CBSE Class 11 Maths Important Questions - Chapter 11 Introduction to Three Dimensional Geometry

1 Marks Questions

1.Name the octant in which the following lie: $(5,2,3)$

Ans: Octant I

2. Name the octant in which the following lie: $(-5,4,3)$

Ans: Octant II

3. Find the image of $(-2,3,4)$ in the y $z$ plane

Ans: $(2,3,4)$

4. Find the image of $(5,2,-7)$ in the x y plane.

Ans: $(5,2,7)$

5. A point lie on $\mathbf{X}$-axis what are co- ordinate of the point

Ans: $(a,0,0)$

6. Write the name of plane in which $x$ axis and $y$- axis are taken together.

Ans: X Y Plane

7. The point $(4,-3,-6)$ lie in which octants

Ans: VIII

8. The point $(2,0,8)$ lie in which plane

Ans: XZ Plane

9. A point is in the $XZ$ plane. What is the value of y co-ordinates?

Ans: $Zero$

10. What is the coordinates of $XY$ plane

Ans: $(x,y,0)$

11. The point $(-4,2,5)$ lie in which octant.

Ans:  Octant II

12. The distance from origin to point $(a,b,c)$ is:

Ans: Distance from origin=$\sqrt{a^2+b^2+c^2}$

4 Marks Questions

1. Given that $P(3,2,-4),Q(5,4,-6)$ and $R(9,8,-10)$ are collinear. Find the ratio in which $Q$ divides PR

Ans: Suppose Q divides PR in the ratio $\lambda :1.$ Then coordinator of $\text{Q}$ are

$\left({\displaystyle \frac{9\lambda +3}{\lambda +1}},{\displaystyle \frac{8\lambda +2}{\lambda +1}},{\displaystyle \frac{-10\lambda -4}{\lambda +1}}\right)$

But, coordinates of $\text{Q}$ are $(5,4,-6).$Therefore

${\displaystyle \frac{9\lambda +3}{2+1}}=5,{\displaystyle \frac{8\lambda +2}{\lambda +1}}=4,{\displaystyle \frac{-10\lambda -4}{2+1}}=6$

These three equations give

$\hat{\alpha }={\displaystyle \frac{1}{2}}$

So Q divides PR in the ratio ${\displaystyle \frac{1}{2}}:1$ or 1:2

2. Determine the points in x y plane which is equidistant from these point A $(2,0,3)$T$\text{B}(0,3,2)$ and $C(0,0,1)$

Ans: Since the z coordinate in the xy plane is zero. So, let P(x, y, 0) be a point in xy- plane, such that PA=PB=PC. $Now,PA=PB$

PA2 = PB2

$\Rightarrow (x-2{)}^2+(y-0{)}^2+(0-3{)}^2=(x-0{)}^2+(y-3{)}^2+(0-2{)}^2$

$2x-3y=0\dots ..(i)$

$PB=PC$

$\Rightarrow P{B}^2=P{C}^2$

$\Rightarrow (x-0{)}^2+(y-3{)}^2+(0-2{)}^2=(x-0{)}^2+(y-0{)}^2+(0-1{)}^2$T

$\Rightarrow -6y+12=0\Rightarrow y=2\dots \dots ..(ii)$

Put $y=2$ in (i) we get $x=3$

Hence the points required are $(3,2,0)$.

3. Find the locus of the point which is equidistant from the point $\text{A}(0,2,3)$ and $\text{B}(2,-2,1)$

Ans: Let $$Q(x,y,z)$$be any point which is equidistant from $A(0,2,3)$ and $B(2,-2,1).$ Then

$$QA=QB$$

Squaring both sides, we get

$$QA$$2 = $$QB$$2

$\Rightarrow \sqrt{{(x-0)}^2+{(y-2)}^2+{(2-3)}^2}=\sqrt{{(x-2)}^2+{(y+2)}^2+{(z-1)}^2}$

