1. ## Vectors: Word Problem

Hi all,

I am currently studying for a first-year Math exam and am working through past exam questions. This one, however, has me totally perplexed (it's to do with planes and vectors and what-not):

A coordinate system is positioned so that the x-axis points North, the y-axis points West and the z-axis points vertically up. A man 2 meters high stands at the origin. The ground is sloping and has the equation x-y+4z = 0. How long is the man's shadow when the sun is due North at an elevation of 45 degrees?

2. Originally Posted by aceOfPentacles
Hi all,
...
A coordinate system is positioned so that the x-axis points North, the y-axis points West and the z-axis points vertically up. A man 2 meters high stands at the origin. The ground is sloping and has the equation x-y+4z = 0. How long is the man's shadow when the sun is due North at an elevation of 45 degrees?
I'm not sure if my reply is really helpful for you because you should have some knowledge about vector geometry if you want to do this problem. Nevertheless here we go:

1. The man is described by the vector $\vec m = (0,0,2)$ (Drawn in black).

2. The direction of the sunlight is described by $\vec d = (-1,0,-1)$.

3. The ray of sunlight which passes over the head of the man is described by the equation of the line l:

$lx,y,z)=(0,0,2)+t \cdot (-1,0,-1)" alt="lx,y,z)=(0,0,2)+t \cdot (-1,0,-1)" />

4. The equation of the plane p can be written as:

$p: x-y+4z=0~\implies~(1,-1,4)\cdot (x,y,z)=0$

5. Now calculate

$p\cap l~\implies~(1,-1,4)\cdot ((0,0,2)+t \cdot (-1,0,-1))=0~\implies~t=-\frac85$

6. The point where the ray hits the plane - that means the point where the shadow of the man ends - is $S\left(-\frac85\ ,\ 0\ ,\ \frac25 \right)$. So the length of the shadow is $|\vec s| = \sqrt{\left(-\frac85 \right)^2 + \left(\frac25 \right)^2}= \frac25 \cdot \sqrt{17} \approx 1.65\ m$ (Drawn in red)

3. Thanks earboth. I am in fact studying vector geometry and that was the sort of comprehensive answer I was hoping for! Although I'm not too sure how you got (-1, 0, -1) as the direction of the sunlight - is it because 45 degrees is similar to the line y = x?

4. Originally Posted by aceOfPentacles
Thanks earboth. I am in fact studying vector geometry and that was the sort of comprehensive answer I was hoping for! Although I'm not too sure how you got (-1, 0, -1) as the direction of the sunlight - is it because 45 degrees is similar to the line y = x?
Correct! The slope of the line y = x is m = 1 which correponds to the angle of 45° between the x-axis and the line.