# Class 11 Maths NCERT solutions of Chapter 12 three dimensional coordinate geometry exercise 12.1

In class VI, VII and VIII you had learned how to show the integers in the number line, which was the beginning of the **co-ordinate geometry** in one dimension. In **class IX and X**, you would have got the idea of two-**dimensional coordinate geometry**. **Two-dimensional geometry** means showing the magnitudes of two physical quantities in the **cartesian planes** made by two number lines intersecting each other at 90โฐ. The vertical line is called **y-axis** and the horizontal line is known as **x-axis**, on the same way in **class XI** we are assigned the elements based on **three-dimensional coordinate geometry** in which **three axis x,y, and z** exist perpendicular to each other in which three physical quantities related to one another are shown.Here you can studyย in free of cost **NCERT solutions,** **CBSE important maths question**s, guess papers, **sample papers** of **maths and science**, articles on **science and maths**, blog post for your carrier etc.

**Class 11 Maths NCERT solutions of Chapter 12 three dimensional coordinate geometry exercise 12.1 **

**Exercise 12.1**

**1- Find the coordinate of the points which divides the line segment joining the points (โ2,3,5) and (1,โ4, 6) in the ratio**

**(i) 2 : 3 internally (ii) 2 : 3 externally**

Solution.

(i) The coordinates of the points C(x,y,z) that divides the line segment

A(โ2,3,5) and B(1,โ4,6) internally in the ratio m: n.

Thus the required coordinates of the point C are

(ii)The coordinates of the points C(x,y,z) that divides the line segment

A(โ2,3,5) and B(1,โ4,6) externally in the ratio m: n.

x = โ 8, y = 17, z = 3

Hence the required coordinates of point C are (โ8,17,3) that divides AB externally in the ratio of 2: 3.

**Q2.Given that P(3,2,โ4) , Q(5,4,โ6) and R(9,8,โ10) are collinear. Find the ratio in which Q divides PR.**

Answer.Let the point Q(5,4,โ6) divide the line segment joining points P(3,2,โ4)ย and R (9,8,โ10) in the ratio of k : 1.

The section formula when the point (x,y,z) divides the line segment joining two points is following

5k + 5 = 9k + 3

4k = 2

The line segment PR is divided by the point Q in the ratio of 1 : 2.

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**Q3. Find the ratio in which the YZ plane divides the line segment formed by joining the points (โ2,4,7) and (3,โ5,8).**

Answer. Let the YZ plane divides the line segment formed by joining the points (โ2,4,7) and (3,โ5,8) in the ratio of k : 1

Since the x coordinates of the point of intersection is 0, so let the coordinates of the point of intersection is (0,y,z)

So YZ plane divides the given line segment in the ratio of 2 : 3

**Q4. Using section formula ,show that the points A(2,โ3,4), B(โ1,2,1) and**

** are collinear.**

Answer.

The given points are A(2,โ3,4), B(โ1,2,1) andย

Let B(โ1,2,1) divides the line segment AC in the ratio of k :1

From each equation we get k = โ3

So, k= โ3 mean B(โ1,2,1) divides AC in the ratio of 3 : 1 and minus sign of k shows that B divides AC externally.

**Q5.Find the coordinates of the points which trisect the line segment joining the points P(4,2,โ6) and Q(10,โ16,6).**

Answer.Let R(a,b,c) and S(d,e,f) are the points which trisect PQ.

As shown in the diagram R(a,b,c) divides the line segment PQ in the ratio ofย 1 : 2, therefore applying the section formula

Substituting the value ofย k= 1/2 and coordinates of points P(4,2,โ6) and Q(10,โ16,6)

(a,b,c) = (6, โ4, โ2)

Hence coordinates of R are (6,โ4, โ2)

S is the midpoint of the points R(6,โ4, โ2) and Q(10,โ16,6), therefore the coordinates of S will be as follows

(d,e,f) = (8, โ10, 2)

Therefore R(6,โ4,โ2) and S(8,โ10, 2)ย trisects the line segment PQ

**Miscellaneous Exercise**

**Q1. Three vertices of a parallelogram ABCD are A(3, โ1,2), B(1,2,โ4) and C(โ1,1,2).Find the coordinates of the fourth vertex.**

