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I have a line $L$ given by $x = 2 -t$, $y = 1 + t$, $z = 1 + 2t$, which intersects a plane $2x + y - z = 1$ at the point $(1,2,3)$. I have to find the angle which the line makes with the plane. I know that, to do this, I should use the following formula: $cos\theta = \frac{\vec{u}\cdot\vec{v}} {||{\vec{u}}||\cdot||{\vec{v}}||}$. However, do I first need to find an equation for the plane using the derivative of $L$ and the point? What would be my $\vec{u}$ and what would be my $\vec{v}$ if this were the case?

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Is the point even necessary to find the angle? Can't I just take the vector of L and the plane and plug it into the formula? Visually speaking, if I moved the point any where on the plane, it seems like it would still be the same angle... –  Dylan May 26 '12 at 4:38
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Yes, that's right, except the angle you get isn't the angle that the line makes with the plane, but its complement. –  joriki May 26 '12 at 4:42
    
Oh I see, but the question is asking to find what angle L makes with the plane. DO you then use the complement to find the angle that L makes with the plane. If so, as the wiki article describes, do I just take 90 degrees minus the complement to find the angle I am looking for? –  Dylan May 26 '12 at 4:50
    
Yes. Try drawing the situation in the plane spanned by $L$ and the normal. The normal and the line where the two planes intersect form a right angle, and $L$ is in between. The angle you get from the calculation is the angle between $L$ and the normal, and the angle you want, between $L$ and the intersection line, is the rest of the right angle. –  joriki May 26 '12 at 5:19

3 Answers 3

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The line can be written as

$$\pmatrix{x\\y\\z}=\pmatrix{2\\1\\1}+t\pmatrix{-1\\1\\2}\;,$$

and the plane can be written as

$$\pmatrix{2\\1\\-1}\cdot\pmatrix{x\\y\\z}=1\;.$$

The angle between the direction vector $\pmatrix{-1\\1\\2}$ of the line and the normal vector $\pmatrix{2\\1\\-1}$ of the plane is complementary to the angle between the line and the plane.

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That is what I thought at first, but I thought for some reason I needed to account for the point and subtract the vector of the plane from the point. But I guess that isn't necessary since visually it doesn't really matter what point it is on the plane, it will be the intersection will result in the same angle. –  Dylan May 26 '12 at 4:43

Confusing question. Maybe deliberately. The point of intersection on the plane is irrelevant, and the point on the line is irrelevant. All that matters is the direction vector of the line and the normal vector of the plane.

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The equation of the line is, $$\frac{x-2}{-1}=\frac{y-1}{1}=\frac{z-1}{2}=t$$

the direction ratios of the line are $(-1,1,2)$

and the direction ratios of the normal vector of the plane are $(2,1,-1)$

the angle between these $2$ vectors gives the angle between the planes.

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