$\Rightarrow 4x-8y-42+4=0$

$\Rightarrow x-2y-2+1=0$

$\Rightarrow x-2y-1=0$

4. Show that the points$$P(2,-1,3),Q(0,1,2),R(2,-1,3)$$  are vertices of an isosceles right angled triangle.

Ans: We have

$PQ=\sqrt{{(2-0)}^2+{(-1-1)}^2{(+3-2)}^2}=\sqrt{4+4+1}=3$

$QR=\sqrt{{(1-2)}^2+{(-3+1)}^2+{(1-3)}^2}=\sqrt{1+4+4}=3$

And $RP=\sqrt{{(1-0)}^2+{(-3-1)}^2+{(1-2)}^2}=\sqrt{1+16+1}=3\sqrt{2}$

Clearly $PQ=QR$ and $PQ$2 + $QR$2 = $RP$2

Hence triangle $PQR$ is an isosceles right angled triangle.

5. Using the section formula, prove that the three points $A(-2,3,5),B(1,2,3)$, and $C(7,0,-1)$ are collinear.

Ans: Assume that the given points are collinear and $C$ divides $AB$ in the ratio $\lambda =1.$

Then coordinates of $C$ are

$\left({\displaystyle \frac{\lambda -2}{2+1}},{\displaystyle \frac{2\lambda +3}{\lambda +1}},{\displaystyle \frac{3\hat{2}+5}{\lambda +1}}\right)$

But the coordinates of $C$ are $(3,0,-1)$ from the above equations we get $\lambda ={\displaystyle \frac{3}{2}}$

Since these equation give the same value of $V_{.}$

∴  the given points are collinear and$C$ divides $AB$ exactly in the ratio 3: 2.

6. Show that coordinator of the centroid of triangle with vertices $A\left(x_1{y}_1{z}_1\right),\text{B}\left(x_2{y}_2{z}_2\right)$,T$\mathrm{and}C\left(x_3{y}_3{z}_3\right)$ is $\left[{\displaystyle \frac{x_1+y_1+z_1}{3}},{\displaystyle \frac{y_1+y_2+y_3}{3}},{\displaystyle \frac{z_1+z_2+z_3}{3}}\right]$

Ans: Let $D$ be the mid-point of $BC$ ,then

Coordinates of $\text{D}$are $\left({\displaystyle \frac{x_2+x_2}{2}},{\displaystyle \frac{y_2+y_3}{2}},{\displaystyle \frac{z_2+z_3}{2}}\right)$. 

Let$\text{G}$ be the centroid of $△ABC$. The $\text{G}$, divides$\text{AD}$ in the ratio 2:  1. So coordinates of $\text{D}$are

$\left({\displaystyle \frac{1.x_1+2{\displaystyle \frac{\left(x_2+x_3\right)}{2}}}{1+2}}\cdot {\displaystyle \frac{{1.2}_1+2\left({\displaystyle \frac{y_2+y_3}{2}}\right)}{1+2}}={\displaystyle \frac{1-z_1+2\left({\displaystyle \frac{z_2+z_3}{2}}\right)}{1+2}}\right)$

$\left({\displaystyle \frac{x_1+x_2+x_3}{3}},{\displaystyle \frac{y_1+y_2+y_3}{3}},{\displaystyle \frac{z_1+z_2+z_3}{3}}\right)$

7. Prove by distance formula that the points $X(1,2,3),Y(-1,-1,-1),Z(3,5,7)$ are collinear.

Ans: The Distance

$|XY|=\sqrt{{(-1-1)}^2+{(-1-2)}^2+{(-1-3)}^2}=\sqrt{4+9+16}=\sqrt{29}$

Distance

$|YZ|=\sqrt{{(3+1)}^2+{(5+1)}^2+{(7+1)}^2}=\sqrt{16+36+64}=2\sqrt{29}$

Distance

$|XZ|=\sqrt{{(3-1)}^2+{(5-2)}^2+{(7-3)}^2}=\sqrt{4+9+16}=\sqrt{29}$

$\therefore |YZ|=|XY|+|XZ|$

The points $X,Y,Z$ are collinear.

