Coordinates in Space Exercise: 9.1 Class 12 Basic Mathematics Solution [NEB UPDATED]

Syllabus: Content to study

1. Co-ordinate in space
2. Distance between two points
3. Locs and Equations
4. Section Formulae: Internal Division and External Division

Exercise: 9.1

1. Find the distance between the points

Solution

a. (-1, 4, 3) and (2,2,-3)

d = $\sqrt {{{\left( {2 + 1} \right)}^2} + {{\left( {2 - 4} \right)}^2} + {{\left( { - 3 - 3} \right)}^2}} $ = $\sqrt {9 + 4 + 36} $ = $\sqrt {49} $ = 7

 

b. (4,-1, 5) and (-4, 3, 6)

d = $\sqrt {{{\left( { - 4 - 4} \right)}^2} + {{\left( {3 + 1} \right)}^2} + {{\left( {6 - 5} \right)}^2}} $ = $\sqrt {64 + 16 + 1} $ = $\sqrt {81} $ = 9.

 

2. a) Show that the point (2, 0, -4), (4, 2, 4) and (10, 2, -2) are the vertices of an equilateral triangle.

Solution

The given vertices are A(2,0,-4), B(4,2,4) and C(10,2,-2).

To show: AB = BC = CA.

Now, AB = $\sqrt {{{\left( {4 - 2} \right)}^2} + {{\left( {2 - 0} \right)}^2} + {{\left( {4 + 4} \right)}^2}} $ = $\sqrt {4 + 4 + 64} $ = $\sqrt {72} $ = 6$\sqrt 2 $.

BC = $\sqrt {{{\left( {10 - 4} \right)}^2} + {{\left( {2 - 0} \right)}^2} + {{\left( { - 2 - 4} \right)}^2}} $ = $\sqrt {36 + 0 + 36} {\rm{\: }}$ = $\sqrt {72} $ = 6$\sqrt 2 $.

And CA = $\sqrt {{{\left( {2 - 10} \right)}^2} + {{\left( {0 - 2} \right)}^2} + {{\left( { - 4 + 2} \right)}^2}} $ = $\sqrt {64 + 4 + 4} {\rm{\: }}$ = $\sqrt {72} $ = 6$\sqrt 2 $.

So, AB = BC = CA ,i.e. $\Delta $ABC is equilateral.

 

b. Show that the point (0, 7, 10), (-1, 6, 6) and (-4, 9, 6) are the vertices of a right-angled isosceles triangle.

Solution

The given vertices are A(0,7,10), B(-1,6,6) and C(-4,9,6).

To show: $\Delta $ABC is an isosceles triangle.

Now, AB = $\sqrt {{{\left( { - 1 - 0} \right)}^2} + {{\left( {6 - 7} \right)}^2} + {{\left( {6 - 10} \right)}^2}} $ = $\sqrt {1 + 1 + 16} $ = $\sqrt {18} $ = 3$\sqrt 2 $.

BC = $\sqrt {{{\left( { - 4 - 1} \right)}^2} + {{\left( {9 - 6} \right)}^2} + {{\left( {6 - 6} \right)}^2}} $ = $\sqrt {9 + 9 + 0} {\rm{\: }}$ = $\sqrt {18} $ = 3$\sqrt 2 $.

And CA = $\sqrt {{{\left( {0 + 4} \right)}^2} + {{\left( {7 - 9} \right)}^2} + {{\left( {10 - 6} \right)}^2}} $ = $\sqrt {16 + 4 + 16} {\rm{\: }}$ = $\sqrt {36} $ = 6.

So, AB² + BC² = 18 + 18 = 36 = CA².

So, AB = BC and AB² + BC² = CA² (i.e. $\angle $ B = 90°)

So, $\Delta $ ABC is an isosceles right angled.

 

3) a. Show that the points (1, 2, 3), (-1, -2, -1), (2, 3, 2) and (4, 7, 6) are the vertices of a parallelogram.

Solution

The given vertices are A(1,2,3), B(-1,-2,-1),C(2,3,2) and D(4,7,6).

To show: ABCD is a parallelogram.

