Vector, parametric, and symmetric equations of a 3D line

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CarissaF

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Sep 22, 2016
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Question:
Write the vector, parametric, and symmetric equations of the line which passes through the intersection of L1 and L2 and is orthogonal to both lines.
L1: x=t, y=-2+2t, z=1+t
L2: x=2+s, y=2-s, z=3+2s

Attempt at solution:
In order to find the equation of the unknown line you need a vector for the line V=<a,b,c> and a point on that line P=(x,y,z)
I found the point on the line to be P=(2,2,3) but am having trouble finding the vector.

I know if the unknown line is orthogonal to L1 and L2 then the dot products should = 0 so I have
a+2b+c=0 <----The dot product of V=<a,b,c> and V1=<1,2,1>
a-b+2c=0 <----The dot product of V=<a,b,c> and V2=<1,-1,2>
This leaves me with 3 unknowns and 2 equations. Is there another equation I am missing so I can finish this problem?

Thank you so much!
 
CarissaF said:
Question:
Write the vector, parametric, and symmetric equations of the line which passes through the intersection of L1 and L2 and is orthogonal to both lines.
L1: x=t, y=-2+2t, z=1+t
L2: x=2+s, y=2-s, z=3+2s

Attempt at solution:
In order to find the equation of the unknown line you need a vector for the line V=<a,b,c> and a point on that line P=(x,y,z)
I found the point on the line to be P=(2,2,3) but am having trouble finding the vector.

I know if the unknown line is orthogonal to L1 and L2 then the dot products should = 0 so I have
a+2b+c=0 <----The dot product of V=<a,b,c> and V1=<1,2,1>
a-b+2c=0 <----The dot product of V=<a,b,c> and V2=<1,-1,2>
This leaves me with 3 unknowns and 2 equations. Is there another equation I am missing so I can finish this problem?

Thank you so much!
Click to expand...
Do you know about cross-product ?
 
CarissaF said:
Question:
Write the vector, parametric, and symmetric equations of the line which passes through the intersection of L1 and L2 and is orthogonal to both lines.
L1: x=t, y=-2+2t, z=1+t
L2: x=2+s, y=2-s, z=3+2s
Click to expand...

The direction vector of \(\displaystyle L_1 \) is \(\displaystyle D_1: <1,2,1> \).
The direction vector of \(\displaystyle L_2 \) is \(\displaystyle D_2: <1,-1,2> \)

Now the direction vector of the wanted line is \(\displaystyle N=~D_1\times D_2 \).

If \(\displaystyle P: (2,2,3) \) is the point on both lines, then the required line is \(\displaystyle P+t\cdot N \).
 
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