Chapter 12 – Introduction To Three Dimensional Geometry

Locate the points

• (i)(2, 3, 4)
• (ii)(–2, –2, 3) in space. Solution (i) To locate the point (2, 3, 4) in space, we move 2 units from O along the positive direction of x-axis. Let this point be A (2, 0, 0). From the point A moves 3 units parallel to +ve direction of y-axis.Let this point be B (2, 3, 0). From the point B moves 4 units along positive direction of z-axis. Let this point be P (2, 3, 4) Fig.(12.3). Fig. 12.3 INTRODUCTION TO THREE DIMENSIONAL GEOMETRY 211 (ii) From the origin, move 2 units along the negative direction of x-axis. Let this point be A (–2, 0, 0). From the point A move 2 units parallel to negative direction of y-axis. Let this point be B (–2, –2, 0). From B move 3 units parallel to positive direction of z - axis. This is our required point Q (–2, –2, 3) (Fig.12.4.) Fig. 12.4

Sketch the plane (i) x = 1 (ii) y = 3 (iii) z = 4 Solution (i) The equation of the plan x = 0 represents the yz-plane and equation of the plane x = 1 represents the plane parallel to yz-plane at a distance 1 unit above yz- plane. Now, we draw a plane parallel to yz- plane at a distance 1 unit above yz- plane Fig.12.5(a). (ii) The equation of the plane y = 0 represents the xz plane and the equation of the plane y = 3 represents the plane parallel to xz plane at a distance 3 unit above xz plane (Fig. 12.5(b)). (iii) The equation of the plane z = 0 represents the xy-plane and z = 3 represents the plane parallel to xy-plane at a distance 3 unit above xy-plane (Fig. 12.5(c)).

• (a)
• (b)
• (c)Fig. 12.5

Let L, M, N be the feet of the perpendiculars drawn from a point P (3, 4, 5) on the x, y and z-axes respectively. Find the coordinates of L, M and N. Solution Since L is the foot of perpendicular from P on the x-axis, its y and z co- ordinates are zero. The coordinates of L is (3, 0, 0). Similarly, the coordinates of M and N are (0, 4, 0) and (0, 0, 5), respectively.

Let L, M, N be the feet of the perpendicular segments drawn from a point P (3, 4, 5) on the xy, yz and zx-planes, respectively. What are the coordinates of L, M and N? Solution Since L is the foot of perpendicular segment from P on the xy-plane, z-coordinate is zero in the xy-plane. Hence, coordinates of L is (3, 4, 0). Similarly, we can find the coordinates of of M (0, 4, 5) and N (3, 0, 5), Fig.12.6.

Let L, M, N are the feet of the perpendiculars drawn from the point P (3, 4, 5) on Fig. 12.6 the xy, yz and zx-planes, respectively. Find the distance of these points L, M, N from the point P, Fig.12.7. Solution L is the foot of perpendicular drawn from the point P (3, 4, 5) to the xy-plane. Therefore, the coordinate of the point L is (3, 4, 0). The distance between the point (3, 4,

Using distance formula show that the points P (2, 4, 6), Q (– 2, – 2, – 2) and R (6, 10, 14) are collinear. Fig. 12.7 Solution Three points are collinear if the sum of any two distances is equal to the third distance. PQ = (–2 – 2) 2 + (–2 – 4) 2 + (–2 – 6) 2 = 16 + 36 + 64 = 116 = 2 29 QR = (6 + 2) 2 + (10 + 2) 2 + (14 + 2) 2 = 64 +144 + 256 = 464 = 4 29 PR = (6 − 2) 2 + (10 − 4) 2 + (14 – 6) 2 = 16 + 36 + 64 = 116 = 2 29 Since QR = PQ + PR. Therefore, the given points are collinear. INTRODUCTION TO THREE DIMENSIONAL GEOMETRY 213

Find the coordinates of a point equidistant from the four points O (0, 0, 0), A (l, 0, 0), B (0, m, 0) and C (0, 0, n). Solution Let P (x, y, z) be the required point. Then OP = PA = PB = PC. Now OP = PA ⇒OP2 = PA2 ⇒ x2 + y2 + z2 = (x – l)2 + (y – 0)2 + (z – 0)2 ⇒ x = m n Similarly, OP = PB ⇒ y = and OP = PC ⇒ z = 2 2 Hence, the coordinate of the required point are ( , , ). 2 2 2

Find the point on x-axis which is equidistant from the point A (3, 2, 2) and B (5, 5, 4). Solution The point on the x-axis is of form P (x, 0, 0). Since the points A and B are equidistant from P. Therefore PA2 = PB2, i.e., (x – 3)2 + ( 0 – 2)2 + (0 – 2)2 = (x – 5)2 + (0 – 5)2 + (0 – 4)2 ⇒ 4x = 25 + 25 + 16 – 17 i.e., x = . Thus, the point P on the x - axis is ( , 0, 0) which is equidistant from A and B.

