3D Coordinates (Q’s & Ans) || V.S.A.Q’S||

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This note is designed by the ‘Basics in Maths’ team. These notes to do help the intermediate First-year Maths students.

Inter Maths – 1B two mark questions and solutions are very useful in IPE examinations.

These notes cover all the topics covered in the intermediate First-year Maths syllabus and include plenty of solutions to help you solve all the major types of Math problems asked in the IPE examinations.  


QUESTION 1

Show that the points A ( – 4, 9, 6), B (– 1, 6, 6), and C(0, 7, 10) right-angled isosceles triangle.

Sol:

Distance between two points P (x1, y1, z1) and Q (x2, y2, z2) is PQ = 3D Geometry 1

3D Geometry 2

AB = BC = 3D Geometry 3

AB2 + BC2  = 18 + 18 = 36 = AC2

∴ ∠ B = 900  

 points A (– 4, 9, 6), B (– 1, 6, 6) and C (0, 7, 10) form a right-angled isosceles triangle

QUESTION 2

Show that the locus of the point whose distance from Y – axis is thrice its distance from (1, 2, – 1) is 8x 2 + 9y2 + 8 z2 – 18x – 36y + 18z + 54 = 0

Sol:

Let P (x, y, z) be the locus of the point

        A (0, y, 0) be any point on Y – axis

         B = (1, 2, – 1)

     Condition is PA = 3PB

           PA2 = (3PB)2

                             z2 = 9[x2 – 2x + 1 + y2 – 4y + 4 + z2 + 2z + 1]

           x2 + z2 = 9x2 – 18x + 9 + 9y2 – 36y + 36 +9 z2 + 18z + 9

∴ 8x 2 + 9y2 + 8 z2 – 18x – 36y + 18z + 54 =0

QUESTION 3

A, B, C are three points on OX, OY and OZ respectively, at distances a, b, c (a≠0, b≠0, c≠0) from the origin ‘O’. Find the coordinate of the point which is equidistant from A, B, C and O  

Sol:

Let P (x, y, z) be the required point

    O = (0, 0, 0)   A = (a, 0, 0), B = (0, b, 0), C = (0, 0, c)

Given, AP = BP = CP = OP

             AP = OP

        ⟹   AP2 =OP2

 (x – a )2 + y2 + z2 = x2 + y2 + z2

  x2 + a2 – 2ax + y2 + z2 = x2 + y2 + z2

  a2 – 2ax = 0

  a (a – 2x) = 0

    a – 2x = 0 (∵ a≠0)

    a = 2x ⟹ a/2

    Similarly, y = b/2 and z = c/2

∴ P = (a/2, b/2, c/2)

QUESTION 4

Show that the points A (3, – 2, 4), B (1, 1, 1) and C (– 1, 4, – 2) are collinear

Sol:

Given points are A (3, – 2, 4), B (1, 1, 1) and C (– 1, 4, – 2)

3D Geometry 4  

AB + BC = 3D Geometry 5

 AB + BC = AC

∴ A B and C are collinear

QUESTION 5

Find x if the distance between (5, – 1, 7), (x, 5, 1) is 9 units.

Sol:

 Let A = (5, – 1, 7), B = (x, 5, 1)

   Given AB = 9

    ⟹ AB2 = 81

  (5 – x)2 + (– 1 – 5)2 + (7 – 1)2 = 81

 (5 – x)2 + 36 + 36 = 81

  (5 – x)2 + 72 = 81

         (5 – x)2 = 81 – 72 = 9

        (5 – x) = ± 3

     5 – x = 3 or 5 – x = – 3

    5 – 3 = x or 5 + 3 = x

     x = 2 or x = 8

QUESTION 6

If the point (1, 2, 3) is changed to the point (2, 3, 1) through the translation of axes. Find a new origin.

Sol:

Given (x, y, z) = (1, 2, 3) and (X, Y, Z) = (2, 3, 1)

x = X + h, y = Y + k, z = Z + l

h = x – X, k = y – Y, l = z – Z

h = 1 – 2, k = 2 – 3, l = 3 – 1

h = – 1, k = – 1, l = 2

New origin is (– 1, – 1, 2)

QUESTION 7

By section formula, find the point which divides the line joining the points (2, – 3, 1) and (3, 4, – 5) in the ratio 1 : 3.

