I have the point $(1,1,1)$ and the plane $2x+2y+z = 0$. I want to find a point that is closest to my point on the plane.
In other words, I want to find a point along the line $(1,1,1)+t(2,2,1)$ but on my plane. Notice that the vector $(2,2,1)$ is my normal vector and therefore I want to find the point parallel to this vector, but from my original point.
I need a nudge to complete this problem! Thank you.
$\endgroup$ 12 Answers
$\begingroup$Hint: A general point on the line has the form $(x,y,z) = (1 + 2t, 1 + 2t, 1 + t)$, and for this to be on the plane, it must satisfy the equation of the plane. Plug those in and solve for $t$.
$\endgroup$ 6 $\begingroup$Hint:
The distance between point $\;(a,b,c)\in\Bbb R^3\;$ and the plane $\;Ax+By+Cz+D=0\;$ is given by
$$\frac{|Aa+Bb+Cc+D|}{\sqrt{A^2+B^2+C^2}}$$
After you do the above (Result: $\;\sqrt3\;$), now just substitute in the euclidean distance formula with a general point on the plane .
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