Vectors a ^ i + b ^ j + ^ k and 2 ^ i - 3 ^ j + 4 ^ k are...

Dot product of two perpendicular vectors will be zero. Therefore,

\left(a\overset{^}{i} + b\overset{^}{j} + \overset{^}{k}\right) \cdot \left(2\overset{^}{i} - 3\overset{^}{j} + 4\overset{^}{k}\right) = 0

$⇒ 2a - 3b + 4 = 0 . . . Eq ( 1 )$

As given in the question,

$3a + 2b = 7 . . . Eq ( 2 )$

From both equations, we get $a = 1 \text{\text{ and } }$ b = 2

Then,

$\dfrac{a}{b} = \dfrac{x}{2} ⇒ x = \dfrac{2a}{b} = \dfrac{2 \times 1}{2}$

$⇒ x = 1$