Find the point on y-axis which is at a distance 10 from the point (1, 2, 3) Solution Let the point P be on y-axis. Therefore, it is of the form P (0, y, 0). The point (1, 2, 3) is at a distance 10 from (0, y, 0). Therefore (1 − 0) 2 + (2 − y ) 2 + (3 − 0) 2 = 10 ⇒ y2 – 4y + 4 = 0 ⇒ (y – 2)2 = 0 ⇒ y = 2 Hence, the required point is (0, 2, 0).

If a parallelopiped is formed by planes drawn through the points (2, 3, 5) and (5, 9, 7) parallel to the coordinate planes, then find the length of edges of a parallelopiped and length of the diagonal. Solution Length of edges of the parallelopiped are 5 – 2, 9 – 3, 7 – 5 i.e., 3, 6, 2. Length of diagonal is 32 + 62 + 22 = 7 units.

Show that the points (0, 7, 10), (–1, 6, 6) and (– 4, 9, 6) form a right angled isosceles triangle. Solution Let P (0, 7, 10), Q (–1, 6, 6) and R (– 4, 9, 6) be the given three points. Here PQ = 1 + 1 + 16 = 3 2 QR = 9+9+0 = 3 2 PR = 16 + 4 + 16 = 6 Now PQ2 + QR2 = (3 2) 2 + (3 2) 2 = 18 + 18 = 36 = (PR)2 Therefore, ∆ PQR is a right angled triangle at Q. Also PQ = QR. Hence ∆ PQR is an isosceles triangle.

Show that the points (5, –1, 1), (7, – 4, 7), (1 – 6, 10) and (–1, – 3, 4) are the vertices of a rhombus. Solution Let A (5, – 1, 1), B (7, – 4, 7), C(1, – 6, 10) and D (– 1, – 3, 4) be the four points of a quadrilateral. Here AB = 4 + 9 + 36 = 7 , BC = 36 + 4 + 9 = 7, CD = 4 + 9 + 36 = 7, DA = 23 + 4 + 9 = 7 Note that AB = BC = CD = DA. Therefore, ABCD is a rhombus.

Find the ratio in which the line segment joining the points (2, 4, 5) and (3, 5, – 4) is divided by the xz-plane. Solution Let the joint of P (2, 4, 5) and Q (3, 5, – 4) be divided by xz-plane in the ratio k:1 at the point R(x, y, z). Therefore 3k + 2 5k + 4 − 4k + 5 x= , y= , z= k +1 k +1 k +1 Since the point R (x, y, z) lies on the xz-plane, the y-coordinate should be zero,i.e., 5k + 4 4 k +1 = 0 ⇒ k = − Hence, the required ratio is – 4 : 5, i.e.; externally in the ratio 4 : 5.

Find the coordinate of the point P which is five - sixth of the way from A (– 2, 0, 6) to B (10, – 6, – 12). INTRODUCTION TO THREE DIMENSIONAL GEOMETRY 215 Solution Let P (x, y, z) be the required point, i.e., P divides AB in the ratio 5 : 1. Then 5 × 10 + 1 × –2 5 × – 6 + 1 × 0 5 ×− 12 + 1 × 6 P (x, y, z) = , , = (8, – 5, – 9) 5 +1 5 +1 5 +1

Describe the vertices and edges of the rectangular parallelopiped with vertex (3, 5, 6) placed in the first octant with one vertex at origin and edges of parallelopiped lie along x, y and z-axes. Solution The six planes of the parallelopiped are as follows: Plane OABC lies in the xy-plane. The z-coordinate of every point in this plane is zero. z = 0 is the equation of this xy-plane. Plane PDEF is parallel to xy-plane and 6 unit distance above it. The equation of the plane is z = 6. Plane ABPF represents plane x = 3. Plane OCDE lies in the yz-plane and x = 0 is the equation of this plane. Plane AOEF lies in the xz-plane. The y coordinate of everypoint in this plane is zero. Therefore, y = 0 is the equation of plane. Plane BCDP is parallel to the plane AOEF at a distance y = 5. Edge OA lies on the x-axis. The x-axis has equation y = 0 and z = 0. Edges OC and OE lie on y-axis and z-axis, respectively. The y-axis has its equation z = 0, x = 0. The z-axis has its equation x = 0, y = 0. The perpendicular distance of the point P (3, 5, 6) from the x- axis is 52 + 6 2 = 61 . The perpendicular distance of the point P (3, 5, 6) from y-axis and z-axis are 32 + 62 = 45 and 32 + 52 =, respectively. The coordinates of the feet of perpendiculars from the point P (3, 5, 6) to the coordinate axes are A, C, E. The coordinates of feet of perpendiculars from the point P on the coordinate planes xy, yz and zx are (3, 5, 0), (0, 5, 6) and Fig. 12.8 (3, 0, 6). Also, perpendicular distance of the point P from the xy, yz and zx-planes are 6, 5 and 3, respectively, Fig.12.8.

