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Lines and Planes in Space Calculator

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Example
Created on 2024-06-20Asked by Aiden King (Solvelet student)
Find the equation of the plane that passes through the point (1,2,3) (1, 2, 3) and is perpendicular to the vector n=4,5,6 \mathbf{n} = \langle 4, 5, 6 \rangle .

Solution

To find the equation of a plane that passes through the point (1,2,3) (1, 2, 3) and is perpendicular to the vector n=4,5,6 \mathbf{n} = \langle 4, 5, 6 \rangle , we use the general form of the equation of a plane: ax+by+cz=d ax + by + cz = d where n=a,b,c \mathbf{n} = \langle a, b, c \rangle . Given n=4,5,6 \mathbf{n} = \langle 4, 5, 6 \rangle , we have: a=4,b=5,c=6 a = 4, \, b = 5, \, c = 6 The point (1,2,3) (1, 2, 3) lies on the plane, so we substitute these coordinates into the equation to find d d : 4(1)+5(2)+6(3)=d 4(1) + 5(2) + 6(3) = d 4+10+18=d 4 + 10 + 18 = d d=32 d = 32 Therefore, the equation of the plane is: 4x+5y+6z=32 4x + 5y + 6z = 32 Solved on Solvelet with Basic AI Model
Some of the related questions asked by Luna Carter on Solvelet
1. Find the equation of the line passing through the points (1,2,3) (1, 2, 3) and (2,1,4) (-2, 1, 4) .2. Determine whether the line with parametric equations x=3t+1x = 3t + 1, y=2t2y = 2t - 2, and z=t+3z = -t + 3 intersects the plane 2x+yz=52x + y - z = 5.
DefinitionLines and planes in space are geometric objects in three dimensions. A line is specifiable by a direction vector and a point,and a plane is specifiable by a normal vector and a point. The positions and orientations of lines and planes are described by their equations. Describe an example: The line through the point (1,2,3) and with direction vector (4,5,6) is given by vector equation r(t)=(1,2,3)+t(4,5,6). The plane passing through (1,2,3) with normal vector (4,5,6) is given by 4(x−1)+5(y−2)+6(z−3)=0.
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