A vector n⃗ of magnitude 8 units is inclined to x-axis at 45°, y-axis at 60° and at an acute angle with z-axis. A plane through the point (√2, -1, 1) is…

Direction cosines of n⃗: l = cos 45° = 1/(√2), m = cos 60° = 1/2, n = cosγ. From l² + m² + n² = 1: n² = 1/4, and since γ is acute, n = 1/2. n⃗ = 8((î)/(√2) + (ĵ)/2 + (k̂)/2) = 4√2î + 4ĵ + 4k̂. The plane passes through a⃗ = √2î - ĵ + k̂, so (r⃗ - a⃗)·n⃗ = 0, giving r⃗·(4√2î + 4ĵ + 4k̂) = 8. Equation of the plane: r⃗·(√2î + ĵ + k̂) = 2

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A vector n⃗ of magnitude 8 units is inclined to x-axis at 45°, y-axis at 60° and at an acute angle…

A vector $\vec{n}$ of magnitude 8 units is inclined to x-axis at $45^\circ$, y-axis at $60^\circ$ and at an acute angle with z-axis. A plane through the point $(\sqrt{2}, -1, 1)$ is normal to $\vec{n}$. Find the equation of the plane in vector form.

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Direction cosines of $\vec{n}$: $l = \cos 45^\circ = \frac{1}{\sqrt{2}}$, $m = \cos 60^\circ = \frac{1}{2}$, $n = \cos\gamma$. From $l^2 + m^2 + n^2 = 1$: $n^2 = \frac{1}{4}$, and since $\gamma$ is acute, $n = \frac{1}{2}$. $\vec{n} = 8\left(\frac{\hat{i}}{\sqrt{2}} + \frac{\hat{j}}{2} + \frac{\hat{k}}{2}\right) = 4\sqrt{2}\hat{i} + 4\hat{j} + 4\hat{k}$. The plane passes through $\vec{a} = \sqrt{2}\hat{i} - \hat{j} + \hat{k}$, so $(\vec{r} - \vec{a})\cdot\vec{n} = 0$, giving $\vec{r}\cdot\left(4\sqrt{2}\hat{i} + 4\hat{j} + 4\hat{k}\right) = 8$. Equation of the plane: $\vec{r}\cdot\left(\sqrt{2}\hat{i} + \hat{j} + \hat{k}\right) = 2$