Given the equation for the plane Π:3x-4y-5z-15 = 0 and P(1, 2, 1) a point in R³ I need to find the parametric equation for the line going through P and perpendicular to the plane Π. Any help on this would be great, I am really struggling with this question. Thanks!
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the standard equation of a plane is:
ax + by + cz + d = 0, where is the normal vector.
so, in the equation 3x - 4y - 5z - 15 = 0, the normal vector is <3,-4,-5>
given the line has the direction of <3,-4,-5> and passes through (1,2,1), the equation that represents this line is:
r(t) = <1,2,1> + <3,-4,-5>t
and from this, the parametric equation is:
x = 1 + 3t
y = 2 - 4t
z = 1 - 5t
I hope this helps （＾＾）/
ax + by + cz + d = 0, where
so, in the equation 3x - 4y - 5z - 15 = 0, the normal vector is <3,-4,-5>
given the line has the direction of <3,-4,-5> and passes through (1,2,1), the equation that represents this line is:
r(t) = <1,2,1> + <3,-4,-5>t
and from this, the parametric equation is:
x = 1 + 3t
y = 2 - 4t
z = 1 - 5t
I hope this helps （＾＾）/