The question ( part (ii) ) above challenges me on my ability to think in vectors. The solution to it gives me some confusion, and I'd appreciate if any light can be shed on it. Here was what my thought process was:
- The shortest distance between two points is a straight line. This must mean in order to solve this problem the normal line for plane 1 must be a scalar of the normal line for plane 2. If this wasn't the case, I don't know how I could solve this.
Thankfully it is, and when I attempted the problem I established the following:
$2\overline n_1 = \overline n_2$
This is true by inspection of the plane equations. Now, my next thought:
- I can create a vector equation for a line to find a point on either plane, using any $n$ as the parallel vector.
- I can arbitrarily pick a point on one of the planes, and use it as $a$ in $$r = a + t(n)$$
- If I set $t = 2$, then I'll get $n_2$ if I solve $r$ now, and so I should perhaps get coordinates for the other point I'll need on plane 2.
I chose the point on plane 1 to be (1, 2, 1). Solving for $r$ with $t(n) = n_2$ gives me (3,6,-3).
Okay, so that's point 2. If I take the difference between those two points, namely $-(1, 2, 1) + (3, 6, -3)$, I'll have the vector I need. Taking the magnitude of it should give me that distance.
When all was said and done, my distance came out to be 6. However, here is a picture of the solution attached to this worksheet:
I must say I'm confused. To find the magnitude, my final answer of 6 was from $\sqrt{4+16+16}$ which is almost identical to their answer, except there's had a $t^2$ on each term. They also used the Cartesian equation of their line for stuff. My questions are as follows:
- Is any of my logic flawed?
- Why is it implied we are solving for $t$ here?
- Where does my answer and my lecturer's diverge? I understand our methodology was clearly different, but I'm not entirely sure about the thought process behind his answer.
Best Answer
There are some formula worth knowing
$ax+by+cz - d = 0$ is the equation of a plane
$\mathbf n = (a,b,c)$ is the vector normal to the plane
$\sqrt {a^2+b^2+c^2} = \| \mathbf n \|$
$d = \mathbf n \cdot \mathbf p$ where $\mathbf p$ is any point on your plane.
if $\mathbf q = (x,y,z)$ is a point not on your plane
$\mathbf u =(\mathbf q - \mathbf p)$ is the vector from $\mathbf p$ on your plane to $\bf q$
$\|\mathbf u\|\cos \theta = \frac {\mathbf u\cdot \mathbf n}{\|\mathbf n\|}$ is the length of the vector $\mathbf u$ pojected onto $\mathbf n$
$\mathbf u\cdot \mathbf n = (\mathbf u\cdot \mathbf q - d) = (ax+by+cz - d)$
And from this we derive the following.
$D = \frac {|ax+by+zc - d|}{\sqrt {a^2+b^2+c^2}}$
if: $ax+by+cz - d = 0$ and $ax+by+cz - e = 0$ are parallel planes
$D = \frac {|d-e|}{\sqrt {a^2+b^2+c^2}}$