Now, AB = $\sqrt {{{\left( { - 1 - 1} \right)}^2} + {{\left( { - 2 - 2} \right)}^2} + {{\left( { - 1 - 3} \right)}^2}} $ = $\sqrt {4 + 16 + 16} $ = $\sqrt {36} $ = 6 .

BC = $\sqrt {{{\left( {2 + 1} \right)}^2} + {{\left( {3 + 2} \right)}^2} + {{\left( {2 + 1} \right)}^2}} $ = $\sqrt {9 + 25 + 9} $ = $\sqrt {41} $.

CD = $\sqrt {{{\left( {4 - 2} \right)}^2} + {{\left( {7 - 3} \right)}^2} + {{\left( {6 - 2} \right)}^2}} $ = $\sqrt {4 + 16 + 16} $ = $\sqrt {36} $ = 6.

And, DA = $\sqrt {{{\left( {1 - 4} \right)}^2} + {{\left( {2 - 7} \right)}^2} + {{\left( {3 - 6} \right)}^2}} $ = $\sqrt {9 + 25 + 9} $ = $\sqrt {41} $.

So, AB = CD and BC = DA.

So, ABCD is a parallelogram.

 

b. Show that the points (1, 1, 1), (-2, 4, 1), (-1, 5, 5) and (2, 2, 5) are the vertices of square.

Solution

The given vertices are A(1,1,1), B(-2,4,1),C(-1,5,5) and D(2,2,5).

To show: ABCD is a square.

Now, AB = $\sqrt {{{\left( { - 2 - 1} \right)}^2} + {{\left( {4 - 1} \right)}^2} + {{\left( {1 - 1} \right)}^2}} $ = $\sqrt {9 + 9 + 0} $ = $\sqrt {18} $ = 3$\sqrt 2 $.

BC = $\sqrt {{{\left( { - 1 + 2} \right)}^2} + {{\left( {5 - 4} \right)}^2} + {{\left( {5 - 1} \right)}^2}} $ = $\sqrt {1 + 1 + 16} $ = $\sqrt {18} $. = 3$\sqrt 2 $.

CD = $\sqrt {{{\left( {2 + 1} \right)}^2} + {{\left( {2 - 5} \right)}^2} + {{\left( {5 - 5} \right)}^2}} $ = $\sqrt {9 + 9 + 0} $ = $\sqrt {18} $ = 3$\sqrt 2 $.

And, DA = $\sqrt {{{\left( { - 1 - 1} \right)}^2} + {{\left( {5 - 1} \right)}^2} + {{\left( {5 - 1} \right)}^2}} $ = $\sqrt {4 + 16 + 16} $ = $\sqrt {36} $. = 6.

Now, AB² + BC² = ${\left( {3\sqrt 2 } \right)^2}$ + ${\left( {3\sqrt 2 } \right)^2}$ = 18 + 18 = 36 = AC².

So, AB = BC = CD = DA and AB² + BC² = AC².

So, the vertices are the vertices of a square.

 

4. Show that the following points are collinear

a. (1, 2, 3), (-2, 3, 4) and (7, 0, 1)

Solution

The given points are A(1,2,3), B(-2,3,4) and C(7,0,1).

To show: A, B, C are collinear.

Now, AB = $\sqrt {{{\left( { - 2 - 1} \right)}^2} + {{\left( {3 - 2} \right)}^2} + {{\left( {4 - 3} \right)}^2}} {\rm{\: }}$= $\sqrt {9 + 1 + 1} $ = $\sqrt {11} $.

BC = $\sqrt {{{\left( {7 + 2} \right)}^2} + {{\left( {0 - 3} \right)}^2} + {{\left( {1 - 4} \right)}^2}} $ = $\sqrt {81 + 9 + 9} $ = $\sqrt {99} $ = 3$\sqrt {11} $.

Now, AB + CA = $\sqrt {11} $ + 2$\sqrt {11} $ = 3$\sqrt {11} $ = BC.

i.e. A,B and C are collinear.

 

b. (-2, 3, 5), (1, 2, 3) and (7,0,-1)

Now, AB = $\sqrt {{{\left( {1 + 2} \right)}^2} + {{\left( {2 - 3} \right)}^2} + {{\left( {3 - 5} \right)}^2}} $ = $\sqrt {9 + 1 + 4} $ = $\sqrt {14} $.

