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 Calculus Calculus Math Forum

 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 Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post BenFRayfield Number Theory 0 October 15th, 2016 11:46 AM maddieline Algebra 0 September 4th, 2013 11:16 AM edunwaigwe@yahoo.com Economics 1 December 26th, 2012 04:57 AM edunwaigwe@yahoo.com Algebra 2 December 26th, 2012 02:42 AM OriaG Calculus 2 September 4th, 2012 12:23 PM

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