8. find the co-ordinate of the point which divides the join of $A(2,-1,4),B(4,3,2)$ in the ratio $2:5(i)$ internally (ii) externally

Ans: Let $C(x,y,z)$ be the required point

i. For internal division

$x={\displaystyle \frac{2\times 4+5\times 2}{2+5}}={\displaystyle \frac{8+10}{7}}={\displaystyle \frac{18}{7}}$

$y={\displaystyle \frac{2\times 3+5\times -1}{2+5}}={\displaystyle \frac{6-5}{7}}={\displaystyle \frac{1}{7}}$

$z={\displaystyle \frac{2\times 2+5\times 4}{2+5}}={\displaystyle \frac{4+20}{7}}={\displaystyle \frac{24}{7}}$

$\therefore$Required point$C\left({\displaystyle \frac{18}{7}},{\displaystyle \frac{1}{7}},{\displaystyle \frac{24}{7}}\right)$

ii. For external division.

$x={\displaystyle \frac{2\times 4-5\times 2}{2-5}}={\displaystyle \frac{8-10}{-3}}={\displaystyle \frac{-2}{-3}}={\displaystyle \frac{2}{3}}$

$y={\displaystyle \frac{2\times 3-5\times -1}{2-5}}={\displaystyle \frac{6+5}{-3}}={\displaystyle \frac{11}{-3}}$

$z={\displaystyle \frac{2\times 2-5\times 4}{2-5}}={\displaystyle \frac{4-20}{-3}}={\displaystyle \frac{-16}{-3}}={\displaystyle \frac{16}{3}}$

$\therefore$Required point $C\left({\displaystyle \frac{2}{3}},{\displaystyle \frac{-11}{3}},{\displaystyle \frac{16}{3}}\right)$

9. Find the co-ordinate of a point equidistant from the four points

$P(0,0,0),Q(a,0,0),R(0,b,0)$ and $S(0,0,c)$.

Ans: let $A(x,y,z)$ be the required point

According to condition

$PA=AQ=AR=AS$

Now $PA=AQ$

$\Rightarrow P{A}^2=A{Q}^2$

$\Rightarrow x^2+y^2+z^2=(x-a{)}^2+(y-0{)}^2+(z-0{)}^2$

$\Rightarrow x^2+y^2+z^2=x^2-2ax+a^2+y^2+z^2$

$2ax=a^2$

$\therefore x={\displaystyle \frac{a}{2}}$

Similarly $PA=AR$

$\Rightarrow y={\displaystyle \frac{b}{2}}$

$Q\left(x,y,z_{,}\right),R\left(x_2,y_2,z_2\right)$ and $S\left(x_3,y_3,z_3\right)D,E$ and $F$ are mid points of side $RS,SQ$ and $QR$ respectively.

Then ${\displaystyle \frac{x_1+x_2}{2}}=-1$

$x_1+x_2=-2\dots ..(1)$

${\displaystyle \frac{y_1+y_2}{2}}=1$

$y_1+y_2=2\dots \dots$(2)

${\displaystyle \frac{z_1+z_2}{2}}=-4$

$z_1+z_2=-8\dots \dots (3)$

${\displaystyle \frac{x_2+x_3}{2}}=1$

$x_2+x_3=2\dots \dots (4)$

${\displaystyle \frac{y_2+y_3}{2}}=2$

$y2+y3=4\dots ..(5)$

${\displaystyle \frac{z_2+z_3}{2}}=-3$

$z_2+z_3=-6\dots \dots$(6)

${\displaystyle \frac{x_1+r_3}{2}}=3$

$x_1+x_3=6\dots \dots (7)$

${\displaystyle \frac{y_1+y_3}{2}}=0$

$y_1+y_3=0\dots \dots (8)$

${\displaystyle \frac{z_1+z_3}{2}}=1$

$z_1+z_3=2\dots \dots (9)$

Adding equation (1), (4) and (7) we get

$2\left(y_1+y_2+y_3\right)=6$

$y_1+y_2+y_3=3\dots \dots (11)$

And $PA=AS$

$\Rightarrow z={\displaystyle \frac{c}{2}}$

Hence co-ordinate of $A({\displaystyle \frac{a}{2}},{\displaystyle \frac{b}{2}},{\displaystyle \frac{c}{2}})$

10. Find the ratio in which the join the $B(2,1,5)$ and $C(3,4,3)$ is divided by the plane$2x+2y-2z=1$. Also find the co-ordinate of the point of division.

