So, the title is: Parallel Vectors: Is AP Parallel to u?

[negation]

The vector PA: <p₁+2, p₂+1,p₃+2>

<p₁+2, p₂+1,p₃+2> = k<1,4,-3>
<p₁+2, p₂+1,p₃+2> = <γ,4γ,3γ>

p₁+2 = k
p₂+1 = 4k
p₃+2 = -3k

Suppose <p1,p2,p3> = <1,2,3>

The k values are inconsistent.

 

[Mark44]

The vector PA: <p₁+2, p₂+1,p₃+2>

This is actually $\vec{AP}$.

$\vec{PA}$ would be <-2 - p₁, -1 - p₂, -2 - p₃>. These two vectors point in the opposite directions, so $\vec{AP} = - \vec{PA}$.

<p₁+2, p₂+1,p₃+2> = k<1,4,-3>
<p₁+2, p₂+1,p₃+2> = <γ,4γ,3γ>

From the previous equation, the right side above should be <k, 4k, -3k>.

p₁+2 = k
p₂+1 = 4k
p₃+2 = -3k

Suppose <p1,p2,p3> = <1,2,3>

$\vec{AP}$ and $\vec{u}$ are parallel, so $\vec{AP} = k\vec{u}$ for some scalar k, with k ≠ 0.

Instead of picking some random values for the coordinates of $\vec{AP}$, pick a value for k.

The k values are inconsistent.

 

[LCKurtz]

The vector PA: <p₁+2, p₂+1,p₃+2>

That isn't PA, it is AP, which is what you want.

<p₁+2, p₂+1,p₃+2> = k<1,4,-3>
<p₁+2, p₂+1,p₃+2> = <γ,4γ,3γ>

p₁+2 = k
p₂+1 = 4k
p₃+2 = -3k

Suppose <p1,p2,p3> = <1,2,3>

The k values are inconsistent.

Big surprise they are inconsistent. Where did <1,2,3> come from? Did you just pick it out of the air?? It's not surprising that guessing values for the p's doesn't work.

To solve the problem, you need to find any $p_1,p_2,p_3$ values and $k$ that do work.

[Edit]Took too long typing this, Mark beat me to it.

 

[negation]

This is actually $\vec{AP}$.

$\vec{PA}$ would be <-2 - p₁, -1 - p₂, -2 - p₃>. These two vectors point in the opposite directions, so $\vec{AP} = - \vec{PA}$.
From the previous equation, the right side above should be <k, 4k, -3k>.
$\vec{AP}$ and $\vec{u}$ are parallel, so $\vec{AP} = k\vec{u}$ for some scalar k, with k ≠ 0.

Instead of picking some random values for the coordinates of $\vec{AP}$, pick a value for k.

let k = 1

p₁+2 = 1 ∴p₁=-1
p₂+1=4 ∴p₂=3
p₃+2=-3 ∴p₃=-5

I'm very sure it doesn't end here.

 

[Mark44]

Way back in the OP, the question was, "Is the vector $\vec{AP}$ parallel to u?" Apparently the coordinates of P were not given (or at least, they weren't shown in this thread). I believe the answer would be showing the conditions on the coordinates of P, something like this:
p₁ = ...
p₂ = ...
p₃ = ...

These three equations would involve the parameter k and the coordinates of u.

It has taken a long while and many posts in this thread to get to this point, so a summary of where you're going might be useful.

1. Two vectors are parallel if either one is a scalar multiple of the other. In light of this problem, the equation is $\vec{AP} = k\vec{u}$, where A is the point (-2, -1, -2) and u is the vector <1, 4, -3>. The point P wasn't given, so we used variables to represent it: P(p₁, p₂, p₃).
2. Write the position vector $\vec{AP}$, which is <p₁ - (-2), p₂ - (-1), p₃ - (-2)>, and set this equal to ku, which is <k, 4k, -3k>.
3. Solve the equation for the unknowns p₁, p₂, and p₃.

 

[LCKurtz]

Way back in the OP, the question was, "Is the vector $\vec{AP}$ parallel to u?" Apparently the coordinates of P were not given (or at least, they weren't shown in this thread). I believe the answer would be showing the conditions on the coordinates of P, something like this:
p₁ = ...
p₂ = ...
p₃ = ...

And, of course, that's assuming we have guessed correctly what the actual statement of the problem was supposed to be n the first place.

 

[Mark44]

And, of course, that's assuming we have guessed correctly what the actual statement of the problem was supposed to be n the first place.

Well, there's that...