If |a| = 3, |b| = 4 and |a + b| = 5, then |a | vedclass

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(B) Given that $|a| = 3$,$|b| = 4$,and $|a + b| = 5$. Using the parallelogram law of vectors,we have the identity: $|a + b|^2 + |a - b|^2 = 2(|a|^2 +

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$5$

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(B) Given that $|a| = 3$,$|b| = 4$,and $|a + b| = 5$. Using the parallelogram law of vectors,we have the identity: $|a + b|^2 + |a - b|^2 = 2(|a|^2 + |b|^2)$ Substituting the given values: $5^2 + |a - b|^2 = 2(3^2 + 4^2)$ $25 + |a - b|^2 = 2(9 + 16)$ $25 + |a - b|^2 = 2(25)$ $25 + |a - b|^2 = 50$ $|a - b|^2 = 50 - 25 = 25$ $|a - b| = \sqrt{25} = 5$. Thus,the correct option is $B$.

If $|a| = 3, |b| = 4$ and $|a + b| = 5,$ then $|a - b| = $

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