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We have two non-parallell vectors $\vec{a}$ and $\vec{b}$ in three dimensional space, together with an arbitrary vector $\vec{c}$. I want to find the orthogonal projection $\vec{c'}$ of $\vec{c}$ on the plane that is spanned by $\vec{a}$ and $\vec{b}$, where $\vec{c'}$ is a linear combination of $\vec{a}$ and $\vec{b}$.

I have tried to approach this problem in a couple of ways.

My first thought was to project c onto the normal vector, $\vec{n}$, of the plane and then take $\vec{c}-proj_{n}\ \vec{c}$ which would give us a vector in the plane, i.e the projection $\vec{c'}$ of $\vec{c}$ onto the plane. However, with this approach, I cannot rewrite it as a linear combination of $\vec{a}$ and $\vec{b}$.

Another approach I have used is project $\vec{c}$ onto $\vec{a}$ and $\vec{b}$, respectively, and then add them together. With this approach I get it as a linear combination of $\vec{a}$ and $\vec{b}$, but this is not the right answer.

Any help and tips would be greatly appreciated!

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  • $\begingroup$ What do you mean with “the projection”? Do you mean the orthogonal projection? $\endgroup$ Oct 21 '19 at 17:40
  • $\begingroup$ The projection of c, that lies in the plane that is spanned by $a$ and $b$. $\endgroup$ Oct 21 '19 at 17:42
  • $\begingroup$ That occurs for every projection and there are infinitely many of them. $\endgroup$ Oct 21 '19 at 17:44
  • $\begingroup$ Yes, I meant the orthogonal projection. $\endgroup$ Oct 21 '19 at 17:44
  • $\begingroup$ Then edit your question and add that hypothesis to it. $\endgroup$ Oct 21 '19 at 17:45
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Since $\vec{a}$ and $\vec{b}$ are non-parallel, you can write $$\vec{c}' = p \vec{a} + q \vec{b}.$$ And since $\vec{c} - \vec{c}'$ is perpendicular to both $\vec{a}$ and $\vec{b}$, $$\vec{c} - \vec{c}' = r \vec{a} \times \vec{b}$$ for some $r$. Thus, $$\vec{c} = p \vec{a} + q \vec{b} + r \vec{a} \times \vec{b}.$$

You can take the dot product with $\vec{a}$, to get $$\vec{a} \cdot \vec{c} = p (\vec{a} \cdot \vec{a}) + q (\vec{a} \cdot \vec{b}).$$ You can also take the dot product with $\vec{b}$, to get $$\vec{b} \cdot \vec{c} = p (\vec{a} \cdot \vec{b}) + q (\vec{b} \cdot \vec{b}).$$ You can then solve the linear system to get $p$ and $q$, which gives you $\vec{c}'$.

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