\n| z<\/td>\n | +<\/td>\n | +<\/td>\n | +<\/td>\n | +<\/td>\n | –<\/td>\n | –<\/td>\n | –<\/td>\n | –<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Question 4.
\nShow that the distance between the points (x1<\/sub>, y1<\/sub>, z1<\/sub>) and (x2<\/sub>, y2<\/sub>, z2<\/sub>) is given by
\n\\(\\sqrt{\\left(x_{2}-x_{1}\\right)^{2}+\\left(y_{2}-y_{1}\\right)^{2}+\\left(z_{2}-z_{1}\\right)^{2}}\\)
\nAnswer:
\nLet P(x1<\/sub>,y1<\/sub>,z1<\/sub>) and Q(x2<\/sub>,y2<\/sub>,z2<\/sub>) be the given points.
\n![\"1st](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/12\/1st-PUC-Maths-Question-Bank-Chapter-12-Introduction-to-Three-Dimensional-Geometry-3.png\")
\nLet PQ be the line not parallel to z-axis. Draw PM and QN perpendiculars to XY-plane. Since XY-plane is two dimensional (2 – D), then the coordinates of M and N are respectively (x1<\/sub> , y1<\/sub>) and (x2<\/sub> , y2<\/sub>).
\n![\"1st](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/12\/1st-PUC-Maths-Question-Bank-Chapter-12-Introduction-to-Three-Dimensional-Geometry-4.png\") <\/span><\/p>\n![\"KSEEB](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/11\/KSEEB-Solutions-300x28.png\") <\/p>\n
Section formula :<\/strong><\/p>\nQuestion 5.
\nIf A(x1<\/sub>,y1<\/sub>,zl<\/sub>) and B(x2<\/sub>,y2<\/sub>,z2<\/sub>) are two distinct points in space and a point P(x,y,z) divides AB in the ratio m : n internally then Prove that.
\n\\(x=\\frac{m x_{2}+n x_{1}}{m+n}, \\quad y=\\frac{m y_{2}+n y_{1}}{m+n}, \\quad z=\\frac{m z_{2}+n z_{1}}{m+n}\\)
\n\\(\\text { i.e., } P=\\left(\\frac{m x_{2}+n x_{1}}{m+n}, \\frac{m y_{2}+n y_{1}}{m+n}, \\frac{m z_{2}+n z_{1}}{m+n}\\right)\\)
\nAnswer:
\nDraw AL, PN and BM perpendiculars to the xy-plane. Clearly the feet of the perpendiculars L, M, N are collinear and L = (x1<\/sub>,y1<\/sub>), M -(x2<\/sub>,y2<\/sub>), N = (x,y). Now, draw a line through P which is parallel to the line LM meets AL produced at C and BM at D. Clearly the triangle PAC and PDB are equiangular and hence similar.
\n![\"1st](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/12\/1st-PUC-Maths-Question-Bank-Chapter-12-Introduction-to-Three-Dimensional-Geometry-5.png\")
\n![\"1st](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/12\/1st-PUC-Maths-Question-Bank-Chapter-12-Introduction-to-Three-Dimensional-Geometry-6.png\") <\/p>\nRemark:
\n(i) It P divides AB externally in the ratio m : n, then the coordinates of the point P are
\n![\"1st](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/12\/1st-PUC-Maths-Question-Bank-Chapter-12-Introduction-to-Three-Dimensional-Geometry-7.png\") <\/p>\n Observations:<\/p>\n \n- The point P divides AB internally in the ratio k : 1 iff k > 0 (i.e., k is + ve)<\/li>\n
- The point P divides AB externally in the ratio |k| : 1 iff k < 0 (i.e., k is – ve)<\/li>\n<\/ul>\n
<\/p>\n Question 6.
\nFind the co-ordinates of the centroid of the triangle whose vertices are (x1<\/sub>,y1<\/sub>,z1<\/sub>) (x2<\/sub>,y2<\/sub>,z2<\/sub>) and (x3<\/sub>,y3<\/sub>,z3<\/sub>)
\nAnswer:
\nNote: Centroid is the point of intersection of the medians of the triangle and it is denoted by G. Centroid divides the median in the ratio 2:1.
\n![\"1st](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/12\/1st-PUC-Maths-Question-Bank-Chapter-12-Introduction-to-Three-Dimensional-Geometry-8.png\")
\nLet A = (x1<\/sub>,y1<\/sub>,z1<\/sub>) , B = (x2<\/sub>,y2<\/sub>,z2<\/sub>) and C = (x3<\/sub>, y3<\/sub>, z3<\/sub>) be the vertices of the triangle ABC. Let AD be the median drawn from A. Then D is the mid-point of BC.
