# NCERT Solutions For Class 11 Maths Chapter 12

## NCERT Solutions Class 11 Maths Introduction to Three Dimensional Geometry

NCERT Solution for class 11 maths Chapter 12 Introduction to Three Dimensional Geometry is one of the most important topic to practice. The NCERT solutions for class 11 maths gives students a basic idea of the types of questions that are usually asked in the examination. NCERT questions are important for the students as practicing these questions are helpful for them to excel in the examination as well in in their competitive examination. Students can also download the NCERT Solution Class 11 Maths Introduction to Three Dimensional Geometry pdf by following the link given. Get more NCERT solutions by visiting our site.

### NCERT Solutions Class 11 Maths Chapter 12 Exercises

In this chapter, student will study about the Coordinate axis and coordinate planes in three dimensional, Coordinate of a point in Space, Distance between two points in a space, Section formula in three dimensional.

Here are some of the important points that needs to be remembered in three dimensional Geometry-

• In three dimensions, the coordinate axes of a rectangular Cartesian coordinate system are three mutually perpendicular lines. The axes are called the x, y and z-axes.
• The three planes determined by the pair of axes are the coordinate planes, called XY, YZ and ZX-planes.
• The three coordinate planes divide the space into eight parts known as octants.
• The coordinates of a point P in three dimensional geometry is always written in the form of triplet like (x, y, z). Here x, y and z are the distances from the YZ, ZX and XY-planes

Three dimensional Geometry holds the various important formulas that needs to be remembered. This chapter has some of the most important formulas, which helps students to have a good understanding for the higher classes. It is essential for all the students to practice questions from this chapter, as it is one of the most important chapter in class 11. Along with its importance, it has various real-life application. Practicing all the questions from the NCERT textbook help students to have a good idea of the various important questions that can be framed in the examination. So we have provided all the solution to the NCERT Chapter Introduction to Three Dimensional Geometry so as to have a good learning experience and clear all doubt.

Below given are the solution for NCERT Class 11 Introduction to Three Dimensional Geometry:

Exercise 12.1

For any given point, the sign of its coordinates determines the octant in which it will lie.

Now, from the following table it can be easily determined in which coordinates the point lies.

Q.1: A point is lying on y – axis. What are its ‘x‘ coordinates and ‘z’ coordinates?

Sol.

‘x’ coordinates and ‘z’ coordinates are zero if any point lies on y –axis.

Q.2: A point is lying on YZ – plane. What are its ‘x‘ coordinates?

Sol.

If any point lies in the YZ plane then its x-coordinates are zero.

Q.3: In which of the octant the following points lie:

(2, 3, 4), (8, -1, -1), (-4, 9, -8), (-1, -2, -3), (4, -5, 6), (7, -1, -4), (-3, -5, 1), (0, 0, -3)

Sol.

$$\Rightarrow$$ The x – coordinates, y – coordinates and z – coordinates of point (2, 3, 4) are all positive. Therefore, this point is lying in octant (I).

$$\Rightarrow$$ The x – coordinates, y – coordinates and z – coordinates of point (8, -1, -1) are positive, negative and negative. Therefore, this point is lying in octant (VII).

$$\Rightarrow$$ The x – coordinates, y – coordinates and z – coordinates of point (-4, 9, -8) are negative, positive and negative. Therefore, this point is lying in octant (VI).

$$\Rightarrow$$ The x – coordinates, y – coordinates and z – coordinates of point (-1, -2, -3) are all negative. Therefore, this point is lying in octant (VII).

$$\Rightarrow$$ The x – coordinates, y – coordinates and z – coordinates of point (4, -5, 6) are positive, negative and positive. Therefore, this point is lying in octant (IV).

$$\Rightarrow$$ The x – coordinates, y – coordinates and z – coordinates of point (7, -1, -4) are positive, negative and negative. Therefore, this point is lying in octant (VII).

$$\Rightarrow$$ The x – coordinates, y – coordinates and z – coordinates of point (-3, -5, 1) are negative, negative and positive. Therefore, this point is lying in octant (III).

$$\Rightarrow$$ The x – coordinates, y – coordinates and z – coordinates of point (-1, 2, 2) are negative, positive and positive. Therefore, this point is lying in octant (II).

