Passage: Let a , b , c be unit vectors. Further, a is perpendicular...

Given $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$ and $\vec{a} \cdot \vec{b} = 0$.

Since $\vec{c}$ makes an angle $\dfrac{\pi}{3}$ with both $\vec{a}$ and $\vec{b}$:

$\vec{c} \cdot \vec{a} = |\vec{c}||\vec{a}|\cos\left(\dfrac{\pi}{3}\right) = \dfrac{1}{2}$

$\vec{c} \cdot \vec{b} = |\vec{c}||\vec{b}|\cos\left(\dfrac{\pi}{3}\right) = \dfrac{1}{2}$

We are given the relation:

$\vec{c} = p\vec{a} + q\vec{b} + r(\vec{a} \times \vec{b})$

Taking the dot product of both sides with $\vec{a}$:

$\vec{c} \cdot \vec{a} = p(\vec{a} \cdot \vec{a}) + q(\vec{b} \cdot \vec{a}) + r((\vec{a} \times \vec{b}) \cdot \vec{a})$

Since $\vec{a} \cdot \vec{a} = 1$, $\vec{b} \cdot \vec{a} = 0$, and $(\vec{a} \times \vec{b}) \cdot \vec{a} = 0$, we get:

$p = \dfrac{1}{2}$

Taking the dot product of both sides with $\vec{b}$:

$\vec{c} \cdot \vec{b} = p(\vec{a} \cdot \vec{b}) + q(\vec{b} \cdot \vec{b}) + r((\vec{a} \times \vec{b}) \cdot \vec{b})$

Since $\vec{b} \cdot \vec{b} = 1$, $\vec{a} \cdot \vec{b} = 0$, and $(\vec{a} \times \vec{b}) \cdot \vec{b} = 0$, we get:

$q = \dfrac{1}{2}$

Therefore, $p + q = \dfrac{1}{2} + \dfrac{1}{2} = 1$.

Answer: $1$