How can I compare the vector equation of a line to a vector in the form (x,y,z)

[RufusDawes]

I am having trouble understanding how the vector equation of a line : (x,y,z) + t(x1,y1,z1) can be used in conjuction with another vector in the form (x,y,z).

I visualise it like this. Points are : O to P and O to Q. So there are two vectors. The vector P to Q is given by Q - P which is the same as (x1,y1,z1).

So that means if someone gives me a vector (x2,y2,z2) to determine if it is perpendicular or parallel I can just use (x1,y1,z1) and do the necessary tests dot product = 0 for perp. and cross product = 0 for parallel ?

Is my understanding correct ?

 

[LCKurtz]

I am having trouble understanding how the vector equation of a line : (x,y,z) + t(x1,y1,z1) can be used in conjuction with another vector in the form (x,y,z).

That isn't an equation since there is no equals sign. The vector equation of a line has the form

<x,y,z> = <x₀,y₀,z₀> + t<d₁,d₂,d₃>

where (x₀,y₀,z₀) is a point on the line and <d₁,d₂,d₃> is a direction vector for the line.

I visualise it like this. Points are : O to P and O to Q. So there are two vectors. The vector P to Q is given by Q - P which is the same as (x1,y1,z1).

Yes. OP is the position vector <x₀,y₀,z₀> and P is the point (x₀,y₀,z₀). And the coordinates of Q-P are the components of the direction vector <d₁,d₂,d₃>.

So that means if someone gives me a vector (x2,y2,z2) to determine if it is perpendicular or parallel I can just use (x1,y1,z1) and do the necessary tests dot product = 0 for perp. and cross product = 0 for parallel ?

Is my understanding correct ?

Yes. Although the test for parallel is easier performed by checking that one vector is a multiple of the other; that's easier than the cross product.

Also note the distinction between the point P = (x₀,y₀,z₀) and the position vector
OP=<x₀,y₀,z₀> It is best not to confuse points with position vectors even though the components of the position vector are the same as the coordinates of the point.