(i). What is the name of a plane determined by the Z-axis and the Y-axis when taken together?

(ii). What is the general form of coordinates of points in the XZ-plane?

(iii). Coordinate plane divides the space into how many octants?

Sol.

(i). The plane determined by the Z-axis and the Y-axis when taken together is known as YZ–plane.

(ii). The general form of coordinates of points in the XZ-plane is ( x, 0 , z ).

(iii). The coordinate plane divides the space into 8 octants.

Exercise: 12.2

Distance Formula:

The distance between point A(x1, y1, z1) and point B(x2, y2, z2) is given by:

AB = $$\sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}+\left ( z_{2}- z_{1}\right )^{2}}$$

Q.1: Find the distance between two points whose coordinates are given below:

(i).     (2, 8, 9) and (4, 5, 8)

(ii).    (-3, 4, 5) and (2, 6, -1)

(iii).   (-6, -4, 1) and (5, -2, 6)

(iv).   (-1, 9, 8) and (6, 5, -3)

Sol.

The distance between point A (x1, y1, z1) and point B (x2, y2, z2) is given by:

AB = $$\sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}+\left ( z_{2}- z_{1}\right )^{2}}\\$$

(i). The distance between point A (2, 8, 9) and point B (4, 5, 8) is:

$$AB = \sqrt{\left ( 4-2 \right )^{2}+\left ( 5-8 \right )^{2}+\left ( 8-9\right )^{2}}\\$$

$$\\\Rightarrow AB = \sqrt{4+9+1}$$

$$\\\Rightarrow AB = \sqrt{14}$$ units

Therefore, The distance between point A (2, 8, 9) and point B (4, 5, 8) is $$\sqrt{14}$$ units

(ii).  The distance between point A (-3, 4, 5) and point B (2, 6, -1) is:

$$AB = \sqrt{\left [ 2-(-3) \right ]^{2}+\left ( 6-4 \right )^{2}+\left ( -1-5\right )^{2}}\\$$

$$\\\Rightarrow$$   $$AB = \sqrt{25+4+36}$$

$$\\\Rightarrow$$  $$AB = \sqrt{65}$$ units

Therefore, The distance between point A (-3, 4, 5) and point B (2, 6, -1) is $$\sqrt{65}$$ units

(iii).  The distance between point A (-6, -4, 1) and point B (5, -2, 6) is:

$$AB = \sqrt{\left [ 5-(-6) \right ]^{2}+ [-2-(-4)]^{2}+\left ( 6-1\right )^{2}}\\$$

$$\\\Rightarrow$$ $$AB = \sqrt{121+4+25}$$

$$\\\Rightarrow$$   $$AB = \sqrt{150}$$ units

Therefore, The distance between point A (-6, -4, 1) and point B (5, -2, 6) is $$5\sqrt{6}$$ units

(iv).  The distance between point A (-1, 9, 8) and point B (6, 5, -3) is:

$$AB = \sqrt{\left [ 6-(-1) \right ]^{2}+ (5-9)^{2}+\left ( -3-8\right )^{2}}\\$$

$$\\\Rightarrow$$   $$AB = \sqrt{49+16+121}$$

$$\\\Rightarrow$$   $$AB = \sqrt{186}$$ units

Therefore, The distance between point A (-1, 9, 8) and point B (6, 5, -3) is $$\sqrt{186}$$ units

Q.2: Show that the points (7, 0, -1), (-2, 3, 5) and (1, 2, 3) are collinear.

Sol.