BC = $\sqrt {{{\left( {7 - 1} \right)}^2} + {{\left( {0 - 2} \right)}^2} + {{\left( { - 1 - 3} \right)}^2}} $ = $\sqrt {36 + 4 + 16} $ = $\sqrt {56} $ = 2$\sqrt {14} $.

And CA = $\sqrt {{{\left( {7 + 2} \right)}^2} + {{\left( {0 - 3} \right)}^2} + {{\left( { - 1 - 5} \right)}^2}} $ = $\sqrt {81 + 9 + 36} $ = $\sqrt {126} $ = 3$\sqrt {14} $.

So, AB + BC + CA.

SO, A, B, C are collinear.

 

5) a. Find the locus of a point which moves such that its distance from the fixed p (1,2,-2) is always 5.

Solution

Let P(x,y,z) be the point whose locus is required and which remains at a distance 5 from the given point A(1,2,-2).

So, PA = 5.

Or, PA² = 25.

Or, (x – 1)² + (y – 2)² + (z + 2)² = 25.

So, x² + y² + z² – 2x – 4y + 4z – 16 = 0 is the required locus.

 

b. Find the locus of a point which moves such that it is equidistant from two fixed po (-1, 2, 3) and (4,-1, 5).

Solution

Let P(x, y, z) be the given point which is at and equal distance from the points A(-1,2,3) and B(4,-1,5).

So, PA = PB.

Or, PA² = PB²

Or, (x + 1)² + (y – 2)² + (z – 3)² = (x – 4)² + (y + 1)² + (z – 5)².

Or, x² + 2x + 1 + y² – 4y + 4 + z² – 6z + 9 = x² – 8x + 16 + y² + 2y + 1 + z² – 10z + 25.

Or, 10x – 6y + 4z – 28 = 0.

So, 5x – 3y + 2z – 14 = 0 is the required locus.

 

6. Find the coordinates of the point which is equidistant from the four points O, A, B and C where O is the origin and A, B and C are the points on the x-, y- and z-axis respectively a distances a, b and c from the origin.

Solution

Let P(x,y,z) be the point equidistant from O (0,0,0), A(a,0,0) , B(0,b,0) and C(0,0,c).

So, OP = AP = BP = CP

Now, OP = AP

Or, OP² = AP²

Or, (x – 0)² + (y – 0)² + (z – 0)² = (x – a)² + (y – 0)² + (z – 0)².

Or, x² + y² + z² = x² – 2ax + a² + y² + z².

Or, 0 = - 2ax + a².

Or, a( -2x + a) = 0

So, - 2x + a = 0  (a ≠ 0).

So, x = $\frac{{\rm{a}}}{2}$.

Similarly, OP = BP à y = $\frac{{\rm{b}}}{2}$ and OP = CP à z = $\frac{{\rm{c}}}{2}$.

So, the required point if P $\left( {\frac{{\rm{a}}}{2},\frac{{\rm{b}}}{2},\frac{{\rm{c}}}{2}} \right)$.

 

7. Find the coordinates of the point which divides the line segment joining each of the following pair of points internally in the ratio 1:2 and externally in the ratio 3:2

a. (1, -2, 3) and (1, 2, 3)

Solution

(x₁, y₁, z₁) = (1,-2,3) and (x₂,y₂,z₂) = (1,2,3).

m₁ : m₂ = 1 : 2.

So, (x,y,z) = $\left( {\frac{{{{\rm{m}}_1}{{\rm{x}}_2} + {{\rm{m}}_2}{{\rm{x}}_1}}}{{{{\rm{m}}_1} + {{\rm{m}}_2}}},\frac{{{{\rm{m}}_1}{{\rm{y}}_2} + {{\rm{m}}_2}{{\rm{y}}_1}}}{{{{\rm{m}}_1} + {{\rm{m}}_2}}},\frac{{{{\rm{m}}_1}{{\rm{z}}_2} + {{\rm{m}}_2} + {{\rm{z}}_1}}}{{{{\rm{m}}_1} + {{\rm{m}}_2}}}} \right)$.

= $\left( {\frac{{1.1 + 2.1}}{{1 + 2}},\frac{{1.2 + 2.\left( { - 2} \right)}}{{1 + 2}},\frac{{1.3 + 2.3}}{{1 + 2}}} \right)$.

