Arrows, planes and vectors

  • Thread starter Anden
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I'm a little uncertain concerning a math-question and would like to know if I've done anything wrong here, I do not have the correct answer which is why I'm asking.

Homework Statement


An arrow is shot in a straight line through the points (3, 4, -1) and (-2, 5, 2) and hits a plane containing the points (2, -1, 3) and (3, -3, 5) and (4, 0, 2). What is the angle between the arrow and the plane?

The Attempt at a Solution


I've done it like this, I first calculated the equation for the plane to be 0x + 5y + 5z - 10 = 0 and the direction of the arrow to (-5, 1, 3).

The arrow hits the plane in (17/4, 15/4, -7/4) and together with the arrow vector and the normal to the plane (0, 5, 5) a new plane can be calculated with help of the determinant. Its equation is 2x - 5y + 5z + 19 = 0.

I then calculated the line which the two planes have in common: x = -9/2-5t, y = 2-t, z = t.

And the last thing I did was to use the scalar product with (-5, -1, 1) and (-5, 1, 3), which gave me the answer that the angle equals about 29 degrees (28,56).

Is this correct?
 

Answers and Replies

  • #2
tiny-tim
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Hi Anden! :smile:
An arrow is shot in a straight line through the points (3, 4, -1) and (-2, 5, 2) and hits a plane containing the points (2, -1, 3) and (3, -3, 5) and (4, 0, 2). What is the angle between the arrow and the plane?

I've done it like this, I first calculated the equation for the plane to be 0x + 5y + 5z - 10 = 0 and the direction of the arrow to (-5, 1, 3).

The arrow hits the plane in (17/4, 15/4, -7/4) and together with the arrow vector and the normal to the plane (0, 5, 5) a new plane can be calculated with help of the determinant. Its equation is 2x - 5y + 5z + 19 = 0. …

eugh! :yuck:

Hint: the angle between the line and the plane is 90º minus the angle between the line and the normal (and you can find the normal directly from the three points). :wink:
 
  • #3
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Hi Anden! :smile:


eugh! :yuck:

Hint: the angle between the line and the plane is 90º minus the angle between the line and the normal (and you can find the normal directly from the three points). :wink:

Ok, LOL (Pardon the expression :smile:)

The scalar product of the normal (0, 5, 5) and the vector (-5, 1, 3) is [tex]4/\sqrt{70} = cos a[/tex] which gives a = 61,439 and 90-a = 28,56.

Thank you very much! :rolleyes:
 
  • #4
tiny-tim
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Hi Anden! :smile:

Just one more tip:

you don't necessarily need to write equations for lines and planes

in this case, all you needed were vectors parallel to the line and to the normal, and for the normal you could just have used the cross product of two vectors in the plane. :wink:
 

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