\n\u2234 D=\\(\\left(\\frac{x_{2}+x_{3}}{2}, \\frac{y_{2}+y_{3}}{2}, \\frac{z_{2}+z_{3}}{2}\\right)\\)
\nLet G be the centroid of the \u0394ABC. median AD in the ratio 2 : 1.
\n\u2234 The coordinates of G are,
\n![\"1st](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/12\/1st-PUC-Maths-Question-Bank-Chapter-12-Introduction-to-Three-Dimensional-Geometry-9.png\") <\/p>\nQuestion 7.
\nFind the ratio in which the
\n(i) xy – plane
\n(ii) yz – plane
\n(iii) zx – plane divides the line segment formed by joining the points (x1<\/sub>,y1<\/sub>,z1<\/sub>) and (x2<\/sub>,y2<\/sub>,z2<\/sub>).
\nAnswer:
\nWe have, the coordinates of a point dividing the line joining the points A (x1<\/sub>,y1<\/sub>,z1<\/sub>) and
\nB (x2<\/sub>,y2<\/sub>,z2<\/sub>) in the ratio k : 1 are,
\n\\(P=\\left(\\frac{k x_{2}+x_{1}}{k+1}, \\frac{k y_{2}+y_{1}}{k+1}, \\frac{k z_{2}+z_{1}}{k+1}\\right)\\)<\/p>\n(i) If xy – plane divides a line segment joining the given points, A, B, then z – coordinate of P is zero.
\n\\(\\text { i.e., } \\frac{k z_{2}+z_{1}}{k+1}=0 \\Rightarrow k z_{2}+z_{1}=0 \\Rightarrow k=\\frac{-z_{1}}{z_{2}}\\)
\n\u2234 xy-plane divides the line segment AB in the ratio -z1<\/sub> : z2
\n<\/sub>Similarly,
\n(ii) The yz-plane in the ratio -x1<\/sub>: x2
\n<\/sub>(iii) The zx-plane in the ratio -y1 : <\/sub>y2<\/sub><\/p>\n<\/p>\n Question 8.
\nA point is on the x-axis. What are its y and z – coordinates?
\nAnswer:
\nSince a point is on the x-axis then it is of the form (x,0,0)
\n\u2234\u00a0 y-coordinate = 0 and z-coordinate = 0.<\/p>\n Question 9.
\nA point is in the xz-plane. What can you say about its y-coordinate?
\nAnswer:
\nSince a point is in the xz-plane, then it is of the form (x,0, z). So its y-coordinate is zero.<\/p>\n Question 10.
\nName the octants in which the following points he. (1, 2, 3), (2, 4, 5), (-3, 1, 2), (-3, 1, -2), (4, -2, 3), (4, -2, -5), (4, 2, -5), (-4, 2, -5), (-4,2, 5), (-3,-1,6), (-2,-4,-7).
\nAnswer:<\/p>\n \n\n\nSl.No<\/strong><\/td>\nPoint<\/strong><\/td>\nOctant<\/strong><\/td>\n<\/tr>\n\n| (i)<\/td>\n | (1,2, 3)<\/td>\n | first octant<\/td>\n<\/tr>\n | \n| (ii)<\/td>\n | (2, 4, 5)<\/td>\n | first octant<\/td>\n<\/tr>\n | \n| (iii)<\/td>\n | (-3,1,2)<\/td>\n | second octant<\/td>\n<\/tr>\n | \n| (iv)<\/td>\n | (-3,1,-2)<\/td>\n | sixth octant<\/td>\n<\/tr>\n | \n| (v)<\/td>\n | (4,-2, 3)<\/td>\n | fourth octant<\/td>\n<\/tr>\n | \n| (vi)<\/td>\n | (4,-2,-5)<\/td>\n | eighth octant<\/td>\n<\/tr>\n | \n| (vii)<\/td>\n | (4, 2,-5)<\/td>\n | fifth octant<\/td>\n<\/tr>\n | \n| (viii)<\/td>\n | (-4, 2, -5)<\/td>\n | sixth octant<\/td>\n<\/tr>\n | \n| (ix)<\/td>\n | (-4, 2, 5)<\/td>\n | second octant<\/td>\n<\/tr>\n | \n| (x)<\/td>\n | (-3, -1,6)<\/td>\n | third octant<\/td>\n<\/tr>\n | \n| (xi)<\/td>\n | (-2, -4, -7)<\/td>\n | seventh octant<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Question 11.
