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The question is; The vectors $a_1 = (1, 1, 0)$ and $a_2 = (1, 1, 1)$ span a plane in $\Bbb R^3$. Find the projection matrix P onto the plane, and find a nonzero vector $b$ that is projected to zero.

I found the matrix $P$ as shown below, but I can't figure out how to find a nonzero vector $b$, I found only the zero vector, help me out?

enter image description here

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  • $\begingroup$ Taking the cross product of $a_1$ and $a_2$ is the easiest way to find such a vector, but another way would be to calculate the null space of the matrix. $\endgroup$ – user137731 Jul 14 '14 at 0:40
  • $\begingroup$ i guess null space is quicker. thanks $\endgroup$ – user124627 Jul 14 '14 at 1:57
  • $\begingroup$ I didn't calculate it, but I would have assumed that the cross product was quicker. The only reason I suggested you could calculate the null space, instead, was that it works in any dimension, while the cross product only works in $\Bbb R^3$ -- so if you later have a similar question about projections onto some subspace of say $\Bbb R^4$ then you'll already know how to answer it. $\endgroup$ – user137731 Jul 14 '14 at 2:08
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You need a vector that is perpendicular to the plane; i.e. perpendicular to the two vectors. Their cross product is perpendicular to the vectors!

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In this case, there's a shortcut:

We observe the vectors $(1,1,0)$ and $(1,1,1)$ have the same first two components. So $$(a,-a,0) \cdot (1,1,0)=0$$ and $$(a,-a,0) \cdot (1,1,1)=0$$ for any $a \in \mathbb{R}$.

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  • $\begingroup$ Just a suggestion, but you might consider editing in that you figured out that $(a,-a,0)$ would be the space of orthogonal vectors because the two vectors having the same 1st two components means that their projections onto the $xy$-plane are on the same line and thus that a vector in that plane orthogonal to that line will be orthogonal to both vectors (or use your own words to make it more intelligible), or else I doubt the OP will be able to see where you got $(a,-a,0)$ from and won't be able to apply this technique elsewhere. $\endgroup$ – user137731 Jul 14 '14 at 2:29

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