## NCERT Solutions Class 11th Maths Chapter 12: Introduction to Three Dimensional Geometry## Exercise 12.1
When a point lies on the x-axis, its y-coordinate and z-coordinate will be 0.
When a point lies in the XZ-plane, its y-coordinate is 0.
The given point has positive coordinates on x, y and z axes. Therefore, it lies in Octant I.
The given point has positive coordinates on x, and z axes and negative coordinate on y-axis. Therefore, it lies in Octant IV.
The given point has positive coordinate on x-axis and negative coordinates on y and z axes. Therefore, it lies in Octant VIII.
The given point has positive coordinates on x, and y axes and negative coordinate on z-axis. Therefore, it lies in Octant V.
The given point has positive coordinate on y-axis and negative coordinates on x and z axes. Therefore, it lies in Octant VI.
The given point has positive coordinates on y, and z axes and negative coordinate on x-axis. Therefore, it lies in Octant II.
The given point has positive coordinate on z-axis and negative coordinates on x and y axes. Therefore, it lies in Octant III.
The given point has positive coordinate on x-axis and negative coordinates on y and z axes. Therefore, it lies in Octant VIII.
## Exercise 12.2
According to the distance formula, Distance between two points = √((x = √((4 - 2) = √((2) = √(4 + 16) = √20 = 2√5 units
According to the distance formula, Distance between two points = √((x = √((2 + 3) = √((5) = √(25 + 9 + 9) = √43 units
According to the distance formula, Distance between two points = √((x = √((1 + 1) = √((2) = √(4 + 36 + 64) = √104 = 2√26 units
According to the distance formula,
= √((-2 - 2) = √((-4) = √(16 + 4) = √20 = 2√5 units
Let the given points be P (-2, 3, 5), Q (1, 2, 3) and R (7, 0, -1). If P, Q and R are collinear then the distance PR will be equal to the sum of distances PQ and QR. Using distance formula, Distance between P and Q = √((x = √((1 + 2) = √(3 = √(9 + 1 + 4) = √14 units Distance between Q and R = √((x = √((7 - 1) = √(6 = √(36 + 4 + 16) = √56 = 2√14 units Distance between P and R = √((x = √((7 + 2) = √(9 = √(81 + 9 + 36) = √126 = 3√14 units PQ + QR = √14 + 2√14 = 3√14 = PR Hence, the points (-2, 3, 5), (1, 2, 3) and (7, 0, -1) are collinear.
If ABC forms an isosceles triangle then any two of its sides must be equal. Using distance formula, Length of side AB = √((x = √((1 - 0) = √(1 = √(1 + 1 + 16) = √18 = 3√2 units Length of side BC = √((x = √((4 - 1) = √(3 = √(9 + 9) = √18 = 3√2 units Sides AB and BC of the triangle are equal. Hence, (0, 7, -10), (1, 6, -6) and (4, 9, -6) are the vertices of an isosceles triangle.
If ABC forms a right angled triangle then the square of longest side must be equal to the sum of squares of the other two sides (Pythagoras Theorem). Length of the side AB = √((x = √((-1 - 0) = √((-1) = √(1 + 1 + 16) = √18 = 3√2 units Length of side BC = √((x = √((-4 + 1) = √((-3) = √(9 + 9) = √18 = 3√2 units Length of side AC = √((x = √((-4 - 0) = √((-4) = √(16 + 4 + 16) = √36 = 6 units AC is the longest side. AC AB = 18 + 18 = 36 = AC ∆ ABC satisfies the Pythagoras Theorem Hence, (0, 7, 10), (-1, 6, 6) and (-4, 9, 6) are the vertices of a right angled triangle.
If ABCD forms a parallelogram then the opposite sides must be equal, i.e., AB = CD and BC = AD. Length of the side AB = √((x = √((1 + 1) = √(2 = √(4 + 16 + 16) = √36 = 6 units Length of side BC = √((x = √((4 - 1) = √(3 = √(9 + 25 + 9) = √43 units Length of the side CD = √((x = √((2 - 4) = √((-2) = √(4 + 16 + 16) = √36 = 6 units Length of the side AD = √((x = √((2 + 1) = √(3 = √(9 + 25 + 9) = √43 units AB = CD = 6 units AND BC = AD = √43 units Hence, (-1, 2, 1), (1, -2, 5), (4, -7, 8) and (2, -3, 4) are the vertices of a parallelogram.
