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?
Best Answer
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.