Let, the coordinates of points A, B and C are (7, 0, -1), (-2, 3, 5) and (1, 2, 3) respectively.

Points A, B and C are collinear if they lie on the same line.

Now, $$AB=\sqrt{(-2-7)^{2}+(3-0)^{2}+[5-(-1)]^{2}}$$

$$\Rightarrow$$   $$AB=\sqrt{81+9+36}$$

$$\Rightarrow AB=\sqrt{126}$$ units

Therefore, AB = $$3\sqrt{14}$$ units

Now, $$BC=\sqrt{[1-(-2)]^{2}+(2-3)^{2}+(3-5)^{2}}$$

$$\Rightarrow$$   $$BC=\sqrt{9+1+4}$$

$$\Rightarrow$$   $$BC=\sqrt{14}$$ units

Therefore, BC = $$\sqrt{14}$$ units

And, $$AC=\sqrt{(1-7)^{2}+(2-0)^{2}+[3-(-1)]^{2}}$$

$$\Rightarrow$$   $$AC=\sqrt{36+4+16}$$

$$\Rightarrow AC=\sqrt{56}$$ units

Therefore, AC = $$2\sqrt{14}$$ units

Since, AC + BC = $$2\sqrt{14}+\sqrt{14}$$ = AB

Therefore, the points (7, 0, -1), (-2, 3, 5) and (1, 2, 3) are collinear.

Q.3: Prove the following statements:

(i). (-4, 9, 6), (0, 7, 10) and (-1, 6, 6) are the vertices of a right angled triangle.

(ii). (4, 9, -6), (0, 7, -10) and (1, 6, -6) are the vertices of an isosceles triangle.

(iii). (2, -3, 4), (1, -2, 5), (-1, 2, 1) and (4, -7, 8) are the vertices of a parallelogram.

Sol.

The distance between points A (x1, y1, z1) and B (x2, y2, z2) is given by:

AB = $$\sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}+\left ( z_{2}- z_{1}\right )^{2}}$$

(i). Let, the coordinates of points A, B and C are (-4, 9, 6), (0, 7, 10) and (-1, 6, 6) respectively.

Now,  $$AB=\sqrt{(0+4)^{2}+(7-9)^{2}+(10-6)^{2}}\\$$

$$\\\Rightarrow$$   $$AB=\sqrt{16+4+16} =\sqrt{36}$$ units.

Therefore, AB = 6 units.

Now,  $$BC=\sqrt{(-1-0)^{2}+(6-7)^{2}+(6-10)^{2}}\\$$

$$\\\Rightarrow$$   $$BC=\sqrt{1+1+16} \;=\sqrt{18}$$ units.

Therefore, BC = $$3\sqrt{2}\\$$ units.

And,  $$AC=\sqrt{(-1+4)^{2}+(6-9)^{2}+(6-6)^{2}}\\$$

$$\\\Rightarrow$$   $$AC=\sqrt{9+9}=\sqrt{18}$$ units.

Therefore, AC = $$3\sqrt{2}$$units.

Now,  AB2 = 36, BC2 = 18 and AC2 = 18

Since, AB2 = AC2 + BC2. Therefore, by Pythagoras theorem, ABC is a right angled triangle.

Hence, the given points (-4, 9, 6), (0, 7, 10) and (-1, 6, 6) are the vertices of a right angled triangle.

(ii). Let, the coordinates of points A, B and C are (4, 9, -6), (0, 7, -10) and (1, 6, -6) respectively.

Now,  $$AB=\sqrt{(0-4)^{2}+(7-9)^{2}+(-10+6)^{2}}\\$$

$$\\\Rightarrow$$   $$AB=\sqrt{36}$$ units.

Therefore, AB = 6 units.

Now,  $$BC=\sqrt{(1-0)^{2}+(6-7)^{2}+(-6+10)^{2}}\\$$

$$\\\Rightarrow$$   $$BC=\sqrt{1+1+16} =\sqrt{18}$$ units.

