If a unit vector r ⃑makes angleπ3 with î̂,π4with ĵ̂ and θ ∈0,π with k̂, then value of θ is ____.

Let unit vector r have (

r_{1}, r_{2}, r_{3}

) components.

r = r_{1} \hat{i} + r_{2} \hat{j} + r_{3} \hat{z}

Since a is a unit vector, ∣r∣=1.

Also,

makes angle \frac{π}{3} with \hat{\hat{i}}, \frac{π}{4} with \hat{\hat{j}} andθ \in (0, π) with \hat{k}

Then, we have:

cosθ = \frac{r_{1}}{|r⃑|}

\cos \frac{π}{3} = \frac{r_{1}}{1}

\frac{1}{2} = r_{1}

cosθ = \frac{r_{2}}{|r⃑|}

\cos \frac{π}{4} = \frac{r_{2}}{1}

\frac{1}{\sqrt{2}} = r_{2}

cosθ = \frac{r_{3}}{|r⃑|}

cosθ = \frac{r_{3}}{1}

cosθ = r_{3}

∣r∣=1

\sqrt{{(r_{1})}^{2} + {(r_{2})}^{2} + {(r_{3})}^{2}} = 1

\sqrt{{(\frac{1}{2})}^{2} + {(\frac{1}{\sqrt{2}})}^{2} + {(\cos cos θ)}^{2}} = 1

\sqrt{\frac{1}{4} + \frac{1}{2} + {(\cos cos θ)}^{2}} = 1

\sqrt{\frac{1 + 2}{4} + {(\cos cos θ)}^{2}} = 1

\sqrt{\frac{3}{4} + {(\cos cos θ)}^{2}} = 1

Taking square root on both the sides

\frac{3}{4} + {(\cos cos θ)}^{2} = {(1)}^{2}

{(\cos cos θ)}^{2} = 1 - \frac{3}{4}

{(\cos cos θ)}^{2} = \frac{4 - 3}{4}

{(\cos cos θ)}^{2} = \frac{1}{4}

cosθ = \sqrt{\frac{1}{4}} = \frac{1}{2}

But

\cos 60^{0} = \frac{1}{2}

∴ θ = 60^{0} = \frac{π}{3}

If a unit vector $r ⃑makes angle \frac{π}{3} with \hat{\hat{i}}, \frac{π}{4} with \hat{\hat{j}} and θ \in (0, π) with \hat{k}, then value of θ is$ ____.