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HELP with this vectors problem!

A photo of merm123 merm123
How do you find distance of a line when you know a point it passes through, the direction angles and you also know a plane it passes through?
Like let's just say it passes through A(1,0,1) and has direction vectors of alpha = 36.87°, β = 84.26,°γ = 53.7° and we know it intersects the plane 3x + y + 5z = 15 at point B. How would I find the distance of AB?

no need to do the question, just help on how to do it would be nice :) thanks guys.
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@merm123 wrote
How do you find distance of a line when you know a point it passes through, the direction angles and you also know a plane it passes through?
Like let's just say it passes through A(1,0,1) and has direction vectors of alpha = 36.87°, β = 84.26,°γ = 53.7° and we know it intersects the plane 3x + y + 5z = 15 at point B. How would I find the distance of AB?

no need to do the question, just help on how to do it would be nice :) thanks guys.


just use picaso's law of theorems and apply it to your vector quantity.
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A photo of merm123 merm123

@arcchit wrote

@merm123 wrote
How do you find distance of a line when you know a point it passes through, the direction angles and you also know a plane it passes through?
Like let's just say it passes through A(1,0,1) and has direction vectors of alpha = 36.87°, β = 84.26,°γ = 53.7° and we know it intersects the plane 3x + y + 5z = 15 at point B. How would I find the distance of AB?

no need to do the question, just help on how to do it would be nice :) thanks guys.


just use picaso's law of theorems and apply it to your vector quantity.



? I don't know what that is..thanks anyhow.
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Don't listen to arcchit, he is going around acting stupid.
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A photo of plato plato

@merm123 wrote
How do you find distance of a line when you know a point it passes through, the direction angles and you also know a plane it passes through?
Like let's just say it passes through A(1,0,1) and has direction vectors of alpha = 36.87°, β = 84.26,°γ = 53.7° and we know it intersects the plane 3x + y + 5z = 15 at point B. How would I find the distance of AB?

no need to do the question, just help on how to do it would be nice :) thanks guys.



I'm not sure I am reading your question correctly... as it's not clear what those direction vectors apply to. Nonetheless, I can point you in the right direction.

To find the distance between point A and the plane (|AB|), start off by finding the parametric equation of a line perpendicular (normal) to the plane. (Hint: cross product of two direction vectors on the plane)

There are two methods to solve from that point.

1. Solve for the intersection B by substituting the components of the parametric equation of the line into the scalar equation of the plane. Once you find the intersection, you can calculate |AB|.

2. Choose any point on the plane P. The projection of PA onto the normal to the plane is is AH, and the magnitude of AH is the distance from point A to the plane. Use |AH|=proj_n (PA)

Hope this helps.

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A photo of merm123 merm123

@plato wrote

@merm123 wrote
How do you find distance of a line when you know a point it passes through, the direction angles and you also know a plane it passes through?
Like let's just say it passes through A(1,0,1) and has direction vectors of alpha = 36.87°, β = 84.26,°γ = 53.7° and we know it intersects the plane 3x + y + 5z = 15 at point B. How would I find the distance of AB?

no need to do the question, just help on how to do it would be nice :) thanks guys.



I'm not sure I am reading your question correctly... as it's not clear what those direction vectors apply to. Nonetheless, I can point you in the right direction.

To find the distance between point A and the plane (|AB|), start off by finding the parametric equation of a line perpendicular (normal) to the plane. (Hint: cross product of two direction vectors on the plane)

There are two methods to solve from that point.

1. Solve for the intersection B by substituting the components of the parametric equation of the line into the scalar equation of the plane. Once you find the intersection, you can calculate |AB|.

2. Choose any point on the plane P. The projection of PA onto the normal to the plane is is AH, and the magnitude of AH is the distance from point A to the plane. Use |AH|=proj_n (PA)

Hope this helps.





I figured it out finally, but thanks anyways! :)

Basically, the direction angles lead you to the direction vector. To find the direction vector you use the equation cos β = b / sqrt of a^2+b^2+c^2 with (a,b,c) being the direction vector. To find the direction vector, you sub sqrt of a^2+b^2+c^2 as any magnitude as direction vector is a scalar quantity and can be many values. So I used magnitude as 10 and I found a,b,c with that (got messy numbers haha). Then I used the parametric equations with the plane and point A for point B and solved for the magnitude of AB, exactly what you said.
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No problem, glad you figured it out on your own :)
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