A line passing through two points

[Guest2]

If the line passing through the points $p = (1,2,-1)$ and $q = (3,1,0)$ contains the point $(a, 4,-3)$ then what's the value of $a$?

I think $a = -3$. but I'm not sure.

$(1,2,-1)x+(3,1,0)y = (a,4,-3)$

$(x,2x,-x)+(3y,y,0) = (a,4,-3)$

$(x+3y, 2x+y, -x) = (a, 4,-3)$

so $x = -3$, $y = -2$ and $a = -3$

 

[Deveno]

Your answer is correct, although the methodology is not as good as you might hope.

Firstly, you have a typo, it should be $x = 3$.

Secondly, although the line going through $p$ and $q$ does lie in the plane generated by $p$ and $q$, it is not true that any point in the plane so generated lies on that line. So you kind of got lucky.

The line through $p$ and $q$ is:

$\{(x,y,z) \in \Bbb R^3: (x,y,z) = p + t(q-p), t \in \Bbb R\}$.

We have:

$q - p = (2,-1,1)$, so our line is:

$(1,2,-1) + t(2,-1,1)$

so: $(2t+1,2-t,t-1) = (a,4,-3)$

Either $2-t = 4$ or $t-1 = -3$ leads to $t = -2$, and consequently $a = 2t + 1 = 2(-2) + 1 = -4 + 1 = -3$.