So, (x,y,z) = $\left( {1, - \frac{2}{3},3} \right)$.

For external division m₁ : m₂ = 3 : 2.

So, (x,y,z) = $\left( {\frac{{{{\rm{m}}_1}{{\rm{x}}_2} - {{\rm{m}}_2}{{\rm{x}}_1}}}{{{{\rm{m}}_1} - {{\rm{m}}_2}}},\frac{{{{\rm{m}}_1}{{\rm{y}}_2} - {{\rm{m}}_2}{{\rm{y}}_1}}}{{{{\rm{m}}_1} - {{\rm{m}}_2}}},\frac{{{{\rm{m}}_1}{{\rm{z}}_2} - {{\rm{m}}_2}{{\rm{z}}_1}}}{{{{\rm{m}}_1} - {{\rm{m}}_2}}}} \right)$.

Or, (x,y,z) = $\left( {\frac{{3.1 - 2.1}}{{3 - 2}},\frac{{3.2 - 2.\left( { - 2} \right)}}{{3 - 2}},\frac{{3.3 - 2.3}}{{3 - 2}}} \right)$

So, (xy,z) = (1,10,3).

 

b. (2, 0, 1) and (4, -2, 5)

(x₁,y₁,z₁) = (2,0,1) and (x₂,y₂,z₂) = (4,-2,5) and m₁ : m₂ = 1 : 2.

So, (x,y,z) = $\left( {\frac{{1.4 + 2.2}}{{1 + 2}},\frac{{1.\left( { - 2} \right) + 2.0}}{{1 + 2}},\frac{{1.5 + 2.1}}{{1 + 2}}} \right)$.

So, (x,y,z) = $\left( {\frac{8}{3}, - \frac{2}{3},\frac{7}{3}} \right)$.

For external division: m₁ : m₂ = 3 : 2.

So, (x,y,z) = $\left( {\frac{{3.4 - 2.2}}{{3 - 2}},\frac{{3.\left( { - 2} \right) - 2.0}}{{3 - 2}},\frac{{3.5 - 2.1}}{{3 - 2}}} \right)$.

So, (x,y,z) = (8,-6,13).

 

8. Find the coordinates of the mid-points of the join of each of the following pair of points

a. (-4, 3, 6) and (2, 1, -3)

Solution

Given points, (x₁,y₁,z₁) = (-4,3,6).

(x₂,y₂z₂) = (6,1,-3)

So, Mid – point = $\left( {\frac{{{{\rm{x}}_1} + {{\rm{x}}_2}}}{2},\frac{{{{\rm{y}}_1} + {{\rm{y}}_2}}}{2},\frac{{{{\rm{z}}_1} + {{\rm{z}}_2}}}{2}} \right)$ = $\left( {\frac{{ - 4 + 6}}{2},\frac{{3 + 1}}{2},\frac{{6 - 3}}{2}} \right)$.

= $\left( {1,2,\frac{3}{2}} \right)$ 

 

b. (2,5,-8) and (4, -1, 6)

Given points are, (x₁,y₁,z₁) = (2,5,-8) and (x₂,y₂,z₂) = (4,-1,6)

Mid – point = $\left( {\frac{{2 + 4}}{2},\frac{{5 - 1}}{2},\frac{{ - 8 + 6}}{2}} \right)$ = (3,2,-1).

 

9. If A(3, 4, 5), B(- 1, 2, 0) and C(- 3, 4, - 2) are the vertices of the triangle ABC, find the length of the median joining the vertex A(3, 4, 5) and middle point of its opposite side.

Solution

Middle point of BC is = P $\left( {\frac{{ - 1 - 3}}{2},\frac{{2 + 4}}{2},\frac{{0 - 2}}{2}} \right)$ = P(-2,3,-1).

So, length of the median, AP = $\sqrt {{{\left( {3 + 2} \right)}^2} + {{\left( {4 - 3} \right)}^2} + {{\left( {5 + 1} \right)}^2}} $ = $\sqrt {25 + 1 + 36} $ = $\sqrt {62} $.

So, AP = $\sqrt {62} $ units.

