How do I find orthogonal projection of a vector $\vec V_1=(2,3,4)^T$ formed with the points $A(0,0,5)$ and $B(2,3,9)$ on $xy$ plane?
$\endgroup$2 Answers
$\begingroup$In general, let $P$ be some plane through the origin defined by perpendicularity to some vector $n$. The orthogonal projection of any vector $v$ to $P$ is given by subtracting from $v$ the component of $v$ perpendicular to $P$, i.e., the component of $v$ in the direction of $n$.
The component of $v$ in the direction of $n$ is the projection of $v$ onto $\mathbb{R}n$. This is $\langle v,n\rangle n$. So the projection of $v$ onto $P$ is $v - \langle v,n\rangle n$.
In your case, $P$ is the $xy$-plane, $n = (0,0,1)$, and $v = (2,3,4)$, so $v - \langle v,n\rangle n = (2,3,4) - \langle (2,3,4),(0,0,1)\rangle (0,0,1) = (2,3,4) - (0,0,4) = (2,3,0)$ as in Agusti Roig's answer.
$\endgroup$ $\begingroup$This way:
$$ (2,3,0) \ . $$
Because
$$ (2, 3, 4) - (2, 3, 0) = (0, 0, 4) $$
is orthogonal to the $xy$-plane.
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