parallel and normal projections
I have a vector $v$ given by $(v_x, v_y, v_z)$ which makes an angle $\theta$ with the $x$-axis. The projection of $v$ onto $x$ is given by the dot product
$$v\cdot x = \cos\theta\sqrt{v_x^2+v_y^2+v_z^2}$$
Say I want to find the projection of $v$ onto the $yz$-plane ($v$ has an angle $\pi/2 - \theta$ relative this plane). Is this simply given by $v\cdot (y+z)$?
EDIT: I know that $v\cdot x = v_x$, so I can find an expression for $\cos \theta$. Is the projection onto the $yz$-plane simply determined by $\sin \theta$, which I can find by $\cos^2 \theta + \sin^2 \theta = 1$?
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$\begingroup$Your notation could be a little confusing. If have the projection $\mathbf{v} \cdot \mathbf{\hat y}$ onto the $y$-axis, and the projection $\mathbf{v \cdot \hat z}$ onto the $z$-axis, then the projection of $v$ onto the $yz$ plane will be $$ \mathbf{v}_{\text{proj}} = (\mathbf{v \cdot \hat y})\, \mathbf{\hat y} + (\mathbf{v \cdot \hat z})\, \mathbf{\hat z}. $$ As you've written it, $v \cdot (y+z)$ appears to be a scalar, which is not what you want (I have given you a 3-vector, which you can reduce to a 2-vector by dropping the 0-term associated with the $x$ component).
This is a little general in case you want to project onto other planes. If you want the projection of a vector onto the $zy$ plane, just drop the $x$-component of the vector, so $\mathbf{v}_\text{proj} = (0, v_y, v_z)$.
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