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I was wondering if anyone could shine some light on the correct thought process for this problem.

The vector equation can be expressed as the following:

$$r = a + t(b - a)$$

If I want to find a point on this line, I look to $r$. In addition, I can relate each projection of $r$ in the $x, y$ and $z$ axes using:

$r_x = a_x + t(n_x)$

$r_y = a_y + t(n_y)$

$r_z = a_z + t(n_z)$

However, how can I tweak these values to satisfy them being 3 units away from A? Do I make $r_x = 3 - a_x$ for each axis? If not, why does that not work? I feel like that would be logical, but I don't know how that could get me anywhere. Would appreciate any guidance on how to express this condition asked for in the problem.

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  • $\begingroup$ Here is an explanation. $\endgroup$ – Théophile May 10 '17 at 1:40
  • $\begingroup$ @Théophile I'm having a bit of trouble understanding it. Can you possibly reiterate? $\endgroup$ – sangstar May 10 '17 at 2:33
  • $\begingroup$ Use a unit vector instead of $b-a$ in your equation. Then the parameter $t$ measures the distance from $a$. $\endgroup$ – amd May 10 '17 at 5:12
  • $\begingroup$ In other words, try using $\mathbf r = \mathbf a + t\frac{\mathbf b - \mathbf a}{\lVert \mathbf b - \mathbf a \rVert}$. $\endgroup$ – Théophile May 10 '17 at 14:06
  • $\begingroup$ Okay, so I'm taking the unit vector of this line, and the unit vector, when scaled to ||$b-a$||, is at $a$, so then is the scalar for this? If this is true, how do I now make it so it's 3 units away? $\endgroup$ – sangstar May 10 '17 at 18:56
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"A direction" of the line $\overleftrightarrow{AB}$ is $B-A=(0,-3,0)$.

(Another direction is $A-B$. But we won't use that.)

So every point on the line $\overleftrightarrow{AB}$ can be expressed as

$$P(t)=A +t(B-A) = (1,2,-2)-t(0,3,0)$$

The distance between the point $P(t)$ and $A$ is $||P(t)-A||=||t(0,3,0)|| = 3|t|.$

So the points on the line that are a distance of $3$ from $A$ satisfy the equation $3|t|=3$. Hence $t\in \{1, -1\}$.

The points are $(1,2,-2) \pm (0,3,0)$

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