Therefore, BC = $$3\sqrt{2}$$ units.

And,  $$AC=\sqrt{(1-4)^{2}+(6-9)^{2}+(-6+6)^{2}}\\$$

$$\\\Rightarrow AC=\sqrt{9+9}=\sqrt{18}$$ units.

Therefore, AC = $$3\sqrt{2}$$ units.

i.e. AC = BC AB

Now, a triangle is said to be an isosceles triangle if two sides of a triangle are equal to each other.

Since, AC = BC. Therefore triangle ABC is an isosceles triangle.

Hence, the given points (4, 9, -6), (0, 7, -10) and (1, 6, -6) are the vertices of an isosceles triangle.

(iii). Let, the coordinates of points A, B, C and D are (1, -2, 5), (4, -7, 8), (2, -3, 4) and (-1, 2, 1) respectively.

Now,  $$AB=\sqrt{(4-1)^{2}+(-7+2)^{2}+(8-5)^{2}}\\$$

$$\\\Rightarrow$$   $$AB=\sqrt{9+25+9} =\sqrt{43}$$ units.

Therefore, AB =$$\sqrt{43}$$ units.

Now,  $$BC=\sqrt{(2-4)^{2}+(-3+7)^{2}+(4-8)^{2}}\\$$

$$\\\Rightarrow$$   $$BC=\sqrt{4+16+16} =\sqrt{36}$$ units.

Therefore, BC = 6 units.

Now,  $$CD =\sqrt{(-1-2)^{2}+(2+3)^{2}+(1-4)^{2}}$$

$$\\\Rightarrow$$   $$AC=\sqrt{9+25+9} =\sqrt{433}$$ units.

Therefore, CD = $$\sqrt{43}$$ units.

And,  $$AD =\sqrt{(-1-1)^{2}+(2+2)^{2}+(1-5)^{2}}\\$$

$$\\\Rightarrow$$   $$AD=\sqrt{4+16+16} =\sqrt{36}$$ units.

Since, BC = AD = 6 units and AB = CD = $$\sqrt{43}$$ units

Since, it is proved that in quadrilateral ABCD opposite sides are equal. Therefore, ABCD is a parallelogram.

Hence, the given points (2, -3, 4), (1, -2, 5), (-1, 2, 1) and (4, -7, 8) are the vertices of a parallelogram.

Q.4: Find the equation of the set of points P which are equidistant from point  A (3, 2, 1) and point B (-1, 2, 3).

Sol.

Let, (x, y, z) be the coordinates of point P, which is equidistant from point A (3, 2, 1) and point B (-1, 2, 3).

Now, according to the given condition:

AP2 = BP2

$$\Rightarrow$$   $$\left [ \sqrt{(x-3)^{2}+(y-2)^{2}+(z-1)^{2}} \;\right ]^{2}=\left [ \sqrt{(x+1)^{2}+(y-2)^{2}+(z-3)^{2}} \;\right ]^{2}$$

$$\\\Rightarrow$$   $$(x-3)^{2}+(y-2)^{2}+(z-1)^{2}=(x+1)^{2}+(y-2)^{2}+(z-3)^{2}$$

$$\\\Rightarrow$$   $$x^{2}+9-6x+y^{2}+4-4y+z^{2}+1-2z=x^{2}+1+2x+y^{2}+4-4y+z^{2}+9-6z$$

$$\\\Rightarrow$$ – 6x – 4y – 2z = +2x – 4y – 6z

$$\\\Rightarrow$$ – 6x – 4y – 2z = + 2x – 4y – 6z

$$\\\Rightarrow$$   8x – 4z = 0

Therefore, the required equation is: 2x – z = 0

Q.5: The sum of distance of point P from point A (3, 0, 0) and point B (-3, 0, 0) is equal to 12 units. Find the equation of the set of points.

Sol.