Let the given points be A (1, 2, 3) and B (3, 2, -1). Let point C (x, y, z) be a point which is equidistant from A and B. Therefore, Distance AC = Distance BC √((x Square both sides (x (x - 1) x -2x - 4y - 6z = -6x - 4y + 2z 4x = 8z x = 2z
Hence, the equation of the set of points which are equidistant from the given points is x - 2z = 0.
Let the given points be A (4, 0, 0) and B (-4, 0, 0). Let C (x, y, z) be a point from the set of points P. Therefore, Distance AC + Distance BC = 10 √((x √((x Square both sides (x (x - 4) x 16x + 100 = 20√(x 4(4x + 25) = 20√(x 4x + 25 = 5√(x Square both sides (4x + 25) 16x 225 = 9x
Hence, the equation of the set of points P, the sum of whose distances from A and B is 10 is 9x ## Exercise 12.3
x = (mx y = (my z = (mz (x (x m : n = 2 : 3 Therefore, x = (2(1) + 3(-2))/(2 + 3) = (2 - 6)/5 = -4/5 y = (2(-4) + 3(3))/(2 + 3) = (-8 + 9)/5 = 1/5 z = (2(6) + 3(5))/(2 + 3) = (12 + 15)/5 = 27/5 Hence, the required point is (-4/5, 1/5, 27/5).
x = (mx y = (my z = (mz (x (x m : n = 2 : 3 Therefore, x = (2(1) - 3(-2))/(2 - 3) = (2 + 6)/(-1) = -8 y = (2(-4) - 3(3))/(2 - 3) = (-8 - 9)/(-1) = 17 z = (2(6) - 3(5))/(2 - 3) = (12 - 15)/(-1) = 3 Hence, the required point is (-8, 17, 3).
Let the ratio in which Q divides PR be k : 1. We know that coordinates of a point which divides the line segment joining the points (x x = (mx y = (my z = (mz (x (x m : n = k : 1 x = (9k + 3)/(k + 1) 5 = (9k + 3)/(k + 1) 5(k + 1) = 9k + 3 5k + 5 = 9k + 3 2 = 4k
Hence, the point Q divides line PR in ratio 1 : 2.
Let the given points be A (-2, 4, 7) and B (3, -5, 8). We know that any point which lies in the YZ-plane will be of the form (0, y, z). Let there be a point C (0, y, z) which divides the line segment AB in ratio k : 1. We know that coordinates of a point which divides the line segment joining the points (x x = (mx y = (my z = (mz (x (x m : n = k : 1 x = (3k - 2)/(k + 1) 0 = (3k - 2)/(k + 1) 0 = 3k - 2 3k = 2
Hence, the YZ-plane divides the line segment formed by joining the points (-2, 4, 7) and (3, -5, 8) in ratio 2 : 3.
Let there be a point P (0, y, z) which divides AB in ratio k : 1. We know that coordinates of a point which divides the line segment joining the points (x x = (mx y = (my z = (mz (x (x m : n = k : 1 x = (-k + 2)/(k + 1) 0 = (-k + 2)/(k + 1) 0 = -k + 2
y = (2k - 3)/(k + 1) = (2(2) - 3)/(2 + 1) = 1/3 z = (k + 4)/(k + 1) = (2 + 4)/(2 + 1) = 6/3 = 2 Therefore, point P (0, 1/3, 2) lies on the line AB and divides it in ratio 2 : 1. Point P (0, 1/3, 2) is same as the point C (0, 1/3, 2). Hence, A, B, and C are collinear.
