A vector vec{n} is inclined to X-axis at 45^{ | vedclass

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(A) Let $\vec{n}$ be inclined at angles $\alpha, \beta, \gamma$ to $X, Y, Z$ axes respectively. Given $\alpha = 45^{\circ}, \beta = 60^{\circ}$. We kn

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$\sqrt{2}x + y + z = 0$

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(A) Let $\vec{n}$ be inclined at angles $\alpha, \beta, \gamma$ to $X, Y, Z$ axes respectively. Given $\alpha = 45^{\circ}, \beta = 60^{\circ}$. We know that $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$. Substituting the values: $\cos^2(45^{\circ}) + \cos^2(60^{\circ}) + \cos^2 \gamma = 1$. $\left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{2}\right)^2 + \cos^2 \gamma = 1$. $\frac{1}{2} + \frac{1}{4} + \cos^2 \gamma = 1$. $\frac{3}{4} + \cos^2 \gamma = 1 \implies \cos^2 \gamma = \frac{1}{4}$. Since $\gamma$ is an acute angle,$\cos \gamma = \frac{1}{2}$,so $\gamma = 60^{\circ}$. The normal vector is $\vec{n} = \cos \alpha \hat{i} + \cos \beta \hat{j} + \cos \gamma \hat{k} = \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{2} \hat{j} + \frac{1}{2} \hat{k}$. Multiplying by $2$ to simplify the normal vector,we can take $\vec{n}' = \sqrt{2} \hat{i} + \hat{j} + \hat{k}$. The equation of the plane passing through $(x_0, y_0, z_0) = (-\sqrt{2}, 1, 1)$ with normal $\vec{n}'$ is $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$. $\sqrt{2}(x - (-\sqrt{2})) + 1(y - 1) + 1(z - 1) = 0$. $\sqrt{2}(x + \sqrt{2}) + y - 1 + z - 1 = 0$. $\sqrt{2}x + 2 + y + z - 2 = 0$. $\sqrt{2}x + y + z = 0$.

$A$ vector $\vec{n}$ is inclined to $X$-axis at $45^{\circ}$,$Y$-axis at $60^{\circ}$ and at an acute angle to $Z$-axis. If $\vec{n}$ is normal to a plane passing through the point $(-\sqrt{2}, 1, 1)$,then the equation of the plane is

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