Ans: Assume the plane$2x+2y-2z=1$ divides $A(2,1,5)$ and $B(3,4,5)$ in the ratio $\lambda :1$ at point $A$

Then the co-ordinate of the point $A$

$\left({\displaystyle \frac{3\lambda +2}{\lambda +1}}\cdot {\displaystyle \frac{4\lambda +1}{\lambda +1}}{\displaystyle \frac{3\lambda +5}{\lambda +1}}\right)$

$\because$ Point $A$ lies on the plane $2x+2y-2z=1$

$\therefore$ Points $A$ must satisfy the equation of plane

$2\left({\displaystyle \frac{3\lambda +2}{\lambda +1}}\right)+2\left({\displaystyle \frac{4\lambda +1}{\lambda +1}}\right)-2\left({\displaystyle \frac{3\lambda +5}{\lambda +1}}\right)=1$

$\Rightarrow 8\lambda -4=\lambda +1$

$\Rightarrow \lambda ={\displaystyle \frac{5}{7}}$

$\therefore$ Required ratio 5:7

11. Find the centroid of a triangle, mid-points of whose sides are $D(1,2,-3),E(3,0,1)$and$F(-1,1,-4)$.

Ans: Suppose the co-ordinate of vertices of $△ABC$ are

Add equation (3), (6) and (9)

$2\left(z_1+z_2+z_3\right)=-8-6+2$

$z_1+z_2+z_3=-6\dots \dots (12)$

Co-ordinate of centroid

$x={\displaystyle \frac{x_1+x_2+x_3}{3}}={\displaystyle \frac{3}{3}}=1$

$y={\displaystyle \frac{y_1+y_2+y_3}{3}}={\displaystyle \frac{3}{3}}=1$

$z={\displaystyle \frac{z_1+z_2+z_3}{3}}={\displaystyle \frac{-6}{3}}=-2$

$(1,1,-2)$

12. The mid points of the sides of $△ABC$ are given by $(-2,3,5),(4,-1,7)$ and $(6,5,3)$ find the co-ordinate of $\text{A},\text{B}$ and$\text{C}$.

Ans: Let us suppose that the co-ordinates of point $A,B$ AND $C$ are $\left(x_1,y_1,z_1\right),\left(x_2,y_2,z_2\right)$and $\left(x_3,y_3,z_3\right)$ respectively. Let $D,E$ and$F$are the mid-points of side $BC,CA$and $AB$respectively.

$x$1 + $x$2 =12T………(1)

${\displaystyle \frac{y_1+y_2}{2}}=5$

$y_1+y_2=10\dots \dots (2)$

${\displaystyle \frac{z_1+z_2}{2}}=3$

$z_1+z_2=6\dots \dots$(3)

${\displaystyle \frac{x_2+x_3}{2}}=-2$

$x_2+x_3=-4\dots \dots (4)$

${\displaystyle \frac{y_2+y_3}{2}}=3$

$y_2+y_3=6\dots \dots (5)$

${\displaystyle \frac{z_1+z_2}{2}}=$

$z_1+z_2=10\dots \dots (6)$

${\displaystyle \frac{x_1+x_3}{2}}=4$

$x_1+x_3=8\dots \dots (7)$

${\displaystyle \frac{y_1+y_3}{2}}=-1$

$y_1+y_3=-2\dots \dots$(8)

${\displaystyle \frac{z_1+z_3}{z}}=7$

$z_1+z_3=14\dots \dots$(9)

Add equation (1), (4) and (7)

$2\left(x_1+x_2+x_3\right)=12-4+8$

$x_1+x_2+x_3={\displaystyle \frac{16}{3}}=8\dots ..(10)$

Similarly,$y_1+y_2+y_3=7\dots \dots (11)$

$z_1+z_2+z_3=15\dots \dots (12)$

Subtract equation (1), (4) and (7) from (10)

$x_3=-4,x_1=12,x_2=0$

Now subtract equation (2), (5) and (8) from (11)

$y_3=-3,y_1=1,y_2=9$

Similarly,$z_3=9,z_1=5,z_2=1$

$\therefore$ co-ordinate of point $A,B$and $C$are

$A(12,0,-4),B(1,9,-3)$ ,and$C(5,1,9)$

13. Find the co-ordinates of the points which trisects the line segment $QP$ formed by joining the point $P(4,2,-6)$ and $Q(10,-16,6)$.

