Calculus III: Find a line perdendicular to XY-plane?

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Homework Statement [/b]
"Find an equation for the line through the point P = (1, 0, −3) and perpendicular
to the xy-plane,"

obviously this includes vector <0, 0, 1>

I am in Calc III and need help understanding how to do this TYPE of problem. Please include step-by-step instructions and any concepts used. Thank you.
 

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  • #2
SteamKing
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Why don't you try drawing a sketch? If a line is perpendicular to the x-y plane, how do the values of x and y change when the z-coordinate varies?
 
  • #3
haruspex
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"Find an equation for the line through the point P = (1, 0, −3) and perpendicular
to the xy-plane,"
obviously this includes vector <0, 0, 1>
It does? Did you mean <1, 0, 0>?
More generally, what can you say about the values of x and y on that line?
I assume it's a vector equation you're after?
 
  • #4
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More generally, what can you say about the values of x and y on that line?
I assume it's a vector equation you're after?
Yes it's a vector equation I am after.

my thought is: r = r0 + tv meaning r=<1,0,-3> + t<0,0,1>

I just don't know if I'm right. Need clarification
 
  • #5
haruspex
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Yes it's a vector equation I am after.

my thought is: r = r0 + tv meaning r=<1,0,-3> + t<0,0,1>
Sure, or just r=<1,0,t-3>. But that's a parametric equation, which might not be what's wanted. Another form might be r x a = b, for some constant vectors a and b.
 
  • #6
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Sure, or just r=<1,0,t-3>. But that's a parametric equation, which might not be what's wanted. Another form might be r x a = b, for some constant vectors a and b.
Still highly confused.
 
  • #7
haruspex
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Can you find constant vectors a and b such that the equation r x a = b implies r is of the form <1,0,*>?
 
  • #8
LCKurtz
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Yes it's a vector equation I am after.

my thought is: r = r0 + tv meaning r=<1,0,-3> + t<0,0,1>

I just don't know if I'm right. Need clarification
Yes, that's exactly right. <1,0,-3> is a position vector to the point and <0,0,1> is a correct direction vector.
 
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