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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?

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2 Answers

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.

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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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@ Agusti roig : how did you get the value of (2,3,0).I need some explanation –  sairaj Mar 1 '12 at 2:51
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Just delete the $4$ from the third coordinate of your vector and put a zero instead. –  a.r. Mar 1 '12 at 3:04
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