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A vector $\vec{n}$ is inclined to $x$-axis at $45^{\circ}$, to $y$-axis at $60^{\circ}$ and at an acute angle to $z$-axis. If $\vec{n}$ is a normal to a plane passing through the point $(\sqrt{2},-1,1)$ then the equation of the plane is :
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Verified Answer
The correct answer is:
$2 x+y+2 z=2 \sqrt{2}+1$
$2 x+y+2 z=2 \sqrt{2}+1$
Direction cosines of $\vec{n}$ are $\frac{1}{2}, \frac{1}{4}, \frac{1}{2}$.
Equation of the plane,
$$
\begin{aligned}
& \frac{1}{2}(x-\sqrt{2})+\frac{1}{4}(y+1)+\frac{1}{2}(z-1)=0 \\
\Rightarrow & 2(x-\sqrt{2})+(y+1)+2(z-1)=0 \\
\Rightarrow & 2 x+y+2 z=2 \sqrt{2}-1+2 \\
\Rightarrow & 2 x+y+2 z=2 \sqrt{2}+1
\end{aligned}
$$
Equation of the plane,
$$
\begin{aligned}
& \frac{1}{2}(x-\sqrt{2})+\frac{1}{4}(y+1)+\frac{1}{2}(z-1)=0 \\
\Rightarrow & 2(x-\sqrt{2})+(y+1)+2(z-1)=0 \\
\Rightarrow & 2 x+y+2 z=2 \sqrt{2}-1+2 \\
\Rightarrow & 2 x+y+2 z=2 \sqrt{2}+1
\end{aligned}
$$
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