If the magnitudes of a, b and a+b are respect | vedclass

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(D) Given,$|a|=3, |b|=4$ and $|a+b|=5$. Since $|a+b|=5$,squaring both sides gives $|a+b|^2=25$. Expanding this,we get $(a+b) \cdot (a+b) = 25$,which s

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$5$

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(D) Given,$|a|=3, |b|=4$ and $|a+b|=5$. Since $|a+b|=5$,squaring both sides gives $|a+b|^2=25$. Expanding this,we get $(a+b) \cdot (a+b) = 25$,which simplifies to $|a|^2 + 2(a \cdot b) + |b|^2 = 25$. Substituting the given values,$3^2 + 2(a \cdot b) + 4^2 = 25$,so $9 + 2(a \cdot b) + 16 = 25$,which implies $2(a \cdot b) = 0$,hence $a \cdot b = 0$. Now,we need to find $|a-b|$. We know that $|a-b|^2 = (a-b) \cdot (a-b) = |a|^2 - 2(a \cdot b) + |b|^2$. Substituting the values,$|a-b|^2 = 3^2 - 2(0) + 4^2 = 9 + 16 = 25$. Therefore,$|a-b| = \sqrt{25} = 5$.

$x$	$1$	$2$	$3$	$4$
$y$	$0.7111$	$0.7222$	$0.7333$	$0.7444$

If the magnitudes of $a, b$ and $a+b$ are respectively $3, 4$ and $5$,then the magnitude of $(a-b)$ is

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