If |a| = |b| = 1 and |a + b| = sqrt{3},then t | vedclass

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(B) Given $|a| = 1$,$|b| = 1$,and $|a + b| = \sqrt{3}$.Squaring the equation $|a + b| = \sqrt{3}$,we get $|a + b|^2 = 3$.Using the property $|a + b|^2

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$-21/2$

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(B) Given $|a| = 1$,$|b| = 1$,and $|a + b| = \sqrt{3}$.Squaring the equation $|a + b| = \sqrt{3}$,we get $|a + b|^2 = 3$.Using the property $|a + b|^2 = |a|^2 + |b|^2 + 2(a \cdot b)$,we have $1^2 + 1^2 + 2(a \cdot b) = 3$.$2 + 2(a \cdot b) = 3 \implies 2(a \cdot b) = 1 \implies a \cdot b = \frac{1}{2}$.Now,expand the expression $(3a - 4b) \cdot (2a + 5b)$:$(3a - 4b) \cdot (2a + 5b) = 6(a \cdot a) + 15(a \cdot b) - 8(b \cdot a) - 20(b \cdot b)$.Since $a \cdot a = |a|^2 = 1$,$b \cdot b = |b|^2 = 1$,and $a \cdot b = b \cdot a$,we get:$6(1) + 7(a \cdot b) - 20(1) = 6 - 20 + 7 \left(\frac{1}{2}\right)$.$= -14 + \frac{7}{2} = \frac{-28 + 7}{2} = -\frac{21}{2}$.

If $|a| = |b| = 1$ and $|a + b| = \sqrt{3}$,then the value of $(3a - 4b) \cdot (2a + 5b)$ is

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