Find parametric eq for a line with point and plane

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SUMMARY

The discussion focuses on deriving parametric equations for a line that passes through the point Po=(3,-1,1) and is perpendicular to the plane defined by the equation 3x+5y-7z=29. The normal vector of the plane, given by the coefficients (3, 5, -7), serves as the direction vector for the line. The resulting parametric equations are x = 3 + 3t, y = -1 + 5t, and z = 1 - 7t, where t is a parameter.

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  • Knowledge of vector operations, specifically cross products
  • Familiarity with the concept of normal vectors in relation to planes
  • Basic algebraic manipulation skills
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marquitos
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Find parametric equations for the line through Po=(3,-1,1) perpendicular to the plane 3x+5y-7z=29.


Hey guys I am slightly confused on where to even start on this one the only way i can think about doing it would be finding two other coordinates on the and then taking the cross product of the two vectors but honestly I am not really sure how to obtain the other two points either. So any help would be nice i might just be over thinking it. Thank you.
 
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well i would go,

|x|...|3| ...|3|
|y|= |-1|+t|5|
|z|...|-1| ..|-7|

then you have x = 3+3t , y= -1+5t, and z = -1-7t

is this of any help?
 
Last edited:
marquitos said:
Find parametric equations for the line through Po=(3,-1,1) perpendicular to the plane 3x+5y-7z=29.


Hey guys I am slightly confused on where to even start on this one the only way i can think about doing it would be finding two other coordinates on the and then taking the cross product of the two vectors but honestly I am not really sure how to obtain the other two points either. So any help would be nice i might just be over thinking it. Thank you.

Remember that the normal to the plane (which is obvous, no?) will do for a direction vector for the line.
 

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