\nFind the distance between the following pairs of points.
\n(i) (1,-3,4) and (-4,1,2)
\n(ii) (2, 3, 5) and (4,3,1)
\n(iii) (-3,7,2) and (2, 4,-1)
\n(iv) (-1,3, -4) and (1,-3,4)
\n(v) (2,-1,3) and (-2,1,3)
\nAnswer:
\n(i) Let A (1, -3, 4) and B = (-4, 1, 2) be the given points.
\nDistance between the points A and B is<\/p>\n ![\"1st](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/12\/1st-PUC-Maths-Question-Bank-Chapter-12-Introduction-to-Three-Dimensional-Geometry-10.png\")
\n![\"1st](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/12\/1st-PUC-Maths-Question-Bank-Chapter-12-Introduction-to-Three-Dimensional-Geometry-11.png\") <\/p>\n
<\/p>\n Question 12.
\nShow that the following points are collinear.
\n(i) (-2,3,5), (1,2,3) and (7, 0,-1)
\n(ii) (1,2,4), (3, 8,8), (7,20,16)
\nAnswer:
\n(i) Let A (-2, 3, 5), B (1, 2, 3) and C (7, 0, -1) be the given points. We know that points are said to be collinear if they lie on a line.
\n![\"1st](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/12\/1st-PUC-Maths-Question-Bank-Chapter-12-Introduction-to-Three-Dimensional-Geometry-12.png\")
\n\\(\\text { Thus } A B+B C=\\sqrt{14}+2 \\sqrt{14}=3 \\sqrt{14}=C A\\)<\/p>\n Question 13.
\nVerify the following:
\n(i) (0, 7, -10), (1, 6, -6) and (4, 9, -6) are the vertices of an isosceles triangle.
\n(ii) (0, 7, 10), (-1, 6, 6) and (-4, 9, 6) are the vertices of a right angled triangle.
\n(iii) (-1, 2, 1), (1, -2, 5), (4, -7, 8) and (2, -3, 4) are the vertices of a parallelogram.
\nAnswer:
\n(i) Let A = (0,7,-10), B = (l,6,-6) and C = (4,9,-6) be vertices of \u0394ABC.
\n![\"1st](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/12\/1st-PUC-Maths-Question-Bank-Chapter-12-Introduction-to-Three-Dimensional-Geometry-13.png\")
\n![\"1st](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/12\/1st-PUC-Maths-Question-Bank-Chapter-12-Introduction-to-Three-Dimensional-Geometry-14.png\")
\ni..e., opposite sides of a quadrilateral ABCD are equal, therefore ABCD is a parallelogram.<\/p>\n Question 14.
\nAre the points A (3, 6, 9), B (10, 20, 30) C (25, -41, 5). The vertices of a right angled triangle.
\nAnswer:
\nBy distance formula,
\nAB2<\/sup> = (10 – 3)2<\/sup> + (20 – 6)2<\/sup> + (30 – 9)2
\n<\/sup>= 49 + 196 + 441 = 686
\nBC2<\/sup> = (25 -10)2<\/sup> + (- 41 – 20)2<\/sup> + (5 – 30)2
\n<\/sup>= 225 + 3721 + 625 = 4571
\nCA2<\/sup> = (3 – 25)2<\/sup> + (6 + 41)2<\/sup> + (9 – 5)2
\n<\/sup>= 484 + 2209 + 16 = 2709
\nWe find that AB2<\/sup> + CA2<\/sup> \u2260 BC2
\n<\/sup>(\u2235 3395 \u2260 4571)
\n\u2234 ABC is not a right angled triangle.<\/p>\nQuestion 15.
\nFind the equation of the set of points which are equidistant from the points (1, 2, 3) and (3,2,-1).
\nAnswer:
\nLet A (1, 2, 3) and B (3, 2, -1) be the given points.
\nLet P(x, y, z) be any point equidistant from A and B, then
\n![\"1st](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/12\/1st-PUC-Maths-Question-Bank-Chapter-12-Introduction-to-Three-Dimensional-Geometry-15.png\")
\n\u21d2 x2<\/sup>-2x + 1+ y2 <\/sup>– 4y + 4 + z2<\/sup> + 2z + 1
\n\u21d2 x2<\/sup>-6x + 9+y2<\/sup>-4y + 4 + z2<\/sup> + 2z + 1
\n\u21d2 4x – 8z = 0
\n\u21d2 x – 2z = 0 is the required equation.
