Find an equation of the plane that passes through the point (1;2;3) and is parallel to the xy-plane. We are given a point in the plane. The normal vector must be perpendicular to the xy-plane, so we can use the direction vector for the z-axis, ~n = h0;0;1i. Thus, an equation of this plane is. Definition. An object h of the data type Plane_3 is. As mentioned earlier r is the sphere's radius; any line from the center to a point on the sphere is also called a radius.. If a radius is extended through the center to the opposite side of the sphere, it creates a diameter.Like the radius, the length of a diameter is also called the diameter, and denoted d.Diameters are the longest line segments that can be drawn between two points on. In this case, we have the points P (−1,−3,−1), Q(−12,12,6) and R(5,0,10) on the plane. Then the vectors PQ and PR are on the plane. Thus: v = PQ = −12 + 1,12+ 3,6+1 = −11,15,7 w = PR = 5+1,0+ 3,10 +1 = 6,3,11 Subsequently, the vectors v and w are parallel to the plane. Then any vector starting from P (x0,y0,z0) = P (−1,−3,−1) on the plane.
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