A vector of magnitude 8 units is inclined to x-axis at , y-axis at and at an acute angle with…

Let the direction angles of $\vec{n}$ be $\alpha = 45^\circ$, $\beta = 60^\circ$, and $\gamma$ (acute). The direction cosines satisfy $l^2 + m^2 + n^2 = 1$. $l = \cos 45^\circ = \frac{1}{\sqrt{2}}$, $m = \cos 60^\circ = \frac{1}{2}$, $n = \cos\gamma$. $\left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{2}\right)^2 + n^2 = 1 \Rightarrow \frac{1}{2} + \frac{1}{4} + n^2 = 1 \Rightarrow \frac{3}{4} + n^2 = 1 \Rightarrow n^2 = \frac{1}{4}$. Since $\gamma$ is acute, $n = \cos\gamma = \frac{1}{2}$. The vector $\vec{n}$ has magnitude 8: $\vec{n} = |\vec{n}|(l\hat{i} + m\hat{j} + n\hat{k}) = 8\left(\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{2}\hat{j} + \frac{1}{2}\hat{k}\right) = 4\sqrt{2}\hat{i} + 4\hat{j} + 4\hat{k}$. The plane passes through $P(\sqrt{2}, -1, 1)$ with position vector $\vec{a} = \sqrt{2}\hat{i} - \hat{j} + \hat{k}$. The vector equation of the plane is $(\vec{r} - \vec{a}) \cdot \vec{n} = 0 \Rightarrow \vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$. $\vec{a} \cdot \vec{n} = (\sqrt{2})(4\sqrt{2}) + (-1)(4) + (1)(4) = 8 - 4 + 4 = 8$. Thus, $\vec{r} \cdot (4\sqrt{2}\hat{i} + 4\hat{j} + 4\hat{k}) = 8$, or dividing by 4: $\vec{r} \cdot (\sqrt{2}\hat{i} + \hat{j} + \hat{k}) = 2$.