\n<\/p>\nQuestion 16.
\nFind 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.
\nAnswer:
\nGiven points are A (4, 0, 0) and B (-4, 0, 0).
\nLet P(x, y, z) be any point such that
\nPA + PB = 10
\n\u21d2 PA = 10 – PB
\n![\"1st](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/12\/1st-PUC-Maths-Question-Bank-Chapter-12-Introduction-to-Three-Dimensional-Geometry-16.png\") <\/p>\n Question 17.
\nFind the equation of set of points P such that PA2<\/sup> + PB2<\/sup> = 2k2<\/sup>, where A and B are the points (3,4,5) and (-1,3, -7), respectively.
\nAnswer:
\nLet P = (x,y,z), B = (3,4,5) and B = (-1,3-7).
\nNow, PB2<\/sup> = (x +1)2<\/sup> + (y – 3)2<\/sup> + (z + 7)2<\/sup> and PA2<\/sup>
\n= (x – 3)2<\/sup> + (y – 4)2<\/sup> + (z – 5)2
\n<\/sup>Given that PA2<\/sup> + PB2<\/sup> = 2k2<\/sup>, then
\n(x-3)2<\/sup> + (y-4)2<\/sup> +(z-5)2<\/sup> + (x + 1)2 <\/sup>+(y-3)2<\/sup>+(z + 7)2<\/sup> = 2k2<\/sup>
\n\u21d2 x2<\/sup> – 6*+ 9+ y2<\/sup> – 8y + 16 + z2<\/sup> – 10z + 22 + x2<\/sup> +2x +1 +
\ny2<\/sup> – 6y + 9 + z2<\/sup> +14z + 49 = 2k.
\n\u21d2 2x2<\/sup> + 2y2<\/sup> + 2z2<\/sup> – 4x – 14y + 4z + 109 = 2k2
\n<\/sup>\u21d2 2x2<\/sup> + 2y2<\/sup> + 2z2<\/sup> – Ax – 14y + 4z = 2k2<\/sup> -109
\n<\/p>\nQuestion 18.
\nFind the coordinates of the point which divides the line segment joining the points
\n(-2, 3, 5) and (1, -4, 6) in the ratio
\n(i) 2 : 3 internally
\n(ii) 2:3 externally.
\nAnswer:
\nThe given points are A (-2, 3, 5) and B (1, -4, 6).
\n![\"1st](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/12\/1st-PUC-Maths-Question-Bank-Chapter-12-Introduction-to-Three-Dimensional-Geometry-17.png\") <\/p>\n Question 19.
\nFind the coordinates of the point which divides the line segment joining the points
\n(1, -2, 3) and (3, 4, -5) in the ratio 2 : 3
\n(i) internally and
\n(ii) externally.
\nAnswer:
\nTry yourself.
\n\\(\\text { (i) }\\left(\\frac{9}{5}, \\frac{2}{5}, \\frac{-1}{5}\\right) \\text { (ii) }(-3,-14,19)\\)<\/p>\n Question 20.
\nGiven that P (3, 2, -4), Q (5, 4, -6) and R (9, 8, -10) are collinear. Find the ratio in which Q divides PR.
\nAnswer:
\nLet Q divides PR in the ratio k : 1, then Q is of the form,
\n\\(Q=\\left(\\frac{k x_{2}+x_{1}}{k+1}, \\frac{k y_{2}+y_{1}}{k+1}, \\frac{k z_{2}+z_{1}}{k+1}\\right)\\)
\n![\"1st](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/12\/1st-PUC-Maths-Question-Bank-Chapter-12-Introduction-to-Three-Dimensional-Geometry-18.png\") <\/p>\n Question 21.
\nFind the ratio in which the yz-plane divides the line segment formed by joining the points (-2,4,7) and (3,-5, 8)
\nAnswer:
\nLet yz – plane divides the line segment joining A (-2, 4, 7) and B (3, -5, 8) at P(x,y,z) in the ratio k : 1. Then the coordinates of P are
\n![\"1st](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/12\/1st-PUC-Maths-Question-Bank-Chapter-12-Introduction-to-Three-Dimensional-Geometry-19.png\") <\/p>\n Question 22.
\nFind the ratio in which the line segment joining the points (4, 8, 10) and (6, 10, -8) is divided by the yz-plane.