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

Now, according to the given condition:

AP + BP = 12

$$\Rightarrow$$   $$\sqrt{(x-3)^{2}+y^{2}+z^{2}}+\sqrt{[x-(-3)]^{2}+y^{2}+z^{2}}=12\\$$

$$\Rightarrow$$   $$\sqrt{(x-3)^{2}+y^{2}+z^{2}}=12-\sqrt{(x+3)^{2}+y^{2}+z^{2}}$$

Now, on squaring both the side, we will get:

$$\Rightarrow$$   $$\left ( \sqrt{(x-3)^{2}+y^{2}+z^{2}} \right )^{2}= \left ( 12-\sqrt{(x+3)^{2}+y^{2}+z^{2}} \right )^{2}\\$$

$$\Rightarrow$$   $$(x-3)^{2}+y^{2}+z^{2}= 144+[(x+3)^{2}+y^{2}+z^{2}]-24\sqrt{(x+3)^{2}+y^{2}+z^{2}}\\$$

$$\Rightarrow$$   $$x^{2}+9-6x+y^{2}+z^{2}=144+x^{2}+9+6x+y^{2}+z^{2}-24\sqrt{(x+3)^{2}+y^{2}+z^{2}}\\$$

$$\Rightarrow$$   $$24\sqrt{(x+3)^{2}+y^{2}+z^{2}}=144+12x\\$$

$$\Rightarrow$$   $$2\sqrt{(x+3)^{2}+y^{2}+z^{2}}=(x+12)$$

Again, on squaring both the side, we will get:

$$\Rightarrow$$   $$4[(x+3)^{2}+y^{2}+z^{2}]=(x+12)^{2}\\$$

$$\Rightarrow$$   $$4x^{2}+4y^{2}+4z^{2}+36+24x=x^{2}+24x+144\\$$

$$\Rightarrow$$   $$3x^{2}+4y^{2}+4z^{2}-108=0$$

Therefore, the required equation is: $$3x^{2}+4y^{2}+4z^{2}-108=0$$

Section Formula:

The coordinates of point P, which divides the line segment joining two points A (a1, b1, c1) and B (a2, b2, c2) internally in the ratio m and n are:

$$\left [ \frac{m.a_{2}\;+\;n.a_{1}}{m\;+\;n},\frac{m.b_{2}\;+\;n.b_{1}}{m\;+\;n},\frac{m.c_{2}\;+\;n.c_{1}}{m\;+\;n} \right ]$$

The coordinates of point Q, which divides the line segment joining two points A (a1, b1, c1) and B (a2, b2, c2) Externally in the ratio m and n are:

$$\left [ \frac{m.a_{2}\;-\;n.a_{1}}{m\;-\;n},\frac{m.b_{2}\;-\;n.b_{1}}{m\;-\;n},\frac{m.c_{2}\;-\;n.c_{1}}{m\;-\;n} \right]$$

Exercise 12.3

Q.1: Find the coordinates of the point which divides the line segment joining the points (-2, 1, 0) and (1, 3, 6) internally in the ratio of 1:4 and externally in the ratio of 1:4.

Sol:

Since, the coordinates of point P, which divides the line segment joining two points A (a1, b1, c1) and B (a2, b2, c2) internally in the ratio m and n are:

$$\left [ \frac{m.a_{2}\;+\;n.a_{1}}{m\;+\;n},\frac{m.b_{2}\;+\;n.b_{1}}{m\;+\;n},\frac{m.c_{2}\;+\;n.c_{1}}{m\;+\;n} \right ]$$

Therefore, the coordinates of point P, which divides the line segment joining two points A(-2, 1, 0) and B (1, 3, 6) internally in the ratio 1 and 4 are:

(x, y, z) = $$\left [ \frac{1\times 1\;+\;4\times -2}{1\;+\;4},\frac{1\times 3\;+\;4\times 1}{1\;+\;4},\frac{1\times 6\;+\;4\times 0}{1\;+\;4} \right ]\\$$

