I have a line $p$ defined below:

$p... \frac {x - 1}{1} = \frac {y}{-2} = \frac {z - 1}{2}$

I have to find the common normal between the line $p$ and the $z-axis$.

I was thinking like this:

I express the z-axis in vector form: $(0,0,0) + \lambda (0,0,1)$

Then, I can use the direction vector of the line ($(1, -2, 2)$) and of z-axis ($(0, 0, 1)$) and compute the vector product to get the normal that is common between the line p and the z-axis.

However, this doesn't yield the correct solution. The authors expressed the solution in terms of intersecting planes (where they intersect, there's where my normal is).

The planes are defined as such:

$\Pi 1 ... 2x - 4y - 5z + 3 = 0$

$\Pi 2 ... x - 2y = 0$

Can someone explain why do I need the 2 planes and where did I go wrong in my approach?

  • 1
    $\begingroup$ You didn't went wrong. The directional vector of the straight line is $(1,-2,2)$, so obviously any multiple of $(-2,-1,0)$ (wich is, btw, both the vectors' cross product and the directional vector of the intersection of the planes) is both perpendicular to the line and the $z$-axis. Please quote the full question. $\endgroup$ – Michael Hoppe Apr 1 '17 at 9:51
  • $\begingroup$ The question is not clear. We can have infinitely many lines orthogonal to the two given lines. $\endgroup$ – Emilio Novati Apr 1 '17 at 10:00
  • $\begingroup$ I think the question is asking for the line which intersects both given lines and the distance between those intersected points is the least. It would then be common normal between those lines. $\endgroup$ – Prajwal Kansakar Apr 1 '17 at 10:00
  • $\begingroup$ Maybe the more accurate translation of the task text would be: "Determine the equation of the normal common to the line p and the z-axis". $\endgroup$ – NumberSymphony Apr 1 '17 at 13:12
  • $\begingroup$ @EmilioNovati - you can have infinitely many lines orthogonal in direction to the two lines, but only one of them also intersects both of those lines. $\endgroup$ – Paul Sinclair Apr 1 '17 at 14:26

Line $p$: $\dfrac{x-1}{1}=\dfrac{y}{-2}=\dfrac{z-1}{2}$ with direction vector $\vec{A}=\vec{i}-2\vec{j}+2\vec{k}$

$z$-axis: $\dfrac{x}{0}=\dfrac{y}{0}=\dfrac{z}{1}$ with direction vector $\vec{B}=\vec{k}$

The common normal has direction vector $\vec{C}=\vec{A}\times \vec{B}=-2\vec{i}-\vec{j}$. The plane containing line $p$ and common normal passes through $(1, 0, 1)$ and is normal to the vector $\vec{A}\times \vec{C}=2\vec{i}-4\vec{j}-5\vec{k}$ so has equation $2x-4y-5z=-3$.

Similarly the plane containing the $z$ axis and common normal passes through $(0, 0, 0)$ and is normal to the vector $\vec{B}\times \vec{C}=\vec{i}-2\vec{j}$ with equation $x-2y=0$. The common line is the intersection of these two planes.

  • $\begingroup$ @NumberSymphony - PJK has it right. You were correct as far as you went, but you didn't go all the way. The direction of the normal is the cross-product, but the question is asking for the full specification of the common normal, so you also need a point on it. The intersection of planes is just the particular method the book used to find both direction and point. $\endgroup$ – Paul Sinclair Apr 1 '17 at 14:29

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