\nAnswer:
\nLet yz – plane divides the line segment joining (4, 8, 10) and (6, 10, -8) at P (x, y, z) in the ratio k : 1, then
\n![\"1st](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/12\/1st-PUC-Maths-Question-Bank-Chapter-12-Introduction-to-Three-Dimensional-Geometry-20.png\")
\n\u2234 yz – plan divided AB externally in the ratio 2 : 3 \\(\\left(\\ k=\\frac{-2}{3}<0\\right)\\)
\n<\/p>\n Question 23.
\nUsing section formula, show that the points A(2,-3,4), B(-1,2,1) and \\(C\\left(0, \\frac{1}{3}, 2\\right)\\)
\nAnswer:
\nA(2,-3,4), B(-1,2,1) and \\(C\\left(0, \\frac{1}{3}, 2\\right)\\) be the ratio k : 1 ,Then coordinates of the points P are,
\n![\"1st](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/12\/1st-PUC-Maths-Question-Bank-Chapter-12-Introduction-to-Three-Dimensional-Geometry-21.png\") <\/p>\n Question 24.
\nFind the coordinates of the points which trisects the line segment joining the points
\nAnswer:
\n![\"1st](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/12\/1st-PUC-Maths-Question-Bank-Chapter-12-Introduction-to-Three-Dimensional-Geometry-22.png\")
\nGiven points P (4, 2, -6) and Q (10, -16, 6).
\nLet A and B be the points of trisection of PQ, then A divides PQ in the ratio 1 : 2 and B divides PQ in the ratio 2:1.
\n![\"1st](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/12\/1st-PUC-Maths-Question-Bank-Chapter-12-Introduction-to-Three-Dimensional-Geometry-23.png\") <\/p>\n Miscellaneous Examples<\/strong><\/p>\nQuestion 1.
\nShow that the points A (1,2, 3), B (-1, -2, -1), C (2, 3, 2) and D (4, 7, 6) are the vertices of a parallelogram ABCD, but it is not a rectangle.
\nAnswer:
\nGiven points are A (1, 2, 3), B (-1, -2, -1), C (2, 3, 2) and D (4, 7, 6).
\nTo show ABCD is a parallelogram we need to show opposite sides are equal.
\nBy distance formula, we have
\n![\"1st](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/12\/1st-PUC-Maths-Question-Bank-Chapter-12-Introduction-to-Three-Dimensional-Geometry-24.png\")
\nSince, AB = CD and BC = DA. ABCD is a parallelogram.
\nNow, it is required to prove that ABCD is not a rectangle. For this, we show that diagonals AC and BD are unequal.
\n![\"1st](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/12\/1st-PUC-Maths-Question-Bank-Chapter-12-Introduction-to-Three-Dimensional-Geometry-25.png\") <\/p>\n Question 2.
\nThree vertices of a parallelogram ABCD are A (3, -1, 2), B (1, 2, -4) and C (-1, 1, 2). Find the co-ordinates of the fourths vertex.
\nAnswer:
\nLet the fourth vertex be D(x1<\/sub> , y1<\/sub>, z1<\/sub>) we know that diagonals of a parallelogram bisect each other. So the mid-point of the diagonal AC is same as the mid-point of the diagonal BD.
\n\u2234 By mid-point formula, we have
\n\\(\\left(\\frac{3+(-1)}{2}, \\frac{-1+1}{2}+\\frac{2+2}{2}\\right)=\\left(\\frac{x_{1}+1}{2}, \\frac{y_{1}+2}{2}, \\frac{z_{1}-4}{2}\\right)\\)
\n![\"1st](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/12\/1st-PUC-Maths-Question-Bank-Chapter-12-Introduction-to-Three-Dimensional-Geometry-26.png\") <\/p>\nQuestion 3.
\nFind the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0, 4, 0) and C (6,0,0).
\nAnswer:
\nGiven (points) vertices A are A (0, 0, 6), B (0, 4, 0) and C (6, 0, 0).
\n![\"1st](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/12\/1st-PUC-Maths-Question-Bank-Chapter-12-Introduction-to-Three-Dimensional-Geometry-27.png\")
\nLet AD, BE and CF be the medians of the \u0394ABC. Then, D, E and F are the mid-points of the sides BC, CA and AB respectively.
\n\u2234 By mid-point formula, we have
\n![\"1st](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/12\/1st-PUC-Maths-Question-Bank-Chapter-12-Introduction-to-Three-Dimensional-Geometry-28.png\") <\/p>\n <\/p>\n Question 4.
