Orthogonal projection of point on line

Question

With use of the dot product, find the orthogonal projection Q of the point P = (-2,4,3) on the line $g=\{(x_1,x_2,x_3)\,|\,(x_1,x_2,x_3)=(1,0,0)+\lambda(2,2,1)\}$

Hint: on what vector should PQ stand vertically ?

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$g$ doesn't pass through the origin, so I guess I should do some kind of "translation" with (1,0,0), but I don't really see what to do exactly.

PS: I don't know why, but I thought about using Householder's matrix.Could you somehow solve this using Householder ?

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EDIT: Question solved, thanks for your help !

Answer 1

HINT

PQ must be orthogonal to v(2,2,1)

Answer 2

I will follow your idea. Let $P^\star=P-(1,0,0)=(-3,4,3)$ and let $g^\star=g-(1,0,0)=\mathbb{R}(2,2,1)$. Let $u=\frac{(2,2,1)}{\bigl\|(2,2,1)\bigr\|}=\left(\frac23,\frac23,\frac13\right)$. Then $g^*$ is a subspace of $\mathbb{R}^3$ and the orthogonal projection of $P^\star$ on $g$ is$$(P^\star.u)u=\left(\frac{10}9,\frac{10}9,\frac59\right).$$So, the point $Q$ is$$\left(\frac{10}9,\frac{10}9,\frac59\right)+(1,0,0)=\left(\frac{19}9,\frac{10}9,\frac59\right)$$