My Math Forum Plane in three-dimensional space

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 April 10th, 2019, 03:26 AM #1 Newbie   Joined: Apr 2019 From: Brazil Posts: 1 Thanks: 0 Plane in three-dimensional space Determine an equation of the plane whose points are equidistant from (2, -1, 1) and (3, 1, 5). -- Answer: x + 2y + 4z = 29/2
 April 10th, 2019, 09:00 AM #2 Member     Joined: Oct 2018 From: USA Posts: 90 Thanks: 62 Math Focus: Algebraic Geometry Since we want the plane to be equidistant, we know the plane passes through the midpoint of the line between the two points : $(2.5,0,3)$ Also, the plane has to be perpendicular to the line between the points, otherwise it won't be equidistant. To find the plane we need the vector between the two points: $\vec{u} = \langle 3-2 , 1-(-1) , 5-1 \rangle = \langle 1,2,4 \rangle$ The coefficients of planar equations are determined by the elements of their perpendicular vectors, so the equation for the plane is $1x + 2y + 4z = ?$. Since we know the midpoint is on the plane we can use it to calculate "?". $2.5 + 2(0) + 4(3) = 14.5 = \frac{29}{2}$. Which gives the full equation $x+2y+4z = \frac{29}{2}$ EDIT: Also, you can also just use the general form for a plane, where $\langle a,b,c \rangle$ is the normal vector and $(x_{0},y_{0},z_{0})$ is the starting point. $a\left( {x - {x_0}} \right) + b\left( {y - {y_0}} \right) + c\left( {z - {z_0}} \right) = 0$ Thanks from topsquark and wniemiec Last edited by Greens; April 10th, 2019 at 09:10 AM. Reason: Additional Info

 Tags plane, planes, plans, space, threedimensional

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