 

10) a. Find the ratio in which the line joining the points (-2, 4, 7) and (3, -5, -8) is divided by the xy-plane.

Solution

The given points are A(x₁,y₁,z₁) = (-2,4,7) and B(x₂,y₂,z₂) = (3,-5,-8).

Any point on the xy – plane is P(a,b,0).

Let P divides AB in the ratio m₁ : m₂.

Then z = $\frac{{{{\rm{m}}_1}{{\rm{z}}_2} + {{\rm{m}}_2}{{\rm{z}}_1}}}{{{{\rm{m}}_1} + {{\rm{m}}_2}}}$.

Or, 0 = $\frac{{ - 8{{\rm{m}}_1} + 7{{\rm{m}}_2}}}{{{{\rm{m}}_1} + {{\rm{m}}_2}}}$.

Or, 8m₁ = 7m₂.

So, m₁ : m₂ = 7 : 8.

 

b. Find the ratio in which the yz-plane divides the line joining the points (4, 6, 7) and (-1, 2, 5). Find also the coordinates of the points on the yz-plane.

Solution

The given points are A(x₁,y₁,z₁) = (4,6,7) and B(x₂,y₂,z₂) = (-1,2,5).

Any point on the yz – plane is P(0,b,c).

Now, let P divide AB in the ratio m₁:m₂.

So, 0 = $\frac{{{{\rm{m}}_1}{{\rm{x}}_2} + {{\rm{m}}_2}{{\rm{x}}_1}}}{{{{\rm{m}}_1} + {{\rm{m}}_2}}}$.

Or, 0 = - m₁ + 4m₂.

So, m₁ : m₂ = 4 : 1.

And, b = $\frac{{{{\rm{m}}_1}{{\rm{y}}_2} + {{\rm{m}}_2}{{\rm{y}}_1}}}{{{{\rm{m}}_1} + {{\rm{m}}_2}}}$ = $\frac{{4.2 + 1.6}}{{4 + 1}}$ = $\frac{{14}}{5}$.

And c = $\frac{{{{\rm{m}}_1}.{{\rm{z}}_2} + {{\rm{m}}_2}.{{\rm{z}}_1}}}{{{{\rm{m}}_1} + {{\rm{m}}_2}}}$ = $\frac{{4.5 + 1.7}}{{4 + 1}}$ = $\frac{{27}}{5}$.

Hence, ratio : 4 : 1 and point $\left( {0,\frac{{14}}{5},\frac{{27}}{5}} \right)$.

 

c. Given three collinear points A(3, 2, - 4) B(5, 4, - 6) and C(9, 8, - 10) find the ratio in which B divides AC.

Solution

Let B(5,4,-6) divides the line AC joining A (3,2,-4) and C(9,8,-10) in the ratio m₁:m₂.

So, 5 = $\frac{{{{\rm{m}}_1}.{{\rm{x}}_2} + {{\rm{m}}_2}.{{\rm{x}}_1}}}{{{{\rm{m}}_1} + {{\rm{m}}_2}}}$

Where, (x₁,y₁,z₁) = (3,2,-4).

(x₂,y₂,z₂) = (9,8,-10)

And (x,y,z) = (5,4,-6)

Or, 5m₁ + 5m₂ = 9m₁ + 3m₂

Or, 4m₁ = 2m₂

Or, $\frac{{{{\rm{m}}_1}}}{{{{\rm{m}}_2}}}$ = $\frac{1}{2}$.

So, m₁ : m₂ = 1 : 2.

 

11) a. Find the point where the line through the points (1, 2, 3) and (4, 4, 9) meets the z- plane.

Solution

Given points are A(x₁,y₁,z₁) = (1,2,3) and B(x₂,y₂,z₂) = (4,-4,9).

Le the line AB cuts the zx – plane at P(a,0,c) in the ratio m₁:m₂.

So, 0 = $\frac{{{{\rm{m}}_1}{{\rm{y}}_2} + {{\rm{m}}_2}{{\rm{y}}_1}}}{{{{\rm{m}}_1} + {{\rm{m}}_2}}}$

Or, 0 = - 4m₁ + 2m₂.

So, m₁ : m₂ = 1 : 2.