(x, y, z) = $$\left [ \frac{-7}{5},\frac{7}{5},\frac{6}{5} \right ]$$

Therefore, the coordinates of point P are: $$\left [ \frac{-7}{5},\frac{7}{5},\frac{6}{5} \right ]\\$$

Now, The coordinates of point Q, which divides the line segment joining two points A (a1, b1, c1) and B (a2, b2, c2) Externally in the ratio m and n are:

$$\left [ \frac{m.a_{2}\;-\;n.a_{1}}{m\;-\;n},\frac{m.b_{2}\;-\;n.b_{1}}{m\;-\;n},\frac{m.c_{2}\;-\;n.c_{1}}{m\;-\;n} \right]$$

Therefore, the coordinates of point Q, which divides the line segment joining two points A(-2, 1, 0) and B (1, 3, 6) externally in the ratio 1 and 4 are:

(x, y, z) = $$\left [ \frac{1\times 1\;-\;4\times -2}{1\;-\;4},\frac{1\times 3\;-\;4\times 1}{1\;-\;4},\frac{1\times 6\;-\;4\times 0}{1\;-\;4} \right ]\\$$

(x, y, z) = $$\left [ \frac{9}{-3},\frac{-1}{-3},\frac{6}{-3} \right ]$$

Therefore, the coordinates of point Q are: $$\left [ -3,\frac{1}{3},-2 \right ]$$

Q.2: Given that the points A (3, 2, -4) B (5, 4, -6) and C (9, 8, -10), are collinear. Find the ratio in which line AC is divided by B.

Sol.

Since, point A, point B and point C are collinear. Therefore, let us assume that point B divides the line segment joining point A and point C in the ratio k : 1.

Hence, by Section Formula, coordinates of point B are:

$$\Rightarrow \left [ \frac{(k\times 9)\;+\;(1\times 3)}{k\;+\;1},\frac{(k\times 8)\;+\;(1\times 2)}{k+1},\frac{(k\times -10)\;+\;(1\times -4)}{k\;+\;1} \right ]\\$$

$$\Rightarrow \left [ \frac{9k\;+\;3}{k\;+\;1},\frac{8k\;+\;3}{k\;+\;1},\frac{-10k\;-\;4}{k\;+\;1} \right ]$$. . . . . . . . . . . . . . (1)

And, the coordinates of point B are: (5, 4, -6) [Given]

Therefore, on comparing it with equation (1) we will get:

$$\Rightarrow \frac{9k\;+\;3}{k\;+\;1}=5\\$$

$$\Rightarrow$$ 9k + 3 = 5k + 5

Therefore, k = $$\frac{1}{2}$$

Hence, point B divides AB in the ratio of 1 : 2

Q.3: Find the ratio in which XZ – plane divides the line segment AB formed by joining the points (3, -5, 8) and (-2, 4, 7).

Sol.

Let the XZ plane divides the line segment PQ in the ratio k : 1

Hence, by Section Formula the coordinates of point of intersection are:

$$\Rightarrow \left [ \frac{(k\times -2)\;+\;(1\times 3)}{k\;+\;1},\frac{(k\times 4)\;+\;(1\times -5)}{k\;+\;1},\frac{(k\times 7)\;+\;(1\times 8)}{k\;+\;1} \right ]\\$$

$$\\\Rightarrow$$   $$\left [ \frac{-2k\;+\;3}{k\;+\;1},\frac{4k\;-\;5}{k\;+\;1},\frac{7k\;+\;8}{k\;+\;1} \right ]$$

Now, in XZ – plane, the y – coordinate of any point is zero.

$$\Rightarrow \frac{4k\;-\;5}{k\;+\;1}=0\\$$

Therefore, $$k = \frac{5}{4}$$

Hence, the XZ plane divides the line segment formed by the joining of given points in the ratio of 2 : 3.

