Class XI Chapter 12 – Introduction to Three Dimensional Geometry Maths ______________________________________________________________________________ 2 2 2 DA 1 2 23 14 925 9 43 Here, AB = CD = 6, BC = AD = 43 Hence, the opposite sides of quadrilateral ABCD, whose vertices are taken in order, are equal. Therefore, ABCD is a parallelogram. Hence, the given points are the vertices of a parallelogram. Question 4: Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, - 1). Solution 4: Let P (x, y, z) be the point that is equidistant from points A(1, 2, 3) and B(3, 2, â€“1). Accordingly, PA = PB 2 2 PA PB x 1 y 2 z 3 x 3 y 2 z 1 2 2 2 2 2 2 ⇒ x 2 â€“ 2x + 1 + y 2 â€“ 4y + 4 + z 2 â€“ 6z + 9 = x 2 â€“ 6x + 9 + y 2 â€“ 4y + 4 + z 2 + 2z + 1 ⇒ â€“2x â€“4y â€“ 6z + 14 = â€“6x â€“ 4y + 2z + 14 ⇒ â€“ 2x â€“ 6z + 6x â€“ 2z = 0 ⇒ 4x â€“ 8z = 0 ⇒ x â€“ 2z = 0 Thus, the required equation is x â€“ 2z = 0. Question 5: Find the equation of the set of points P, the sum of whose distances from A (4, 0, 0) and B (-4, 0, 0) is equal to 10. Solution 5: Let the coordinates of P be (x, y, z). The coordinates of points A and B are (4, 0, 0) and (â€“4, 0, 0) respectively. It is given that PA + PB = 10. 2 2 2 2 2 2 x 4 y z x 4 y z 10 4 10 4 2 2 2 2 2 2 x y z x y z On squaring both sides, we obtain Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.

Class XI Chapter 12 – Introduction to Three Dimensional Geometry Maths ______________________________________________________________________________ 4 100 20 4 4 2 2 2 2 2 2 2 2 2 x y z x y z x y z x 8x 16 y z 100 20 x 8x16 y z x 8x16 y z 2 2 2 2 2 2 2 2 2 2 2 2 20 x 8x 16 y z 100 16x 2 2 2 5 x 8x 16 y z 25 4x On squaring both sides again, we obtain 25 (x 2 + 8x + 16 + y 2 + z 2 ) = 625 + 16x 2 + 200x ⇒ 25x 2 + 200x + 400 + 25y 2 + 25z 2 = 625 + 16x 2 + 200x ⇒ 9x 2 + 25y 2 + 25z 2 â€“ 225 = 0 Thus, the required equation is 9x 2 + 25y 2 + 25z 2 â€“ 225 = 0. Exercise 12.3 Question 1: Find the coordinates of the point which divides the line segment joining the points (-2, 3, 5) and (1, -4, 6) in the ratio (i) 2:3 internally, (ii) 2:3 externally. Solution 1: (i) The coordinates of point R that divides the line segment joining points P (x1, y1, z1) and Q (x2, y2, z2) internally in the ratio m: n are mx2 nx1 my2 ny1 mz2 nz1 , , mn mn mn Let R (x, y, z) be the point that divides the line segment joining points (â€“2, 3, 5) and (1, â€“4, 6) internally in the ratio 2:3 21 32 24 33 2635 x , y ,andz= 23 23 23 -4 1 27 i.e., x= , y , and z= 5 5 5 4 1 27 Thus, the coordinates of the required point are - , , 5 5 5 (ii) The coordinates of point R that divides the line segment joining points P (x1, y1, z1) and Q (x2, y2, z2) externally in the ratio m: n are mx2 nx1 my2 ny1 mz2 nz1 , , mn mn mn Let R (x, y, z) be the point that divides the line segment joining points(â€“2, 3, 5) and (1, â€“4, Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.

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