Let → a and → b be two vectors such that | → a + → b | ^ 2 = | → a...

Given,

${\left|\overset{\to}{a} + \overset{\to}{b}\right|}^{2} = {\left|\overset{\to}{a}\right|}^{2} + 2{\left|\overset{\to}{b}\right|}^{2} \text{ and } \overset{\to}{a} \cdot \overset{\to}{b} = 3$

We know that

${\left|\overset{\to}{a} + \overset{\to}{b}\right|}^{2} = {\left|\overset{\to}{a}\right|}^{2} + {\left|\overset{\to}{b}\right|}^{2} + 2\overset{\to}{a} \cdot \overset{\to}{b}$

Now using the above formula in ${\left|\overset{\to}{a} + \overset{\to}{b}\right|}^{2} = {\left|\overset{\to}{a}\right|}^{2} + 2{\left|\overset{\to}{b}\right|}^{2}$ we get,

${\left|\overset{\to}{a}\right|}^{2} + {\left|\overset{\to}{b}\right|}^{2} + 2\overset{\to}{a} \cdot \overset{\to}{b} = {\left|\overset{\to}{a}\right|}^{2} + 2{\left|\overset{\to}{b}\right|}^{2}$

Now on comparing both side we get, ${\left|\overset{\to}{b}\right|}^{2} = 2\overset{\to}{a} \cdot \overset{\to}{b} = 6 . . . . . . . . . \left(1\right)$

Also given ${\left|\overset{\to}{a} \times \overset{\to}{b}\right|}^{2} = 75$

$⇒ {\left|\overset{\to}{a}\right|}^{2}{\left|\overset{\to}{b}\right|}^{2}\sin^{2}\theta = 75$

$⇒ {\left|\overset{\to}{a}\right|}^{2}{\left|\overset{\to}{b}\right|}^{2}\left(1 - \cos^{2}\theta\right) = 75$

$⇒ {\left|\overset{\to}{a}\right|}^{2}{\left|\overset{\to}{b}\right|}^{2} - {\left|\overset{\to}{a}\right|}^{2}{\left|\overset{\to}{b}\right|}^{2}\cos^{2}\theta = 75$

$⇒ {\left|\overset{\to}{a}\right|}^{2}{\left|\overset{\to}{b}\right|}^{2} - {\left(\overset{\to}{a} \cdot \overset{\to}{b}\right)}^{2} = 75$

Now putting the value of ${\left|\overset{\to}{b}\right|}^{2} = 2\overset{\to}{a} \cdot \overset{\to}{b} = 6$ we get, $6{\left|\overset{\to}{a}\right|}^{2} - 9 = 75 ⇒ {\left|\overset{\to}{a}\right|}^{2} = 14$