Find the point on the plane x-y+z=4 that is closest to the point (1, 2, 3).

Question

boronganfh · Accepted Answer

The normal vector of the plane is also the directional vector of the perpendicular line.
v=(1,-1,1)
The equation of the line passing through a point P and parallel to the direction v is given by L(t)=P+tv where t ranges over real values.
L(t)=(1,2,3)+t(1,-1,1)=(1+2t, 2-t, 3+t)
Now that our point is located on both a line and a plane, we can use the equation of the line to replace the coordinate values in the equation of the plane and find the value of t.
x-y+z=4
(1+t)-(2-t)+(3+t)=4
2+3t=4
$t = \frac{2}{3}$
On substituting it in the equation we get the coordinate of the point.
$x = 1 + t = \frac{5}{3}$
$y = 1 - t = \frac{4}{3}$
$z = 3 + t = \frac{11}{3}$
Therefore, the point of intersection of the plane and the line is $Q (\frac{5}{3}, \frac{4}{3}, \frac{11}{3})$ . This location is also the one that is most near Point P.
Result:
$Q (\frac{5}{3}, \frac{4}{3}, \frac{11}{3})$

Find the point on the plane x-y+z=4 that is closest

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