A vector has magnitude same as that of → A = 3 ^ j + 4 ^ j and is...

Magnitude of vector $\left|\overset{\to}{A}\right| = \sqrt{{3}^{2} + {4}^{2}} = 5$

Now, unit vector along \overset{\to}{B} = \dfrac{( 4\overset{^}{i} + 3\overset{^}{j} )}{\sqrt{{4}^{2} + {3}^{2}}} = \dfrac{4}{5}\overset{^}{i} + \dfrac{3}{5}\overset{^}{j}

Now, \overset{\to}{N} = \left|\overset{\to}{A}\right|\overset{^}{B} = 5 \times \left(\dfrac{4}{5}\overset{^}{i} + \dfrac{3}{5}\overset{^}{j}\right) = 4\overset{^}{i} + 3\overset{^}{j}

$∴ x = 4$