Let A (3, 2, 0), B (5, 3, 2), C (– 9, 6, – 3) be three points forming a triangle. AD, the bisector of ∠ BAC, meets BC in D. Find the coordinates of the point D. Solution Note that AB = (5 – 3) 2 + (3 − 2) 2 + (2 − 0) 2 = 4 +1 + 4 = 3 AC = (–9 – 3) 2 + (6 − 2) 2 + (−3 − 0) 2 = 144 +16 + 9 = 13 BD AB 3 Since AD is the bisector of ∠ BAC,We have = = DC AC 13 i.e., D divides BC in the ratio 3 : 13. Hence, the coordinates of D are 3( − 9) +13(5) 3(6) +13(3) 3( − 3) +13(2) 19 57 17 , , = , , 3 + 13 3 + 13 3 + 13 8 16 16

Determine the point in yz-plane which is equidistant from three points A (2, 0 3) B (0, 3, 2) and C (0, 0, 1). Solution Since x-coordinate of every point in yz-plane is zero. Let P (0, y, z) be a point on the yz-plane such that PA = PB = PC. Now PA = PB ⇒ (0 – 2)2 + (y – 0)2 + (z – 3)2 = (0 – 0)2 + (y – 3)2 + (z – 2)2 , i.e. z – 3y = 0 and PB = PC ⇒ y2 + 9 – 6y + z2 + 4 – 4z = y2 + z2 + 1 – 2z , i.e. 3y + z = 6 Simplifying the two equating, we get y = 1, z = 3 Here, the coordinate of the point P are (0, 1, 3).

The length of the foot of perpendicular drawn from the point P (3, 4, 5) on y-axis is

• (A)10
• (B)34
• (C)113
• (D)5 2 Solution Let l be the foot of perpendicular from point P on the y-axis. Therefore, its x and z-coordinates are zero, i.e., (0, 4, 0). Therefore, distance between the points (0, 4, 0) and (3, 4, 5) is 9 + 25 i.e., 34 . INTRODUCTION TO THREE DIMENSIONAL GEOMETRY 217

What is the perpendicular distance of the point P (6, 7, 8) from xy-plane?

• (A)8
• (B)7
• (C)6
• (D)None of these Solution Let L be the foot of perpendicular drawn from the point P (6, 7, 8) to the xy- plane and the distance of this foot L from P is z-coordinate of P, i.e., 8 units.

L is the foot of the perpendicular drawn from a point P (6, 7, 8) on the xy- plane. What are the coordinates of point L?

• (A)(6, 0, 0)
• (B)(6, 7, 0)
• (C)(6, 0, 8)
• (D)none of these Solution Since L is the foot of perpendicular from P on the xy-plane, z-coordinate is zero in the xy-plane. Hence, coordinates of L are (6, 7, 0).

L is the foot of the perpendicular drawn from a point (6, 7, 8) on x-axis. The coordinates of L are

• (A)(6, 0, 0)
• (B)(0, 7, 0)
• (C)(0, 0, 8)
• (D)none of these Solution Since L is the foot of perpendicular from P on the x- axis, y and z-coordinates are zero. Hence, the coordinates of L are (6, 0, 0).

What is the locus of a point for which y = 0, z = 0?

• (A)equation of x-axis
• (B)equation of y-axis
• (C)equation of z-axis
• (D)none of these Solution Locus of the point y = 0, z = 0 is x-axis, since on x-axis both y = 0 and z = 0.

L, is the foot of the perpendicular drawn from a point P (3, 4, 5) on the xz plane. What are the coordinates of point L ?

• (A)(3, 0, 0)
• (B)(0, 4, 5)
• (C)(3, 0, 5)
• (D)(3, 4, 0) Solution Since L is the foot of perpendicular segment drawn from the point P (3, 4, 5) on the xz-plane. Since the y-coordinates of all points in the xz-plane are zero, coordinate of the foot of perpendicular are (3, 0, 5). Fill in the blanks in Examples 24 to 28.

A line is parallel to xy-plane if all the points on the line have equal _____. Solution A line parallel to xy-plane if all the points on the line have equal z-coordinates.