Q.4: By using the section formula; Show that the points P (-1, 2, 1), Q (0, $$\frac{1}{3}$$, 2) and R (2, -3, 4) are collinear.

Sol.

Let, point A divides line PQ in the ratio k : 1

Hence, by Section Formula the coordinates of point A are:

$$\Rightarrow \left [ \frac{(k\times 0)\;+\;(1\times -1)}{k\;+\;1},\frac{(k\times \frac{1}{3})\;+\;(1\times 2)}{k\;+\;1},\frac{(k\times 2)\;+\;(1\times 1)}{k\;+\;1} \right ]\\$$

$$\Rightarrow \left [ \frac{-1}{k\;+\;1},\frac{\frac{k}{3}\;+\;2}{k\;+\;1},\frac{2k\;+\;1}{k\;+\;1} \right ]\\$$

Therefore, the coordinates of point A are: $$\left [ \frac{-1}{k\;+\;1},\frac{k\;+\;6}{3k\;+\;3},\frac{2k\;+\;1}{k\;+\;1} \right ]$$

Now, the value of ‘k’ for which point A coincides with point R:

$$\Rightarrow \frac{-1}{k+1}= 2\\$$

$$\Rightarrow -1= 2k+2\\$$

Therefore, k = $$\frac{-3}{2}\\$$

Now, for k = $$\frac{-3}{2}$$, the coordinates of point A are:

$$\\\Rightarrow$$    $$\left [ \frac{-1}{\left ( \frac{-3}{2} \right )\;+\;1}\;,\;\frac{\left ( \frac{-3}{2} \right )\;+\;6}{3\times \left ( \frac{-3}{2} \right )\;+\;3}\;,\;\frac{2\times \left ( \frac{-3}{2} \right )\;+\;1}{\left ( \frac{-3}{2} \right )\;+\;1} \right ]\\$$

$$\\\Rightarrow$$   $$\left [ \frac{-1\times 2}{-3\;+\;2}\;,\;\frac{-3\;+\;12}{-9\;+\;6}\;,\;\frac{-6\;+\;2}{-3\;+\;2} \right ]\\$$

Therefore, the coordinates of point A are:  (2, -3, 4)

Hence, the coordinates of point A coincides with the coordinates of point R.

$$\Rightarrow$$R (2, -3, 4) is a point that divides PQ externally and is same as point A.

Hence, the points P, Q and R are colinear.

Q.5: The line segment joining points A (5, 3, -6) and B (9, 15, 7) is trisected by the points P and Q, Find the coordinates of points P and Q.

Sol.

Point P divides the line segment AB in the ratio 1 : 2 and point Q divides the line segment AB in the ratio 2 : 1.

Therefore, by section formula the coordinates of point P and point Q are:

For Point P:

$$\Rightarrow \left [ \frac{(1\times 9)\;+\;(2\times 5)}{1\;+\;2},\frac{(1\times 15)\;+\;(2\times 3)}{1\;+\;2},\frac{(1\times 7)\;+\;(2\times -6)}{1\;+\;2} \right ]\\$$

$$\Rightarrow \left [ \frac{19}{3},\frac{21}{3},\frac{-5}{3} \right ]\\$$

Therefore, the coordinates of point P are:$$\left [ \frac{19}{3},7,\frac{-5}{3} \right ]$$

For Point Q:

$$\Rightarrow \left [ \frac{(2\times 9)\;+\;(1\times 5)}{2\;+\;1},\frac{(2\times 15)\;+\;(1\times 3)}{2\;+\;1},\frac{(2\times 7)\;+\;(1\times -6)}{2\;+\;1} \right ]\\$$

$$\Rightarrow \left [ \frac{23}{3},\frac{33}{3},\frac{8}{3} \right ]\\$$

Therefore, the coordinates of point Q are: $$\left [ \frac{23}{3},11,\frac{8}{3} \right ]$$