AB is a rod which is held such that $A=(1,-2,3)$ and $B=(2,3,-4)$ . A source of light is at the origin. Find the length of the shadow of the rod on a plane screen whose equation is $x+y+2z=1$

I figured out that origin and point B are on one side of given plane and point A is on other side. I found projection points of A and B and calculated distance between them but that is not matching with answer. Please help.


3 Answers 3


If what you say is correct, that the points A and B are on opposite sides of the plane, then the you answer will be wrong indeed, what you are looking for is not the projection of a vector representing the rod on the plane, but the shadow of the rod. Shadows are used to explain projections, but its a very particular analogy requiring specific circumstances. Draw it out for yourself and you'll see that actually the lenght you're looking for is the distance between the projection point of B on the plane, and the point where the the rod actually intersects the plane, as the part on the opposite side of the plane (opposite of the light source) won't cast a shadow on it


The rod can be described by the parametric curve $R(t)$ defined on $0\leq t \leq 1$: $$\begin{eqnarray*}R(t)&=&A+(B-A)t\\&=&\left(t+1,5t-2,-7t+3\right)\end{eqnarray*}$$ Now fix some $t_0\in [0,1]$. The light ray emanating from the origin that interacts with $R(t_0)$ has equation $tR(t_0):t\geq 0$ and intersects $x+y+2z=1$ whenever $t=\frac{1}{5-8t_0}$. This means the shadow will trace out some portion of the parametric curve $$\frac{1}{5-8t}R(t):t\in [0,1],t\neq \frac{5}{8}$$ The exact portion of this parametric curve can be determined by finding the values of $t$ for which $\frac{1}{5-8t}\geq 0$ and $\|R(t)\|\leq \|\frac{1}{5-8t}R(t)\|$ which gives us $t\in\left[\frac{1}{2},\frac{5}{8}\right)$. In other words, $$\frac{1}{5-8t}(t+1,5t-2,-7t+3):t\in \left[\frac{1}{2},\frac{5}{8}\right)$$ is your shadow.

  • $\begingroup$ Hi... This method is new to me... But seems good... Can you please give some references so that I can understand it better. $\endgroup$ May 31 at 8:54

You're correct!

O(origin) and B(2,3,-4) are on same side of the plane and A(1,-2,3) is on the opposite side of the plane.

Let us assume that the rod AB meets the plane at point X, and the line joining OB is extended to a point Y such that Y lies on the given plane.

If you try to draw it, you would find that the shadow is the line segment joining X and Y.

Now everything is simplified...we shall use vectors to find X and Y.

The general point on the line OB is represented as:


If we put this point on the equation of the plane, we will get the point of intersection of the line OB and the given plane, i.e. the point Y.

Thus,$$2\lambda+3\lambda-8\lambda=1$$ $$\implies\lambda=\frac{-1}{3}$$

Putting the value of $\lambda$ we get $\text{Y}=(-\frac{2}{3},-1,\frac{4}{3})$.

Similarly, with the general equation of line AB: $$\vec{r}=(1,-2,3)+\mu(2,3,-4)$$

Put this point on the equation of plane and we will get the coordinates of the point X.

Thus, $$1+2\mu-2+3\mu+6-8\mu=1$$ $$\implies \mu=\frac{4}{3}$$

Putting $\mu$ in the general point of AB, we get $\text{X}=(\frac{11}{3},2,\frac{-7}{3})$.

Now simply find the distance between X and Y to obtain the length of the shadow.Image for reference


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