Ans: Let $R\mathrm{\&}S$ be the points of trisection of the line  segment $QP$ . Then

$\Rightarrow 2PR=RQ$

$\Rightarrow {\displaystyle \frac{PQ}{RQ}}={\displaystyle \frac{1}{2}}$

∴ $S$divide $QP$in the ratio 1: 2

$\therefore$ Co-ordinates of point

$S\left[{\displaystyle \frac{1(10)+2\times 4}{1+2}},{\displaystyle \frac{1(-16)+2\times 2}{1+2}},{\displaystyle \frac{1\times 6+2(-6)}{1+2}}\right]$

$S(6,-4,-2)$

Similarly,$PS=2SQ$

$\Rightarrow {\displaystyle \frac{PS}{SQ}}={\displaystyle \frac{2}{1}}$

$\therefore$$R$ divide the line segment $QP$ in the ratio 2:1

$\therefore$ co-ordinates of point $R$

$R\left[{\displaystyle \frac{2(10)+1(4)}{1+2}},{\displaystyle \frac{2(-16)+1(2)}{1+2}},{\displaystyle \frac{2(6)+1(-6)}{1+2}}\right]$

$\therefore R(8,-10,2)$

14. Show that the point $A(1,2,3),B(-1,-2,-1),C(2,3,2)$ and $D(4,7,6)$ taken in order form the vertices of a parallelogram. Do these form a rectangle?

Ans: Mid-point of $AC$ is $\left({\displaystyle \frac{1+2}{2}},{\displaystyle \frac{2+3}{2}},{\displaystyle \frac{3+2}{2}}\right)$

i.e. $\left({\displaystyle \frac{3}{2}},{\displaystyle \frac{5}{2}},{\displaystyle \frac{5}{2}}\right)$

also the mid-point of $BD$ is $\left({\displaystyle \frac{-1+4}{2}},{\displaystyle \frac{-2+7}{2}},{\displaystyle \frac{-1+6}{2}}\right)$

i.e.$\left({\displaystyle \frac{3}{2}},{\displaystyle \frac{5}{2}},{\displaystyle \frac{5}{2}}\right)$

Then$AC$and  $BD$ have same mid-points

$\therefore$$AC$ and $BD$ bisect each other , It is a Parallelogram.

Now 

$AC=\sqrt{{(2-1)}^2+{(3-2)}^2+{(2-3)}^2}=\sqrt{3}$ and

$BD=\sqrt{{(4+1)}^2+{(7+2)}^2+{(6+1)}^2}=\sqrt{155}$

$\therefore AC\ne BD$ diagonals are not equal

PQRS is not a rectangle.

15. A point $C$with $x$ co-ordinates 4 lies on the line segment joining the points$A(2,-3,4)$ and $B(8,0,10)$ find the co-ordinates of the point $C$.

Ans: let the point $C$ divide the line segment joining the point $A$ and $B$in the ratio $\lambda =1$, Then co-ordinates of Point $\text{R}$

$\left[{\displaystyle \frac{8\lambda +2}{\lambda +1}},{\displaystyle \frac{-3}{\lambda +1}},{\displaystyle \frac{10\lambda +4}{\lambda +1}}\right]$

The $x$ co-ordinates of point $C$is 4

$\Rightarrow {\displaystyle \frac{8\lambda +2}{\lambda +1}}=4=\lambda ={\displaystyle \frac{1}{2}}$

$\therefore$ co-ordinates of point $\text{R}$

$\left[4,{\displaystyle \frac{-3}{{\displaystyle \frac{1}{2}}+1}}\cdot {\displaystyle \frac{10\times {\displaystyle \frac{1}{2}}+4}{{\displaystyle \frac{1}{2}}+1}}\right]$ i.e. $(4,-2,6)$

