If a vector of magnitude 2 units makes an angle (π)/(3) with 2 i ,...

Let, $\vec{a}$ have $\left(a_1, a_2, a_3\right)$ components $$\begin{array}{ll} \Rightarrow & a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k} \\ \because & |\vec{a}|=2 \end{array}$$Also, It is given that \vec{a} make angle$with$\frac{\pi}{3} with$2 \hat{i}$, \frac{\pi}{4}$with 3 \hat{j} and$an acute angle \theta with$4 \hat{k}$ Then we have, $$\begin{aligned} \vec{a}\cdot2 \hat{i} & =|\vec{a}||2 \hat{i}| \cos \frac{\pi}{3}\\ \left( a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\right) \cdot2 \hat{i} & =2.2 \times \frac{1}{2}\\ 2 a_1 & =2 \Rightarrow a_1=1 \\ \therefore \quad \vec{a} \cdot3 \hat{j} & =|\vec{a}||3 \hat{j}| \cos \frac{\pi}{4}\end{aligned}$$$and$3 a_2=2 \times3 \times \frac{1}{\sqrt{2}} \Rightarrow$a_2=$\sqrt{2}$$\begin{aligned} \vec{a} \cdot4 \hat{k} & =|\vec{a}||4 \hat{k}| \cos\theta\\ 4 a_3 & =2 \times4 \cos \theta \end{aligned}$$\Rightarrow \quad $\cos \theta=\frac{a_3}{2} N$ow, $\quad|\vec{a}|$=2$$\begin{array}{ll} \Rightarrow& \sqrt{a_1^2+a_2^2+a_3^2}=2 \\ \Rightarrow & a_1^2+a_2^2+a_3^2=4 \end{array}$$ Put, a_1=1 and $a_2=\sqrt{2}(1)^2+(\sqrt{2})^2+a_3^2=4 \Rightarrow a_3^2= \pm 1$ When, $a_3=-1$ then, $\cos \theta=-\frac{1}{2}\Rightarrow \quad \theta=\frac{2 \pi}{3} > \frac{\pi}{2} ($contradict) $\because \theta is acute angle with$4 \hat{k}\therefore \quad a_3=1$Hence, the components of \vec{a}=(1, \sqrt{2}, 1)$.