt{63}} \\|\overrightarrow a|=\large\frac{8\sqrt2}{\sqrt{63}},\:|\overrightarrow b|=\large\frac{\sqrt 8}{\sqrt{63}} \\ |\overrightarrow a|=\:8\sqrt{\large\frac{8}{62}},\:|\overrightarrow b|=\sqrt{\large\frac{8}{62}} \end{array} $

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1 Answer

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Toolbox:
  • $(\overrightarrow a+\overrightarrow b).(\overrightarrow a-\overrightarrow b)=|\overrightarrow a|^2-|\overrightarrow b|^2$
Step 1:
Given : $(\overrightarrow a+\overrightarrow b).(\overrightarrow a-\overrightarrow b)=8$
$|\overrightarrow a|=8|\overrightarrow b|$--------(1)
We know $(\overrightarrow a+\overrightarrow b).(\overrightarrow a-\overrightarrow b)=|\overrightarrow a|^2-|\overrightarrow b|^2$
Therefore $|\overrightarrow a|^2-|\overrightarrow b|^2=8$
Since $|\overrightarrow a|=8|\overrightarrow b|$,substitute for $\mid \overrightarrow a\mid$
Step 2:
Therefore we get $|8\overrightarrow b|^2-|\overrightarrow b|^2=8$
$\Rightarrow 64b^2-b^2=8$
$63b^2=8$
$|\overrightarrow b|=\sqrt{\large\frac{8}{63}}$
$|\overrightarrow b|=\large\frac{2\sqrt 2}{\sqrt {63}}$
Step 3:
Substitute the value of $|\overrightarrow b|$ in equ(1) we get
$|\overrightarrow a|=8\times \large\frac{2\sqrt 2}{\sqrt {63}}$
$\qquad=\large\frac{16\sqrt 2}{\sqrt {63}}$
Step 4:
Therefore $|\overrightarrow a|= \large\frac{16\sqrt 2}{\sqrt {63}}$
$\qquad\;\;\;\;\;\; |\overrightarrow b|=\large\frac{2\sqrt 2}{\sqrt {63}}$
answered May 20, 2013 by sreemathi.v
edited May 20, 2013 by sreemathi.v
 

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