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If $|\overrightarrow{\mathrm{a}}|=\sqrt{2},|\overrightarrow{\mathrm{b}}|=\sqrt{3}$ and $|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|=\sqrt{6}$, then what is $|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|$
equal to?
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equal to?
Solution:
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Verified Answer
The correct answer is:
2
Consider $|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|^{2}+|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|^{2}=2\left(|\overrightarrow{\mathrm{a}}|^{2}+|\overrightarrow{\mathrm{b}}|^{2}\right)$
By putting the values of $|\overrightarrow{\mathrm{a}}|,|\overrightarrow{\mathrm{b}}|$ and $|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|$, we get
$6+|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|^{2}=2(2+3)$
$\Rightarrow \quad|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|=2$
By putting the values of $|\overrightarrow{\mathrm{a}}|,|\overrightarrow{\mathrm{b}}|$ and $|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|$, we get
$6+|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|^{2}=2(2+3)$
$\Rightarrow \quad|\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}|=2$
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