The equation x = b represents a plane parallel to _____ plane. Solution Since x = 0 represent yz-plane, therefore x = b represent a plane parallel to yz -plane at a unit distance b from the origin.

Perpendicular distance of the point P (3, 5, 6) from y-axis is ________ Solution Since M is the foot of perpendicular from P on the y-axis, therefore, its x and z-coordinates are zero. The coordinates of M is (0, 5, 0). Therefore, the perpendicular distance of the point P from y-axis 32 + 62 = 45 .

L is the foot of perpendicular drawn from the point P (3, 4, 5) on zx- planes. The coordinates of L are ________. Solution Since L is the foot of perpendicular from P on the zx-plane, y-coordinate of every point is zero in the zx-plane. Hence, coordinate of L are (3, 0, 5).

The length of the foot of perpendicular drawn from the point P (a, b, c) on z-axis is _____. Solution The coordinates of the foot of perpendicular from the point P (a, b, c) on z- axis is (0, 0,c). The distance between the point P (a, b, c) and (0, 0, c) is a 2 + b2 . Check whether the statements in Example from 30 to 37 are True or False

The y-axis and z-axis, together determine a plane known as yz-plane. Solution True

The point (4, 5, – 6) lies in the VIth octant. Solution False, the point (4, 5, – 6) lies in the Vth octant,

The x-axis is the intersection of two planes xy-plane and xz plane. Solution True.

Three mutually perpendicular planes divide the space into 8 octants. Solution True.

The equation of the plane z = 6 represent a plane parallel to the xy-plane, having a z-intercept of 6 units. Solution True.

The equation of the plane x = 0 represent the yz-plane. Solution True.

The point on the x-axis with x-coordinate equal to x0 is written as (x0, 0, 0). Solution True.

x = x0 represent a plane parallel to the yz-plane. Solution True. INTRODUCTION TO THREE DIMENSIONAL GEOMETRY 219 Match each item given under the column C1 to its correct answer given under column C2.

Column C1 Column C2 (a) If the centriod of the triangle is

• (i)Parallelogram origin and two of its vertices are (3, – 5, 7) and (–1, 7, – 6) then the third vertex is (b) If the mid-points of the sides of
• (ii)(–2, –2, –1) triangle are (1, 2, – 3), (3, 0, 1) and (–1, 1, – 4) then the centriod is (c) The points (3, – 1, – 1), (5, – 4, 0),
• (iii)as Isosceles right-angled triangle (2, 3, – 2) and (0, 6, – 3) are the vertices of a (d) Point A(1, –1, 3), B (2, – 4, 5) and
• (iv)(1, 1, – 2) C (5, – 13, 11) are (e) Points A (2, 4, 3), B (4, 1, 9) and (v) Collinear C (10, – 1, 6) are the vertices of Solution (a) Let A (3, – 5, 7), B (– 1, 7, – 6), C (x, y, z) be the vertices of a ∆ ABC with centriod (0, 0, 0) 3 −1+ x −5 + 7 + y 7 − 6 + z x+2 y+2 Therefore, (0, 0, 0) = , , . This implies =0 , =0 , 3 3 3 3 3 z +1 =0 . Hence x = – 2, y = – 2, and z = – 1.Therefore (a) ↔ (ii) (b) Let ABC be the given ∆ and DEF be the mid-points of the sides BC, CA, AB, respectively. We know that the centriod of the ∆ ABC = centriod of ∆ DEF. 1+ 3 −1 2 + 0 +1 −3 +1− 4 Therefore, centriod of ∆ DEF is , , = (1, 1, – 2) 3 3 3 Hence (b) ↔ (iv) 3 + 2 −1+ 3 –1– 2 5 −3 (c) Mid-point of diagonal AC is , , = ,1, 2 2 2 2 2 5+0 −4+6 0− 3 5 −3 Mid-point of diagonal BD is , , = ,1, 2 2 2 2 2 Diagonals of parallelogram bisect each other. Therefore (c) ↔ (i) (d) AB = (2 −1) 2 + (− 4 +1) 2 + (5 − 3) 2 = 14 BC = (5 − 2) 2 + (−13 + 4) 2 + (11− 5) 2 = 3 14 AC = (5 −1) 2 + (−13 +1) 2 + (11 − 3) 2 = 4 14 Now AB + BC = AC . Hence Points A, B, C are collinear. Hence (d) ↔ (v) (e) AB = 4 + 9 + 36 = 7 BC = 36 + 4 + 9 = 7 CA = 64 + 25 + 9 = 7 2 Now AB2 +BC2 = AC2 . Hence ABC is an isosceles right angled triangle and hence (e) ↔ (iii)