Now, a = $\frac{{{{\rm{m}}_1}{{\rm{x}}_2} + {{\rm{m}}_2}{{\rm{x}}_1}}}{{{{\rm{m}}_1} + {{\rm{m}}_2}}}$ = $\frac{{4.1 + 2.1}}{{1 + 2}}$ = 2

c = $\frac{{{{\rm{m}}_1} + {{\rm{z}}_2} + {{\rm{m}}_2}{{\rm{z}}_1}}}{{{{\rm{m}}_1} + {{\rm{m}}_2}}}$ = $\frac{{1.9 + 2.3}}{{1 + 2}}$ = 5.

So, the required point on the z-x plane is (2,0,5).

 

b. Find the point where the line joining the points (2, -3, 1) and (3, -4, -5) cuts the plane 2x + y + z = 7

Solution

The given plane is 2x + y + z = 7 ….(1)

And the points are A(2,-3,1) and B(3,-4,-5).

Let the plane (1) divide the line AB in the ratio m₁ : m₂ at the point P(a,b,c).

So, (a,b,c) = $\left( {\frac{{{{\rm{m}}_1}.{{\rm{x}}_2} + {{\rm{m}}_2}.{{\rm{x}}_1}}}{{{{\rm{m}}_1} + {{\rm{m}}_2}}},\frac{{{{\rm{m}}_1}{{\rm{y}}_2} + {{\rm{m}}_2}{{\rm{y}}_1}}}{{{{\rm{m}}_1} + {{\rm{m}}_2}}},\frac{{{{\rm{m}}_1}{{\rm{z}}_2} + {{\rm{m}}_2}{{\rm{z}}_1}}}{{{{\rm{m}}_1} + {{\rm{m}}_2}}}} \right)$.

Or, (a,b,c) = $\left( {\frac{{3{{\rm{m}}_1} + 2{{\rm{m}}_2}}}{{{{\rm{m}}_1} + {{\rm{m}}_2}}},\frac{{ - 4{{\rm{m}}_1} - 3{{\rm{m}}_2}}}{{{{\rm{n}}_1} + {{\rm{m}}_2}}},\frac{{ - 5{{\rm{m}}_1} + {{\rm{m}}_2}}}{{{{\rm{m}}_1} + {{\rm{m}}_2}}}} \right)$ …(2).

The point (a,b,c) lies on (1) so,

Or, 2 $\left( {\frac{{3{{\rm{m}}_1} + 2{{\rm{m}}_2}}}{{{{\rm{m}}_1} + {{\rm{m}}_2}}}} \right){\rm{\: }}$+ $\left( {\frac{{ - 4{{\rm{m}}_1} - 3{{\rm{m}}_2}}}{{{{\rm{m}}_1} + {{\rm{m}}_2}}}} \right)$ + $\left( {\frac{{ - 5{{\rm{m}}_1} + {{\rm{m}}_2}}}{{{{\rm{m}}_1} + {{\rm{m}}_2}}}} \right)$ = 7.

Or, 6m₁ + 4m₂ – 4m₁ – 3m₂ – 5m₁ + m₂ = 7m₁ + 7m₂.

Or, - 3m₁ + 2m₂ = 7m₁ + 7m₂

Or, - 10m₁ = 5m₂à m₁ : m₂ = - 1 : 2.

Thus, the required point is,

(a,b,c) = $\left( {\frac{{ - 3 + 4}}{{1 + 2}},\frac{{4 - 6}}{{ - 1 + 2}},\frac{{5 + 2}}{{ - 1 + 2}}} \right)$ = (1,-2,7).

 

12. Show that the following points represent the vertices of a parallelogram

a. (3, 0, 1), (2, 2, 2), (-1, 3, 3) and (0, 1, 2)

Solution

Given points are A(3,0,1), B(2,2,2), C(-1,3,3) and D(0,1,2).

Now, AB = $\sqrt {{{\left( {2 - 3} \right)}^2} + {{\left( {2 - 0} \right)}^2} + {{\left( {2 - 1} \right)}^2}} $ = $\sqrt {1 + 4 + 1} $ = $\sqrt 6 $.

BC = $\sqrt {{{\left( { - 1 - 2} \right)}^2} + {{\left( {3 - 2} \right)}^2} + {{\left( {3 - 2} \right)}^2}} $ = $\sqrt {9 + 1 + 1} $ = $\sqrt {11} $.