Answer. Let the ย coordinates of the fourth vertex (D) are (x,y,z)

As we know the diagonals of a parallelogram bisect each other

โด The midpoint of AC = Midpoint of BD

The coordinates of midpoint O from the coordinates of A and C

………..(i)

The coordinates of midpoint O from the coordinates of B and D

……………..(ii)

From (i) and (ii)

x = 1, y = โ2, z = 8

Hence the coordinates of the fourth vertex D are (1,โ2, 8)

**Q2. Find the lengths of the medians of the triangle with vertices A(0,0,6), B(0,4,0) and C(6,0,0).**

Answer. Let the ฮABC has the medians DC, AE and BF

D, E and F will be the midpoints of AB, BC and AC respectively

โด The coordinates of D, E and F will be as follows

Therefore the coordinates of D , E and F are (0,2,3), (3,2,0) and (3,0,3) respectively and we are given the coordinates A(0,0,6), B(0,4,0) and C(6,0,0) ,so lengths ofย medians DC, AE and BF are calculated as follows.

Applying the distance formula

Hence the length of the medians of the given triangle are 7unit,7 unit and โ34 unit.

**Q3. If the origin is the centroid of the triangle PQR with vertices P(2a,2,6),Q(โ4,3b,โ10) and R(8,14,2C), then find the value of a, b and c.**

Answer. The coordinates of the centroid of the triangle whose vertes are given by the following formula

We are given that the origin(0,0,0) is the centroid

Thus,the value of a,b and c are respectively.

**Q4. Find the coordinates of a point on the y-axis which are at a distance of 5โ2ย from point P(3,โ2,5).**

Answer. If a point is on the y-axis then its x and z coordinates will be zero, so let the coordinates of the point are (0,a,0).

We are given distance 5โ2 from the point P(3,โ2,5) from the point(0, a,0) on the y-axis, so applying the distance formula.

Squaring both sides

50 = aยฒ + 4a + 38

aยฒ + 4a โ 12 = 0

aยฒ + 6aโ 2aโ 12 =0

a (a + 6)โ 2(a + 6) = 0

(a + 6)(a โ 2) = 0

a = โ6, 2

Therefore the coordinates of the point on the y-axis are (0,โ6,0) or (0,2,0)

**Q5. A point R with x coordinate 4 lies on the line segment joining the points P(2,โ3,4) and Q(8,0,10). Find the coordinates of R.**

Answer. Let R divides the line segment PQ in the ratio of k : 1

The coordinates of the point (x,y,z) dividing the line segment joining the pointsย are given by

We are given here x coordinate only i.e 4 and the points P(2,โ3,4) and Q(8,0,10)

**Q6. If A and B be the two points (3,4,5) and (โ1,3,7) respectively, find the equation of set of points P such that PAยฒ + PBยฒ = kยฒ, where k is a constant.**

Answer. Let (x,y,z) are the coordinates of the point P satisfying the given condition PAยฒ + PBยฒ = kยฒ.

Applying the distance formula

PAยฒ = (xโ3)ยฒ +ย (yโ4)ยฒ +(zโ5)ยฒ = xยฒ + yยฒ + zยฒ โ6x โ8y โ10z + 50

PAยฒ =xยฒ + yยฒ + zยฒ โ6x โ8y โ10z + 50………(i)

PBยฒ = (x+1)ยฒ +ย (yโ3)ยฒ +(z+7)ยฒ=xยฒ + yยฒ + zยฒ +2x โ6y+14z +59

PBยฒ = xยฒ + yยฒ + zยฒ +2x โ6y+14z +59……….(ii)

PAยฒ + PBยฒ =kยฒ

xยฒ + yยฒ + zยฒ โ6x โ8y โ10z + 50 + xยฒ + yยฒ + zยฒ +2x โ6y+14z +59 =kยฒ

2xยฒ + 2yยฒ + 2zยฒ โ4x โ14y +4z+109 =kยฒ

2xยฒ + 2yยฒ + 2zยฒ โ4x โ14y +4z = kยฒ โ109

Dividing the equation by 2

So,this is the required equation

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