I'm trying to prepare for my calculus 3 class coming up this fall and doing some practice problems. I'm having a hard time visualizing some of these 3D coordinates.

$D(0,-5,5)$ $E(1,-2,4)$ $F(3,4,2)$

The answer is yes and I drew out the coordinates to try and understand why this answer is a yes and although I can't 'see' why it's a yes I imagine it is because all the points appear to be on the same plane, yz?

Is this correct and do you have any suggestions that can help visualizing or is it just a matter of practice?


If you're comfortable with the definition of vectors in $3\text D$, then you may create vectors $\vec{DE} = (1,3,-1)$ and $\vec{EF} = (2,6,-2)$. If $\vec{DE}$ is a scalar multiple of $\vec{EF}$, then visually, you'd have two vectors pointing in the same (or complete opposite) direction. Remember that geometrically, a vector is a straight arrow with magnitude and direction.

So find $c$ such that $$c(1,3,-1) = (2,6,-2).$$ The solution is then $c = 2$ because $$2(1,3,-1) = (2,6,-2).$$ This means that $\vec{DE}$ points in exactly the same direction as $\vec{EF}$ which means that they lie on the same line in space.

  • $\begingroup$ Ahhh your answer is bringing back some linear algebra nightmares for me. This makes perfect sense. $\endgroup$ – hax0r_n_code Jul 23 '15 at 1:09

To show the three points are on the same line, you can form vectors with their coordinates and check that the vectors are parallel. It is not enough to show they are on the same plane, because three points may be on the same plane and not be on the same line (in fact, any three non-colinear points determine a plane).

For example, the points (0,0,1), (0,1,0), and (1,0,0) are on the same plane, but you can see that they are not on the same line.

Now, in your example, you can form the following vectors:

$$DE=(1-0,-2-(-5),4-5)=(1,3,-1)\\ DF=(3-0,4-(-5),2-5)=(3,9,-3)$$

By definition,two vectors $u,v$ are parallel if you can find a real number $k$ such that $u=kv$.

Can you find a $k$ such that $(3,9,-3)=k(1,3,-1)$?

  • 1
    $\begingroup$ This is a great explanation! Thank you for taking the time to explain this. $\endgroup$ – hax0r_n_code Jul 22 '15 at 0:19
  • $\begingroup$ @inquisitor thanks, and you're welcome :) $\endgroup$ – coldnumber Jul 22 '15 at 0:22

One way to check that all the points are co-linear is to verify that this determinant: $$\begin{vmatrix} 0 & - 5 & 5 \\ 1 & - 2 & 4 \\3 & 4 & 2\end{vmatrix} =0 .$$

  • $\begingroup$ Now that I read over your answer and the question again, I see what's going on. The title of the question asks whether the points given are in a line - for that reason I thought you probably wanted to say co-linear. Now reading it over, I see that there was no mention of "points" in the body of the question, and your answer (correctly) covers the case where $D$, $E$, and $F$ are vectors, not points. Sorry for the confusion. $\endgroup$ – Peter Woolfitt Jul 22 '15 at 0:22
  • $\begingroup$ @PeterWoolfitt Eventually, it seems you were right in the first place, because the OP refers to points in the second paragraph and I guess he means the points $D,E,F$. $\endgroup$ – thanasissdr Jul 22 '15 at 0:30

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