Writing parallel vector equation

[Physics345]

Homework Statement

Write a vector equation for the line that passes through the point P(–1, 0, 3) and is parallel to the y-axis.

Homework Equations

(x,y)=(x_0,y_0)+t(a,b)

The Attempt at a Solution

u ⃗=(0,1,0)
(x,y,z)=(-1,0,3)+t(0,1,0)

 

[Orodruin]

Did you have a question? You seem to have solved it already.

 

[Physics345]

I was just confirming it. Thanks for letting me know, I guess I was unsure for no reason, basically in this case y can equal anything except for 0 for it to be parallel to the y-axis?

 

[Orodruin]

basically in this case y can equal anything except for 0 for it to be parallel to the y-axis?

Do you mean the y-component of the tangent vector? In that case yes. It would only be a reparametrisation of the line.

As for the y-value of the line, it can take any value - including 0 - depending on the value of the curve parameter.

 

[Physics345]

Do you mean the y-component of the tangent vector? In that case yes. It would only be a reparametrisation of the line.

As for the y-value of the line, it can take any value - including 0 - depending on the value of the curve parameter.

Good to know, and yes I am referring to the y-component of the tangent vector. I was just confirming my understanding.

 

[Tonyb24]

if it passes through that point, then why not use that point as the support point?
Then, what is a point on the Y axis that you can multiply to reach any y-axis point?
So now, use your support point P and add to it a vector that can be multiplied by t where t E R, that will allow you to reach anypoint on the line (like a unit vector or something). L:(x,y,z)=P+t(x,y,z);t E R.

 

[Orodruin]

if it passes through that point, then why not use that point as the support point?

It is exactly what he did.

Then, what is a point on the Y axis that you can multiply to reach any y-axis point?

Again, it is exactly what he did.

So now, use your support point P and add to it a vector that can be multiplied by t where t E R, that will allow you to reach anypoint on the line (like a unit vector or something). L:(x,y,z)=P+t(x,y,z);t E R.

Which is exactly what he got:

(x,y,z)=(-1,0,3)+t(0,1,0)