16. If the points $A(1,0,-6)=B(-3,P,q)$ and $C(-5,9,6)$are collinear, find the values of $\mathbf{P}$and $\mathbf{q}$

Ans: Given points 

$A(1,0,-6)=B(-3,P,q)$ and $C(-5,9,6)$ are collinear

Let point $B$ divide $AC$ in the ratio K:1

$\therefore$  co-ordinates of point $A\left({\displaystyle \frac{1-5K}{K+1}}\cdot {\displaystyle \frac{0+9K}{K+1}},{\displaystyle \frac{-6+6K}{K+1}}\right)$

$B(-3,P,q)$

${\displaystyle \frac{1-5K}{K+1}}=-3$

$1-5K=-3K-3$

$-2K=-4$

$K={\displaystyle \frac{-4}{-2}}$

$K=2$

$\therefore$ The value of $\text{P}$ and $\text{q}$ are 6 and 2.

17. Three consecutive vertices of a parallelogram $PQRS$ are $P(3,-1,2),Q(1,2,-4)$ and $R(-1,1,2)$. Find fourth vertex $S$.

Ans: Given vertices of $Paralle\log ramPQRS$

$P(3,-1,2),Q(1,2,-4),R(-1,1,2)$

Suppose co-ordinates of fourth vertex $S(x,y,z)$

Mid-point of $PR\left({\displaystyle \frac{3-1}{2}},{\displaystyle \frac{-1+1}{2}},{\displaystyle \frac{2+2}{2}}\right)$

$=(1,0,2)$

Mid-point of $QS\left({\displaystyle \frac{x+1}{2}},{\displaystyle \frac{y+2}{2}},{\displaystyle \frac{-4+z}{2}}\right)$

Mid-point of $PR$ = mid-point of $QS$

${\displaystyle \frac{x+1}{2}}=1\Rightarrow x=1$

${\displaystyle \frac{y+2}{2}}=0\Rightarrow y=-2$

${\displaystyle \frac{-4+z}{2}}=2\Rightarrow z=8$

Co-ordinates of point $S(1,-2,8)$.

18. If $P$ and $Q$be the points $(3.4,5)$ and $(-1,3,7)$ respectively. Find the eq. of the set points $A$ such that $A{P}^2+A{Q}^2=K^2$ where $\mathbf{K}$ is a constant.

Ans: Let co-ordinates of point P be$(x,y,z)$

$A{P}^2=(x-3{)}^2+(y-4{)}^2+(z-5{)}^2$

$=x^2-6x+9+y^2-8y+16+z^2-10z+25$

$=x^2+y^2+z^2-6x-8y-10z+50$

$A{Q}^2=(x+1{)}^2+(y-3{)}^2+(z-7{)}^2$

$=x^2+2x+1+y^2-6y+9+z^2-14+49$

$=x^2+y^2+z^2+2x-6y-14z+59$

$A{P}^2+A{Q}^2=K^2$

$2\left(x^2+y^2+z^2\right)-4x-14y-24z+109=K^2$

$x^2+y^2+z^2-2x-7y-12z={\displaystyle \frac{K^2-109}{2}}$

6 Marks Questions

1. Prove that the lines joining the vertices of a tetrahedron to the centroids of the opposite faces are concurrent.

Ans: Let $PQRS$ be tetrahedron such that the coordinates of its vertices are $P\left(x_1,y_1,z_1\right)$, $Q\left(x_2,y_2,z_2\right),R\left(x_2,y_3,z_3\right)$ and $S(x_4,y_4,z_4)$.

The coordinates of the centroids of faces $PQR,SPQ,SQR$ and $SRP$respectively.

