Finding Points That Lie On A 3D Line?

Question

Which of the points $P(1, 2, 0), Q(−5, 1, 5), R(−4, 2, 5)$ lie on the line $l : r(t) = (i + 2j) + t(6i + j − 5k)?$

$l3 : r3(ν) = (6i − j) − ν(2i − 4j + 6k),$

$l4 : r4(w) = (1/2+ 1/2w)i − wj − (1 + 2/3w)k.$

What I have tried so far is to plug in numbers into the points but that has not resullted in any success towards the answer. What are some hints to figuring out what to do in this problem? Also what line do you plug in these coordinates?

Answer 1

Step 1: Convert the points $P$, $Q$, and $R$ into the vectors $\overrightarrow {OP}$, $\overrightarrow {OQ}$, and $\overrightarrow {OR}$. That's very easy to do.

Example: $$P(1,2,0) \longrightarrow \overrightarrow {OP} = i+2j$$

Step 2: Plug each of the $3$ vectors we found in step 1 into $r(t)$ and see if there is some $t$ that makes the equation hold. If there is, then that point is on the line $l$. If not, then the point is not on the line $l$.

Example: $$\begin{align}i+2j & \stackrel{?}= (i+2j)+t(6i+j−5k) \\ &= (1+6t)i+(2+t)j +(-5t)k\end{align}$$

We can immediately see here that setting $t=0$ will make this equation hold. So clearly $P$ is a point on line $l$.

Step 3: Repeat step 2 for lines $l_3$ and $l_4$.

Answer 2

For the first one:

$r(t) = (1,2,0) + t(6,1,-5) = (6t+1, t+2, -5t)$

Therefore, $x = 6t+1$, $y = t+2$ and $z=-5t$

Solving for $t$ gives $\frac{x-1}{6} = y-2 = \frac{-z}{5}$.

Now just plug the values for $x$, $y$ and $z$ into the equation to see that $P$ and $Q$ are on this line but $R$ is not.