# Tricky minimum distance vector problem

## Homework Statement

The line intersects a plane at an angle alpha=2pi/3. The line is defined by rn, and the plane by r.m=0, with n and m unit vectors. Calculate the shortest distance from the plane to the point on the line with μ=2.

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## The Attempt at a Solution

I want to find the value of μ at the point of intersection. Then it will be easy to find the distance, as d=(2-μ at intersection)sin(2pi/3). However, I don't know how to find μ. Any hints?

I'm still stuck. Any ideas would be much appreciated!

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Hi Lucy Yeats! To find μ at the point of intersection you need to substitute the equation of the line into the equation of the plane.
That is (μn).m=0. From this it follows that μ=0.

That's funny!!
Yes, the origin is on the line and the origin is also in the plane.

Btw, the regular way to calculate the distance of a point to a plane, is to project any vector from the plane to the point, onto the unit normal vector.
You can do this with the dot product.

If n and m are perpendicular, surely mu could be anything? How do you know that they aren't perpendicular?

I'm guessing that you have to use the angle 2pi/3 in some way?

Thanks for helping, btw!

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Homework Helper
If n and m are perpendicular, surely mu could be anything? How do you know that they aren't perpendicular?

Yes.
What would happen (geometrically speaking) with the point of intersection if n and m are perpendicular?

I'm guessing that you have to use the angle 2pi/3 in some way?

Thanks for helping, btw!

Yes.
But first you need the formula for the distance of a point to a plane:
distance = (any vector from plane to point) . (normal vector of plane)

Last edited:
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Actually, your distance formula d=(2-μ at intersection)sin(2pi/3) also works.

Just fill in the μ we just found for the intersection.

Btw, did you know that n.m=sin(2pi/3)?

Ah, I see, they can't be perpendicular because then the angle wouldn't be 2pi/3. so mu has to be zero.

Brilliant, thank you very much!

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Homework Helper
You're welcome! 