Equation of a line perpendicular to two vectors

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SUMMARY

The discussion centers on finding the parametric and symmetric equations for a line that passes through the point (2,1,0) and is perpendicular to the vectors <1,1,0> and <0,1,1>. The cross product of these vectors yields the direction vector <1,-1,1>, which is used to derive the parametric equations: x = 2 + t, y = 1 - t, z = t. The symmetric equation is established as x-2 = y-1 = z, with a note to verify the signs in the symmetric form.

PREREQUISITES
  • Understanding of vector operations, specifically cross products
  • Familiarity with parametric equations of a line
  • Knowledge of symmetric equations in three-dimensional space
  • Basic skills in coordinate geometry
NEXT STEPS
  • Study the properties of cross products in vector calculus
  • Learn how to derive parametric equations from vector equations
  • Explore the relationship between parametric and symmetric equations
  • Practice solving problems involving lines in three-dimensional space
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Students studying vector calculus, geometry enthusiasts, and anyone looking to strengthen their understanding of three-dimensional line equations.

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Homework Statement


Find the parametric and symmetric equations for the line through the point (2,1,0) and perpendicular to both <1,1,0> and <0,1,1,>


Homework Equations


Parametric equ:
x = x_0 + at
y = y_0 + bt
z = z_0 + ct


The Attempt at a Solution


Cross product of <1,1,0> and <0,1,1> = <1,-1,1> = <a,b,c>

Parametric equation:
x = 2 + t
y = 1 - t
z = t

Symmetric equation:
x-2 = y-1 = z

I'm not sure if I did this correctly. Is the cross product of the two vectors <a,b,c>?
 
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Looks pretty much OK. Your cross product and parametric equations are correct. Check your signs on the symmetric version.
 
Thanks for catching that.
 

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