Sol:

If a point P divides the line segment joining the points (x1, y1, z1), (x2, y2, z2) in the ratio, then P =3D Geometry 6

 Let P divides the line joining the points (2, – 3, 1) and (3, 4, – 5) in the ratio 1 : 3.

3D Geometry 7

QUESTION 8

Find the ratio in which the line joining two points (7, 0, – 1) and (– 2, 3, 5) is divided by the point (1,2,3).

Sol:

Let A = (7, 0, – 1), B = (– 2, 3, 5) and P = (1,2,3)

Suppose P divides AB in the ratio k : 1

P = 3D Geometry 8

  (1,2,3) = 3D Geometry 9

     1 = 3D Geometry 10

    7 – 2k = k + 1

    7 – 1 = k + 2k

     6 = 3k

     k = 2

 ∴ P divides AB in the Ratio 1 : 2

QUESTION 9

Using section formula, verify whether the points A (2, –4, 3), B (–4, 5, 6), C (4, –7, 2) are collinear or not.

Sol:

Given Points are A (2, –4, 3), B (–4, 5, 6), C (4, –7, 2)

Let C divides AB in the ratio k : 1

   C = 3D Geometry 11

 (4, –7, 2) =3D Geometry 12

     4 =3D Geometry 13

 k = 4 (k + 1)

 2 – 4k = 4k + 4

– 4k– 4k = 4 – 2

– 8k = 2

K = -1/4

C divides AB in the Ratio 1 : 4 externally

∴ A, B, C are collinear

QUESTION 10

Find the centroid of the triangle whose vertices are (5, 4, 6), (1, –1, 3) and (4, 3, 2)

Sol:

The centroid of the triangle whose vertices are (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) is 3D Geometry 14

3D Geometry 15

∴ The centroid of the triangle = 3D Geometry 16

QUESTION 11

Find the centroid of the tetrahedron whose vertices are (2, 3, –4), (–3, 3, –2), (–1, 4, 2) and (3, 5, 1)

Sol:

The centroid of the triangle whose vertices are (x1, y1, z1), (x2, y2, z2) (x3, y3, z3), and (x4, y4, z4) is 3D Geometry 17

 the centroid of the tetrahedron whose vertices are (2, 3, –4), (–3, 3, –2), (–1, 4, 2) and (3, 5, 1) is 3D Geometry 18

= 3D Geometry 19

QUESTION 12

Find the ratio in which the YZ-plane divides the line joining A(2, 4, 5) and B (3, 5, – 4)

Sol:

Let P be any point on the YZ-plane

   P = (0, y, z)

Let P divides AB in the ratio k:1

P =3D Geometry 20

⟹ (0, y, z) =3D Geometry 20

3D Geometry 21 = 0

      3k + 2 = 0

    3k =– 2

      k = 3D Geometry 22 

YZ-plane divides AB in the ratio – 2:3

QUESTION 13

Find the fourth vertex of the parallelogram whose consecutive vertices are (2, 4, – 1), (3, 6, – 1) and (4, 5, 1).

Sol:

 let A = (2, 4, – 1), B = (3, 6, – 1), C = (4, 5, 1) and D = (x, y, z)

 ABCD is a parallelogram

 The midpoint of AC = Midpoint of BD

   3D Geometry 23         

  ⟹3D Geometry 24  

   ⟹ 3 + x = 6, 6 + y = 9, – 1 + z = 0

   x = 3, y = 3, z = 1

    ∴ The fourth vertex D = (3, 3, 1)

QUESTION 14

A (5, 4, 6), B (1, –1, 3), C (3, 3, 1) are three points. Find the coordinates of the point in which the bisector of ∠BAC meets the side BC.

Sol:

We know that the bisector of ∠BAC divides BC in the ratio AB:AC

3D Geometry 30

AB =3D Geometry 25

3D Geometry 26

AC =3D Geometry 27

  3D Geometry 28

AB:AC = 5 3D Geometry 29: 33D Geometry 29

            = 5:3

If D is a point where the bisector of ∠BAC meets BC

 ⟹ D divides BC in the ratio 5:3

3D Geometry 31

QUESTION 15

If M (α, β, γ) is the midpoint of the line joining the points (x1, y1, z1) and B, then find B

Sol:

Let B (x, y, z) be the required point

 M is the midpoint of AB

⟹ (α, β, γ) = 3D Geometry 32

  ⟹ 2 α = x + x1; 2 β = y + y1; 2 γ = z + z1

         x =2 α – x1; x =2 β – y1; x =2 γ – z1

    ∴ B = (2 α – x1, 2 β – y1, 2 γ –)

QUESTION 16

If H, G, S and I respectively denote orthocenter, centroid, circumcenter and incentre of a triangle formed by the points (1, 2, 3), (2, 3, 1) and (3, 1, 2), then find H, G, S, I.

