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 $(a+b) \cdot (a+b) = 25$, whi

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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, $|a-b| = \sqrt{25} = 5$.

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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