# Finding the orthogonal projection $c'$ of $c$ on a plane that is spanned by $\vec{a}$ and $\vec{b}$

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!

• What do you mean with “the projection”? Do you mean the orthogonal projection? Oct 21, 2019 at 17:40
• The projection of c, that lies in the plane that is spanned by $a$ and $b$. Oct 21, 2019 at 17:42
• That occurs for every projection and there are infinitely many of them. Oct 21, 2019 at 17:44
• Yes, I meant the orthogonal projection. Oct 21, 2019 at 17:44
• Then edit your question and add that hypothesis to it. Oct 21, 2019 at 17:45

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 $$\vec{c}'$$.