Sol:

Let A = (1, 2, 3), B= (2, 3, 1), C = (3, 1, 2)

AB =3D Geometry 33

      =3D Geometry 34

      =3D Geometry 35

BC =3D Geometry 36

      =3D Geometry 37

      =3D Geometry 35

AC=3D Geometry 38

      =

      =3D Geometry 35

AB = BC = AC

 ⟹ ∆ ABC is an equilateral triangle

 We know that, in an equilateral triangle orthocenter, centroid, circumcenter and incentre are same

Centroid G =3D Geometry 40

                      = (2, 2, 2)

∴ H = (2, 2, 2), S = (2, 2, 2), I = (2, 2, 2)

QUESTION 17

Find the incentre of the triangle formed by the points (0, 0, 0), (3, 0, 0) and (0, 4, 0).

Sol:

Let A = (0, 0, 0), B = (3, 0, 0) and C = (0, 4, 0)

3D Geometry 41

=

     a = 5

3D Geometry 42

   b = 4

ATTACHMENT DETAILS 3D-Geometry-43.jpg February 15, 2021 13 KB 426 by 74 pixels Edit Image Delete permanently Alt Text Describe the purpose of the image(opens in a new tab). Leave empty if the image is purely decorative.

     c = 3

incentre =3D Geometry 45

  ⟹ I =3D Geometry 45

          =3D Geometry 46

∴ I = (1, 1,0)

QUESTION 18

Find the ratio in which the point P (5, 4, – 6) divides the line segment joining the points A (3, 2, – 4) and B (9, 8, –10). Also, find the harmonic conjugate of P

Sol:

Harmonic Conjugate: If P divides AB in the ratio m: n, then the Harmonic Conjugate of P (i.e., Q) divides AB in the ratio –m: n.

Given points are A (3, 2, – 4), B (9, 8, –10) and P (5, 4, – 6)

P divides AB in the ratio is x2 – x : x – x2

                                                     = 3 – 5 : 5 – 9

                                                      = 1 : 2 (internally)

Let Q be the harmonic conjugate of P

  ⟹ Q divides AB in the ratio –1: 2

      Q = 3D Geometry 48

         = (–3, –4, –2)

Q (–3, –4, –2) is the harmonic conjugate of P (5, 4, – 6)

QUESTION 19

If (3, 2, – 1), (4, 1,1) and (6, 2, 5) are three vertices and (4, 2, 2) is the centroid of the tetrahedron, find the fourth Vertex.

Sol:

Given vertices of Tetrahedron are A (3, 2, – 1), B (4, 1,1), C (6, 2, 5) and centroid G = (4, 2, 2)

Let the fourth vertex is D = (x, y, z)

 The centroid of the tetrahedron whose vertices are (x1, y1, z1), (x2, y2, z2) (x3, y3, z3), and (x4, y4, z4) is 3D Geometry 17

(4, 2, 2) = 3D Geometry 49

(4, 2, 2) =3D Geometry 50

13 + x = 16, 5 + y = 8, 5 + z = 2

x = 3, y = 3, z = 3

∴ the fourth vertex D = (3, 3, 3)  

QUESTION 20

Show that the points A(3, 2, –4), (5, 4, –6) and C(9, 8, –10) are collinear and find the ratio in which B divides AC.

Sol:

Given points are A(3, 2, –4), (5, 4, –6) and C(9, 8, –10)

AB =3D Geometry 51

      3D Geometry 52

  BC =3D Geometry 55

  3D Geometry 54

      AC=3D Geometry 55

3D Geometry 56

     AB + BC = 3D Geometry 57  = AC

∴ the points A (3, 2, –4), (5, 4, –6) and C(9, 8, –10) are collinear

B divides AB in the ratio is AB:BC =  3D Geometry 58  = 1:2


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