Find the point on the plane x-y+z=4 that is closest to the point (1,2,3).
.... Not sure how to go about this. Thanks!
Hello, jlt1209!
Find the point on the plane $\displaystyle x-y+z\:=\:4$ that is closest to the point $\displaystyle (1,2,3).$
The normal direction of the plane is: .$\displaystyle \vec{n} \:=\:\langle 1,-1,1\rangle $
The line through $\displaystyle (1,2,3)$ with $\displaystyle \vec{n} = \langle1,-1,1\rangle$ has equations: .$\displaystyle \begin{array}{ccc}x &=& 1 + t \\ y &=& 2 - t \\ z &=&3 + t \end{array}$
Substitute into the equation of the plane:
. . $\displaystyle (1+t) - (2-t) + (3+t) \:=\:4 \quad\Rightarrow\quad t \:=\:\tfrac{2}{3}$
Then: .$\displaystyle \begin{Bmatrix}x &=& 1 + \frac{2}{3} &=&\frac{5}{3} \\ \\[-4mm] y &=& 2 - \frac{2}{3} &=& \frac{4}{3} \\ \\[-4mm] z &=& 3 + \frac{2}{3} &=& \frac{11}{3} \end{Bmatrix} \quad\Rightarrow\quad \left(\frac{5}{3},\:\frac{4}{3},\:\frac{11}{3}\rig ht)
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