Find parametric equations for the line through (2, 4, 6) that is perpendicular t

Reeves

Reeves

Answered question

2021-09-28

Find parametric equations for the line through (2, 4, 6) that is perpendicular to the plane x-y+3z=7

Answer & Explanation

Roosevelt Houghton

Roosevelt Houghton

Skilled2021-09-29Added 106 answers

Remember that <a,b,c> is a normal vector to the plane with the scalar equation ax+by+cz=d
Therefore, the vector perpendicular to the given plane is <1,1,3>
If the passes through the point with position vector a and is parallel to b, then its equation is r=a+tb
Since the line passes through (2,4,6) and is parallel to <1,-1,3>, the vector equation of the line can be written as
r=<2,4,6>+t<1,1,3>
Simplify
r=<2,4,6>+<t,t,3t>
r=<2+t,4t,6+3t>
Therefore, the parametric equations are
x=2+t,y=4t,z=6+3t
Result: x=2+t,y=4t,z=6+3t

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