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

Given $\vec{a}$, $\vec{b}$, $\vec{c}$ are unit vectors, so $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$.

Since $\vec{a}$ is perpendicular to $\vec{b}$, $\vec{a} \cdot \vec{b} = 0$.

The vector $\vec{c}$ makes an angle of $\dfrac{\pi}{3}$ with both $\vec{a}$ and $\vec{b}$, giving $\vec{c} \cdot \vec{a} = 1 \cdot 1 \cdot \cos\left(\dfrac{\pi}{3}\right) = \dfrac{1}{2}$ and $\vec{c} \cdot \vec{b} = 1 \cdot 1 \cdot \cos\left(\dfrac{\pi}{3}\right) = \dfrac{1}{2}$.

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

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

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

$\dfrac{1}{2} = p(1) + q(0) + r(0) \Rightarrow p = \dfrac{1}{2}$

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

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

$\dfrac{1}{2} = p(0) + q(1) + r(0) \Rightarrow q = \dfrac{1}{2}$

Taking the dot product of $\vec{c}$ with itself:

$|\vec{c}|^2 = p^2|\vec{a}|^2 + q^2|\vec{b}|^2 + r^2|\vec{a} \times \vec{b}|^2$

Since $\vec{a}$ and $\vec{b}$ are orthogonal unit vectors, $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\left(\dfrac{\pi}{2}\right) = 1$.

Substituting the values:

$1 = p^2 + q^2 + r^2$

$1 = \left(\dfrac{1}{2}\right)^2 + \left(\dfrac{1}{2}\right)^2 + r^2$

$1 = \dfrac{1}{4} + \dfrac{1}{4} + r^2$

$r^2 = 1 - \dfrac{1}{2} = \dfrac{1}{2}$

Answer: $\dfrac{1}{2}$