Let A and B be the points which trisect the line segment joining the points P (4, 2, -6) and Q (10, -16, 6).
x = (mx y = (my z = (mz (x (x A divides PQ in the ratio 1 : 2. Therefore, x = (10 + 8)/3 = 18/3 = 6 y = (-16 + 4)/3 = -12/3 = -4 z = (6 - 12)/3 = -6/3 = -2 So, A is (6, -4, -2). B divides PQ in the ratio 2 : 1. Therefore, x = (20 + 4)/3 = 24/3 = 8 y = (-32 + 2)/3 = -30/3 = -10 z = (12 - 6)/3 = 6/3 = 2 So, B is (8, 10, -2). Hence, the coordinates of the points which trisect the line segment joining the points P (4, 2, -6) and Q (10, -16, 6) are (6, -4, -2) and (8, 10, -2). ## Miscellaneous Exercise
Let the fourth vertex of the parallelogram be D (x, y, z). It is given that ABCD is a parallelogram. Therefore, the diagonals AC and BD will bisect each other, i. e., their mid-points will be the same. We know that coordinates of the mid-point of the line segment joining the points (x x = (x y = (y z = (z Mid-point of AC: x y z Mid-point of BD: x y z Therefore, 1 = (1 + x)/2 2 = 1 + x
0 = (2 + y)/2 0 = 2 + y
2 = (-4 + z)/2 4 = -4 + z
Hence, the coordinates of fourth vertex of parallelogram ABCD is D (1, -2, 8).
Let D, E and F be mid-points of the sides AB, BC, and AC respectively. Then, CD, AE, BF are the medians. According to the mid-point formula, coordinates of the mid-point of the line segment joining the points (x x = (x y = (y z = (z Coordinates of D: x y z D is (0, 2, 3) Coordinates of E: x y z E is (3, 2, 0) Coordinates of F: x y z F is (3, 0, 3) Using Distance Formula, we have: Length of median CD = √((0 - 6) = √((-6) = √49 = 7 units Length of median AE = √((3 - 0) = √(3 = √49 = 7 units Length of median BF = √((3 - 0) = √(3 = √34 units Hence, the length of the medians of given triangle to the sides AB, BC and AC are 7, 7 and √34 units respectively.
It is given that the origin is centroid of the given triangle PQR. Therefore, the centroid is at (0, 0, 0). We know that the coordinates of the centroid of the triangle, whose vertices are (x x = (x y = (y z = (z Centroid of triangle with vertices P (2a, 2, 6), Q (-4, 3b, -10) and R(8, 14, 2c): x = (2a - 4 + 8)/3 0 = (2a + 4)/3 0 = 2a + 4 2a = -4
y = (2 + 3b + 14)/3 0 = (3b + 16)/3 3b + 16 = 0 3b = -16
z = (6 - 10 + 2c)/3 0 = (2c - 4)/3 2c - 4 = 0 2c = 4
Let the point on the y-axis be A (0, y, 0). Using the distance formula, We know that the distance between two points (x1, y1, z1) and (x2, y2, z2) is given by Distance = √[(x So, the distance between the points A (0, y, 0) and P (3, -2, 5) is given by Distance of AP = √[(3 - 0) It is given that the distance between the points A (0, y, 0) and P (3, -2, 5) is 5√2. Therefore, 5√2 = √[(3 - 0) 5√2 = √[3 5√2 = √[(-2 - y) 5√2 = √[(-2 - y) Square both sides, (-2 - y) (-2 - y) 4 + y y y y(y + 6) - 2(y + 6) = 0 (y + 6)(y - 2) = 0 y = -6, y = 2 Hence, the points (0, 2, 0) and (0, -6, 0) are the required points on the y-axis.
Let the coordinates of the required point be R (4, y, z). Let the ratio in which point R (4, y, z) divides the line segment joining the points P (2, -3, 4) and Q (8, 0, 10) be k : 1. Using the section formula,
x = (mx y = (my z = (mz Therefore, 4 = (8k + 2)/(k + 1) 4(k + 1) = 8k + 2 4k + 4 = 8k + 2 2 = 4k k = 1/2 Now, y = (0 - 3)(1/2 + 1) y = (-3)/(3/2) 3y/2 = -3 3y = -6 y = -2 z = (10(1/2) + 4)/(1/2 + 1) z = (5 + 4)/(3/2) 3z/2 = 9 3z = 18 z = 6 Hence, coordinates of the required point are (4, -2, 6).
Let the point be P (x, y, z). Using the distance formula, We know that the distance between two points (x Distance = √[(x Therefore, PA = √((3 - x) and PB = √((-1 - x) It is given that PA ((3 - x) ((9 + x 9 + x 2x 2x 2(x (x Hence, the required equation is (x |