Let a unit vector which makes an angle of $60^{\circ}$ with $2 \hat{i}+2 \hat{j}-\hat{k}$ and an angle of $45^{\circ}$ with $\hat{i}-\hat{k}$ be $\vec{C}$. Then $$\vec{C}+\left(-\frac{1}{2} \hat{i}+\frac{1}{3 \sqrt{2}} \hat{j}-\frac{\sqrt{2}}{3} \hat{k}\right)$$ is:

<p>$$\begin{aligned} & \text { Let } \vec{C}=a \hat{i}+b \hat{j}+c \hat{k} \\ & (a \hat{i}+b \hat{j}+c \hat{k}) \cdot(2 \hat{i}+2 \hat{j}-\hat{k})=1 \times 3 \times \frac{1}{2} \\ & 2 a+2 b-c=\frac{3}{2} \qquad \text{... (1)}\\ & (a \hat{i}+b \hat{j}+c \hat{k}) \cdot(\hat{i}-\hat{k})=1 \times \sqrt{2} \times \frac{1}{\sqrt{2}} \\ & a-c=1 \quad \text{... (2)}\\ & a^2+b^2+c^2=1 \quad \text{... (3)}\\ \end{aligned}$$</p> <p>Solving (1), (2) and (3)</p> <p>$$\begin{aligned} & a+2 b=\frac{1}{2} \\ & a^2+b^2+(a-1)^2=1 \\ & 2 a^2-2 a+b^2=0 \\ & 2 a^2-2 a+\left(\frac{2 a-1}{4}\right)^2=0 \\ & 32 a^2-32 a+4 a^2-4 a+1=0 \\ & 36 a^2-36 a+1=0 \\ & a=\frac{36 \pm \sqrt{(36)^2-4(36)}}{2 \times 36} \\ & =\frac{1}{2} \pm \frac{\sqrt{2}}{3} \\ & b=\frac{1-2 a}{4} \Rightarrow b=\frac{1 \pm \frac{2 \sqrt{2}}{3}-1}{4} \\ & =\mp \frac{1}{3 \sqrt{2}} \\ & C=-\frac{1}{2} \pm \frac{\sqrt{2}}{3} \end{aligned}$$</p> <p>$$\begin{aligned} & C+\left(\frac{-1}{2} \hat{i}+\frac{1}{3 \sqrt{2}} \hat{j}-\frac{\sqrt{2}}{3} \hat{k}\right) \\ & =\frac{\sqrt{2}}{3} \hat{i}-\frac{1}{2} \hat{k} \end{aligned}$$</p>