CD = $\sqrt {{{\left( {0 + 1} \right)}^2} + {{\left( {1 - 3} \right)}^2} + {{\left( {2 - 3} \right)}^2}} $ = $\sqrt {1 + 4 + 1} $ = $\sqrt 6 .$

And DA = $\sqrt {{{\left( {3 - 0} \right)}^2} + {{\left( {0 - 1} \right)}^2} + {{\left( {1 + 2} \right)}^2}} $ = $\sqrt {9 + 1 + 1} $ = $\sqrt {11} $.

So, AB = CD and BD = DA,

Hence, A,B,C,D are the vertices of a parallelogram.

 

 

b. (1, 3, 4), (-1, 6, 10), (-7, 4, 7) and (-5, 1, 1)

Solution

Given points are A(1,3,4), B(-1,6,10), C(-7,4,7) and D(-5,1,1).

Now, AB = $\sqrt {{{\left( { - 1 - 1} \right)}^2} + {{\left( {6 - 3} \right)}^2} + {{\left( {10 - 4} \right)}^2}} $ = $\sqrt {4 + 9 + 36} $ = 7.

BC = $\sqrt {{{\left( { - 7 + 1} \right)}^2} + {{\left( {4 - 6} \right)}^2} + {{\left( {7 - 10} \right)}^2}} $ = $\sqrt {36 + 4 + 9} $  = 7.

CD = $\sqrt {{{\left( { - 5 + 7} \right)}^2} + {{\left( {1 - 4} \right)}^2} + {{\left( {1 - 7} \right)}^2}} $ = $\sqrt {4 + 9 + 36} $ = 7.

And DA = $\sqrt {{{\left( {1 + 5} \right)}^2} + {{\left( {3 - 1} \right)}^2} + {{\left( {4 - 1} \right)}^2}} $ = $\sqrt {36 + 4 + 9} $ = 7.

So, AB = BC = CD = DA

Hence, A,B,C,D is a parallelogram.

 

13. Two vertices of a triangle ABC are A(2,-4, 3) and B(3, - 1, - 2) and its centroid is (1, 0,3) Find the third vertex C.

Solution

Given vertices are A(x₁,y₁,z₁) = (2,-4,3), B(x₂,y₂,z₂) = (3,-1,-2)

And centroid (x,y,z) = (1,0,3). Let the third vertex be C(α,β,γ) = (x₃,y₃,z₃).

We have, x = $\frac{{{{\rm{x}}_1} + {{\rm{x}}_2} + {{\rm{x}}_3}}}{3}$, y = $\frac{{{{\rm{y}}_1} + {{\rm{y}}_2} + {{\rm{y}}_3}}}{3}$, z = $\frac{{{{\rm{z}}_1} + {{\rm{z}}_2} + {{\rm{z}}_3}}}{3}$

Or, 1 = $\frac{{2 + 3 + \alpha }}{3}$, 0 = $\frac{{ - 4 - 1 + \beta }}{3}$, 3 = $\frac{{3 - 2 + \gamma }}{3}$

So, α= - 2, β= 5, γ = 8.

So, (α,β,γ) = (-2,5,8).

 

14. Three vertices of a parallelogram ABCD are A(- 5, 5, 2) B(- 9, - 1, 2) and C( - 3 , -3, 0) Find the coordinates of the fourth vertex.

Solution

Given vertices are A(-5,5,2), B(-9,-1,2) and C(-3,-3,0).

Let (α,β,γ) be the fourth vertex of the parallelogram ABCD.

Now, mid – point of AC = $\left( {\frac{{ - 5 - 3}}{2},\frac{{5 - 3}}{2},\frac{{2 + 0}}{2}} \right)$ = (-4,1,1).

Since, the diagonal of a parallelogram bisects each other so,

Mid – point of BD = (-4,4,1).

Now, using mid – point formula for BD, we have,

Or, - 4 = $\frac{{ - 9 + \alpha }}{2}$, 1 = $\frac{{1 + \beta }}{2}$, 1 = $\frac{{2 + \gamma }}{2}$.

So, α= 1, β = 3, γ= 0.

So, fourth vertex = D(1,3,0).