(a) The cube can be described as a set of points in 3-dimensional space where each coordinate (x, y, z) satisfies -1 ≤ x ≤ 1, -1 ≤ y ≤ 1, and -1 ≤ z ≤ 1.
(b) Let O be the origin of the coordinate system. Then,The vector −→OE can be written as OE = 1i + 0j + 0k.The vector −−→OD can be written as OD = 0i + 1j + 0k.The vector −−→OF can be written as OF = 0i + 0j + 1k.The vector −−→OG can be written as OG = 1i + 1j + 1k.
(c) The center point of the cube can be found by taking the average of the coordinates of opposite corners. Let A = (-1, -1, -1) be one corner of the cube and B = (1, 1, 1) be the opposite corner. Then the coordinates of the center point C are:C = ((-1 + 1)/2)i + ((-1 + 1)/2)j + ((-1 + 1)/2)k = 0i + 0j + 0k = (0, 0, 0)
(d) Let D, E, and F be the vertices of the cube that are adjacent to A, B, and C respectively. Then the point P that is the vector sum of −→OA, −−→OB and −−→OC is:P = A + OD + OF = (-1, -1, -1) + (0, 1, 0) + (0, 0, 1) = (-1, 0, 0)
(e) The plane of all linear combinations of −→OA and −→OA + −−→OB is the plane that passes through points A, O, and A + B. The normal vector of this plane is the cross product of vectors OA and (OA + OB), which is:OA x (OA + OB) = (-1i - 1j - 1k) x (0i + 0j + 2k) = 2i - 2jTherefore, the equation of the plane is:2x - 2y + z = 0This plane passes through the origin O, which is the point (0, 0, 0) in xyz space.