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I have to find the equation of the plane that is perpendicular to the line $\overline{l}(t)=(10, 0, 4)t+(6, -2, 2)$ and passes through $(10, -2, 0)$.

We know that a plane that has a perpendicular vector $(A, B, C)$ and passes through a point $P=(x_0, y_0, z_0)$ is given by $$A(x-x_0)+B(y-y_0)+C(z-z_0)=0$$

We have that the plane passes through a point $P$ and we have to find a perpendicular vector.

Since $\overrightarrow{l}$ is perpendicular to the plane, so the vector that is parallel to the line is also perpendicular to the plane.

So can we use as $(A, B, C)$ the vector $(10, 0, 4)$ ??

But how can we justify it??

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To justify it you simply take $2$ points on this line. Choose $t=0$ for the first point, and $t=1$ for the second point. The point $C = l(0)=(6,-2,2)$, and $D = l(1) = (10,0,4)+(6,-2,2)$. Thus $\vec{CD} = (10,0,4)$ is the vector that is normal to the given plane.

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  • $\begingroup$ I seee... Thank you so much!!! :-) $\endgroup$ – Mary Star Mar 1 '15 at 3:31
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    $\begingroup$ You are welcome, Mary. $\endgroup$ – DeepSea Mar 1 '15 at 4:08
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your answer is correct. the plane is $10x + 4z = 100$ that satisfies both requirements. i don't what it means to justify other than verify both requirements.

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  • $\begingroup$ Ok... Thank you!!! :-) $\endgroup$ – Mary Star Mar 1 '15 at 3:32

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