Equations for a Line Passing Through a Point with Given Direction Angles

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SUMMARY

The discussion focuses on determining the vector, parametric, and symmetric equations of a line defined by direction angles of 60°, 45°, and 60°, passing through the point (1, -2, 5). The vector equation is expressed as (1/2)i + (√2/2)j + (1/2)k. The correct parametric equations are x = (1/2)t + 1, y = (√2/2)t - 2, and z = (1/2)t + 5. The symmetric equation is given by (x-1)/0.5 = (y+2)/(√2/2) = (z-5)/0.5, with a noted correction needed for the y-component.

PREREQUISITES
  • Understanding of vector equations in three-dimensional space
  • Familiarity with parametric equations of lines
  • Knowledge of symmetric equations of lines
  • Basic trigonometry, particularly direction angles
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  • Study the derivation of vector equations from direction angles
  • Learn about parametric equations in three-dimensional geometry
  • Explore symmetric equations and their applications in line representation
  • Investigate the relationship between direction angles and line orientation
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Students studying geometry, particularly those focusing on vector calculus and three-dimensional line equations, as well as educators teaching these concepts in mathematics courses.

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Homework Statement



A line has direction angles 60°, 45°, 60° and passes through the point (1, -2, 5). Determine vector, parametric and symmetric equations of this line.

Homework Equations





The Attempt at a Solution



1/2 i + [root 2]/2 j +1 /2 k
parametric equation will be [1 ,-2, 5]
x= 1/2t+1
y=[root 2] /2 t+2
z=1/2 t+5

[x,y]= [1,2] + t[1/2,root2/2]

symmetrical
[x-1]/.5= {y-2]/[root2/2]}=[z-5]/.5

thats my final answer... please check if i have done it the proper way...
 
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f22archrer said:
x= 1/2t+1
y=[root 2] /2 t+2
z=1/2 t+5
Hey archer. I hope this reply is not too late for you. Anyway, you are very close to the parametric equation of the line. Remember the line passes through the point (1,-2,5) So your equation has one slight mistake in it.

f22archrer said:
[x,y]= [1,2] + t[1/2,root2/2]

symmetrical
[x-1]/.5= {y-2]/[root2/2]}=[z-5]/.5
Again, you are really close, the x and z parts are correct, but you have made the same slight mistake with the y part.

f22archrer said:
1/2 i + [root 2]/2 j +1 /2 k
This is part of the answer for the vector equation of the line, but there is more. hint: will this vector equation take you from the origin to any general point on the line?
 
thanks
 

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