\nIf the origin is the centroid of the triangle PQR with vertices P (2a, 2, 6),
\nQ (-4, 3b, -10)and R (8, 14, 2c), then find the values of a, b and c.
\nAnswer:
\nGiven vertices are P (2a, 2, 6), Q (-4, 3b, -10) and R (8, 14, 2c).
\nLet G be the centroid of the \u0394PQR then G = (0, 0, 0) (given).
\n![\"1st](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/12\/1st-PUC-Maths-Question-Bank-Chapter-12-Introduction-to-Three-Dimensional-Geometry-29.png\") <\/p>\n Question 5.
\nThe centroid of a triangle ABC is at the point (1, 1, 1). If the coordinates of A and B are (3, -5, 7) and (-1, 7, -6) respectively, find the coordinates of the point C.
\nAnswer:
\nLet C = (x,y,z) and centroid G = (1,1,1). Then
\n![\"1st](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/12\/1st-PUC-Maths-Question-Bank-Chapter-12-Introduction-to-Three-Dimensional-Geometry-30.png\") <\/p>\n Question 6.
\nFind the coordinates of a point on the y-axis which are at a distance of \\(5 \\sqrt{2} \\)from the point P (3, -2,5).
\nAnswer:
\nGiven point is P (3, -2, 8)
\nLet Q be the point on the y-axis then Q be (0, y, 0).
\n![\"1st](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/12\/1st-PUC-Maths-Question-Bank-Chapter-12-Introduction-to-Three-Dimensional-Geometry-31.png\")
\n\u2234 Required point on the y – axis is either (0,2,0) or (0,-6, 0)<\/p>\n Question 7.
\nA point R with x-coordinate 4 lies on the line segment joining the points P (2, -3, 4) and Q (8, 0, 10). Find the coordinates of the point R.
\nAnswer:
\nSince R lies on the line PQ, then R divides PQ in some ratio say k : 1.
\n\u2234 By section formula R is of the form,
\n![\"1st](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/12\/1st-PUC-Maths-Question-Bank-Chapter-12-Introduction-to-Three-Dimensional-Geometry-32.png\")
\n<\/p>\n Question 8.
\nIf A and B be the points (3, 4, 5) and (-1, 3, -7) respectively, find the equation of the set of points P such that PA2<\/sup> +PB2<\/sup> =k2 <\/sup>where k is constant.
\nAnswer:
\nGiven points are A (3, 4, 5) and B (-1, 3, -7).
\nLet P = (x,y,z)
\nGiven: PA2<\/sup> + PB2<\/sup> = k2
\n<\/sup>\u21d2 (x-3)2 <\/sup>+ (y-4)2<\/sup> + (z-5)2<\/sup> +(x + 1)2<\/sup> + (y – 3)2<\/sup> + (z + 7)2<\/sup> =k2
\n<\/sup>\u21d2 2x1<\/sup> – 6x + 9 + y2<\/sup> -8y + 16 + z2<\/sup> -10z + 25 + x2<\/sup> + 2x +
\n1+y2<\/sup> – 6y + 9 + z2<\/sup> +14z + 49 = k2
\n<\/sup>\u21d2 2x2<\/sup> + 2y2<\/sup> + 2z2<\/sup> – 4x – 14y + 4z +109 = k2
\n<\/sup>\u2234 Required equations of the set of points P is,
\n2x2<\/sup> + 2y2<\/sup> +2z2 <\/sup>– 4x – 14y + 4z + 109 – k2<\/sup> =0<\/p>\nQuestion 9.
\nFind the equation of the set of points P such that PA = PB, where A (3, 4, -5) and
\nB (-2,1,4).
\nAnswer:
\nTry yourself. (10x – 6y – 18z – 29 = 0)<\/p>\n","protected":false},"excerpt":{"rendered":" Students can Download Maths Chapter 12 Introduction to Three Dimensional Geometry Questions and Answers, Notes Pdf, 1st PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 1st PUC Maths Question Bank Chapter 12 Introduction to Three Dimensional Geometry Question 1. …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[81],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/15591"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=15591"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/15591\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=15591"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=15591"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=15591"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}} | | | | |
![\"1st](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/12\/1st-PUC-Maths-Question-Bank-Chapter-12-Introduction-to-Three-Dimensional-Geometry-1.png\")
![\"1st](\"https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2019\/12\/1st-PUC-Maths-Question-Bank-Chapter-12-Introduction-to-Three-Dimensional-Geometry-2.png\")
<\/p>\n