$G_1\left[{\displaystyle \frac{x_1+x_2+x_3}{3}},{\displaystyle \frac{y_1+y_2+y_3}{3}},{\displaystyle \frac{z_1+z_2+z_3}{3}}\right]$

$G_2\left[{\displaystyle \frac{x_1+x_2+x_4}{3}},{\displaystyle \frac{y_1+y_2+y_4}{3}}\cdot {\displaystyle \frac{z_1+z_2+z_4}{3}}\right]$

$G_3\left[{\displaystyle \frac{x_2+x_3+x_4}{3}},{\displaystyle \frac{y_2+y_3+y_4}{3}},{\displaystyle \frac{z_2+z_3+z_4}{3}}\right]$

$G_4\left[{\displaystyle \frac{x_4+x_3+x_1}{3}},{\displaystyle \frac{y_4+y_3+y_1}{3}}\cdot {\displaystyle \frac{z_4+z_3+z_1}{3}}\right]$

Now, coordinates of point $G$ dividing $S{G}_1$ in the ratio 3: 1are

$\left[{\displaystyle \frac{1.x_4+3\left({\displaystyle \frac{x_1+x_2+x_3}{3}}\right)}{1+3}}={\displaystyle \frac{1.y_4+3\left({\displaystyle \frac{y_1+y_2+y_3}{3}}\right)}{1+3}}={\displaystyle \frac{1-z_4+3\left({\displaystyle \frac{z_1+z_2+z_3}{3}}\right)}{1+3}}\right]$

$=\left[{\displaystyle \frac{x_1+x_2+x_3+x_4}{4}},{\displaystyle \frac{y_1+y_2+y_3+y_4}{4}},{\displaystyle \frac{z_1+z_2+z_3+z_4}{4}}\right]$

Similarly the point dividing $R{G}_2,P{G}_3,Q{G}_4$ and $S{G}_1$ in the ratio 3:1 has the same coordinates.

Hence the point $G\left[{\displaystyle \frac{x_1+x_2+x_3+x_4}{4}},{\displaystyle \frac{y_1+y_2+y_3+y_4}{4}},{\displaystyle \frac{z_1+z_2+z_3+z_4}{4}}\right]$ is common to

$S{G}_1$, $R{G}_2,P{G}_3$ and $Q{G}_4$

Hence, they are concurrent.

2. The midpoints of the sides of a triangle are $(1,5,-1),(0,4,-2)$ and $(2,3,4)$. Find its vertices.

Ans: Let the vertices of triangle be

$A(x_1,y_1,z_1),B(x_2,y_2,z_2)$ and $C(x_3,y_3,z_3)$

Mid-point of $AC$ is $E$

$\therefore ({\displaystyle \frac{x_1+x_3}{2}},{\displaystyle \frac{y_1+y_3}{2}},{\displaystyle \frac{z_1+z_3}{2}})\equiv (0,4,-2)$

So, $C(x_3,y_3,z_3)\equiv C(-x_1,8-y_1,-4-z_1)$

(i)

Mid-point of $AB$ is $F$

$\therefore ({\displaystyle \frac{x_1+x_2}{2}},{\displaystyle \frac{y_1+y_2}{2}},{\displaystyle \frac{z_1+z_2}{2}})\equiv (2,3,4)$

So, $B(x_2,y_2,z_2)\equiv B(4-x_1,6-y_1,8-z_1)$

(ii) Mid-point of BC is

 $\therefore {\displaystyle \frac{-x_1+4-x_1}{2}}=1,$${\displaystyle \frac{8-y_1+6-y_1}{2}}=5$

 ${\displaystyle \frac{-4-z_1+8-z_1}{2}}=-1$

 $\Rightarrow x_1=1,y_1=2\text{ and }{z}_1=3$

 $\therefore A\equiv (1,2,3)$

 So, $B\equiv (3,4,5)$ [Using (ii)]

 and $C\equiv (-1,6,-7)$ [Using (i)]

 Centroid, $G\equiv ({\displaystyle \frac{1+3-1}{3}},{\displaystyle \frac{2+4+6}{3}},{\displaystyle \frac{3+5-7}{3}})$

 $\equiv (1,4,{\displaystyle \frac{1}{3}})$

3. Let $P\left(x_1\cdot y_1,z_1\right)$ and $Q\left(x_2\cdot y_2,z_2\right)$ be two points in space find co- ordinate of point $R$which divides$P$and $Q$ in the ratio $m_1:m_2$ by geometrically.