Miscellaneous

Chapter 11 Class 11 - Intro to Three Dimensional Geometry
Serial order wise

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Misc 1 Three vertices of a parallelogram ABCD are A (3, β1, 2), B (1, 2, β4) and C (β1, 1, 2). Find the coordinates of the fourth vertex. In parallelogram, diagonals bisect each other Hence AO = OC & BO = OD We can say that O is Midpoint of AC and O is Midpoint of BD Finding co-ordinates of O Now, O is the mid-point of AC Here, A (3, β1. 2 ) & C ( β1 , 1, 2) We know that If O (x, y, z) is the mid point of A (x1, y1, z1) & B (x2, y2, z2) , then coordinates of O O (x, y, z) = ((γ π₯γ_1 +γ π₯γ_2)/2,(γ π¦γ_1 + γ π¦γ_2)/2,(π§_1 +γ π§γ_2)/2) Here, x1 = 3, y1 = β1, z1 = 2 x2 = β1 , y2 = 1, z2 = 2 Putting values, O = ((3 + (β1))/2, (β1 + 1)/2, (2 + 2)/2) = ((3 β 1)/2, 0/2, 4/2) = (2/2, 0, 2) = (1, 0, 2) Thus, the co-ordinates of O is (1, 0, 2) Now, O is also the mid-point of BD , Finding coordinates of O, Here, B (1, 2, β4) , O (1, 0, 2) Let D Be (x, y, z) Now Co-ordinates of O = ((1 + π₯)/2,(2 + π¦)/2, (β4 + π§)/2) (1, 0, 2) = ((1 + π₯)/2, (2 + π¦)/2, (β4 + π§)/2) Thus, x = 1 , y = -2 , z = 8 β΄ Coordinates of point D are (1, β2, 8) Thus, Coordinates of the fourth vertex are (1, β2, 8) (1 + π₯)/2=1 1 + x = 2 x = 2 β 1 x = 1 (2 + π¦)/2=0 2 + y = 0 Γ 2 2 + y = 0 y = β 2 (β4 + π§)/2=2 β 4 + z = 2 Γ 2 z = 4 + 4 z = 8