Find the vectors that have a given projection on the $xy$ plane

Question

I want to determine all the $x$ vectors that belong to $\mathbb R^3$ which have a projection on the $xy$ plane of $w=(1,1,0)$ and so that $||x||=3$.

I know the formula to find a projection of two vectors:

$$p_v(x)=\frac{\langle x, v\rangle}{\langle v, v\rangle}\cdot v$$

So I have the projection so I should be able to fill that in:

$$(1, 1, 0)=\frac{\langle x, v\rangle}{\langle v, v\rangle}\cdot v$$

Now I consider a generic vector $x = (x_1, x_2, x_3)$ and I calculate the dot products, though I don't exactly understand what $v$ is. I know it's a vector that it should be parallel to the projection of the vectors $x$, but not necessarily of the same length.

Any hints on how to proceed from here or if I'm doing the right thing? Should I use any formulas?

Answer 1

I assume you mean vectors $v$ which have the same projection $w = (1,1,0)$ onto the $xy$-plane:

$$P_{xy}v = w = (1,1,0)$$

So, you look for all $$w_t = (1,1,t) \mbox{ with } ||w_t|| = 3 \Leftrightarrow 1+1+t^2 = 9$$

Can you take it from there?

Answer 2

What you’ve given in your question is a formula for computing an orthogonal projection onto the vector $v$, but in this problem you’re projecting onto a plane, so you can’t use that formula, at least not directly.

Think about what it means geometrically to orthogonally project a vector $v$ onto the $x$-$y$ plane: You draw a line through through $v$ that’s perpendicular to the plane and then see where this line intersects the plane. Reversing this process, it should be clear that all of the vectors that have $w$ as their projection lie on a line through $w$ that’s perpendicular to the plane. You know that the $z$-axis is perpendicular to the $x$-$y$ plane, therefore this line can be given parametrically as $v = w+t(0,0,1)$. Set the length of this vector to $3$ and solve for $t$.