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

Last updated date: 07th Sep 2024
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## CBSE Class 11 Maths Chapter-11 Important Questions - Free PDF Download

Free PDF download of Important Questions with solutions for CBSE Class 11 Maths Chapter 11 - Introduction to Three Dimensional Geometry prepared by expert Maths teachers from latest edition of CBSE(NCERT) books. Register online for Maths tuition on Vedantu.com to score more marks in your Examination.

Also, check CBSE Class 11 Maths Important Questions for other chapters:

 CBSE Class 11 Maths Important Questions Sl.No Chapter No Chapter Name 1 Chapter 1 Sets 2 Chapter 2 Relations and Functions 3 Chapter 3 Trigonometric Functions 4 Chapter 4 Principle of Mathematical Induction 5 Chapter 5 Complex Numbers and Quadratic Equations 6 Chapter 6 Linear Inequalities 7 Chapter 7 Permutations and Combinations 8 Chapter 8 Binomial Theorem 9 Chapter 9 Sequences and Series 10 Chapter 10 Straight Lines 11 Chapter 11 Conic Sections 12 Chapter 12 Introduction to Three Dimensional Geometry 13 Chapter 13 Limits and Derivatives 14 Chapter 14 Mathematical Reasoning 15 Chapter 15 Statistics 16 Chapter 16 Probability
Competitive Exams after 12th Science

## Study Important Questions for Class 11 Mathematics 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

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

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

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

These three equations give

$\hat \alpha = \dfrac{1}{2}$

So Q divides PR in the ratio $\dfrac{1}{2}:1$ 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

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

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

$PB = PC$

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

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

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

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

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

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

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$\operatorname{and} C\left( {{x_3}{y_3}{z_3}} \right)$ is $\left[ {\dfrac{{{x_1} + {y_1} + {z_1}}}{3},\dfrac{{{y_1} + {y_2} + {y_3}}}{3},\dfrac{{{z_1} + {z_2} + {z_3}}}{3}} \right]$

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

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

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

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

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

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

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

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

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

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

ii. For external division.

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

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

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

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

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 $PA = AQ$

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

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

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

$2ax = {a^2}$

$\therefore x = \dfrac{a}{2}$

Similarly $PA = AR$

$\Rightarrow y = \dfrac{b}{2}$

$Q\left( {x,y,{z_,}} \right)\quad ,R\left( {{x_2},{y_2},{z_2}} \right)$ 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 $\dfrac{{{x_1} + {x_2}}}{2} = - 1$

${x_1} + {x_2} = - 2 \ldots ..(1)$

$\dfrac{{{y_1} + {y_2}}}{2} = 1$

${y_1} + {y_2} = 2 \ldots \ldots$(2)

$\dfrac{{{z_1} + {z_2}}}{2} = - 4$

${z_1} + {z_2} = - 8 \ldots \ldots (3)$

$\dfrac{{{x_2} + {x_3}}}{2} = 1$

${x_2} + {x_3} = 2 \ldots \ldots (4)$

$\dfrac{{{y_2} + {y_3}}}{2} = 2$

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

$\dfrac{{{z_2} + {z_3}}}{2} = - 3$

${z_2} + {z_3} = - 6 \ldots \ldots$(6)

$\dfrac{{{x_1} + {r_3}}}{2} = 3$

${x_1} + {x_3} = 6 \ldots \ldots (7)$

$\dfrac{{{y_1} + {y_3}}}{2} = 0$

${y_1} + {y_3} = 0 \ldots \ldots (8)$

$\dfrac{{{z_1} + {z_3}}}{2} = 1$

${z_1} + {z_3} = 2 \ldots \ldots (9)$

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

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

${y_1} + {y_2} + {y_3} = 3 \ldots \ldots (11)$

And $PA = AS$

$\Rightarrow z = \dfrac{c}{2}$

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

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 $A$

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

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

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

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

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

$\Rightarrow \lambda = \dfrac{5}{7}$

$\therefore$ 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 $\vartriangle ABC$ are

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

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

${z_1} + {z_2} + {z_3} = - 6 \ldots \ldots (12)$

Co-ordinate of centroid

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

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

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

$(1,1, - 2)$

12. The mid points of the sides of $\vartriangle ABC$ 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)

$\dfrac{{{y_1} + {y_2}}}{2} = 5$

${y_1} + {y_2} = 10 \ldots \ldots (2)$

$\dfrac{{{z_1} + {z_2}}}{2} = 3$

${z_1} + {z_2} = 6 \ldots \ldots$(3)

$\dfrac{{{x_2} + {x_3}}}{2} = - 2$

${x_2} + {x_3} = - 4 \ldots \ldots (4)$

$\dfrac{{{y_2} + {y_3}}}{2} = 3$

${y_2} + {y_3} = 6 \ldots \ldots (5)$

$\dfrac{{{z_1} + {z_2}}}{2} =$

${z_1} + {z_2} = 10 \ldots \ldots (6)$

$\dfrac{{{x_1} + {x_3}}}{2} = 4$

${x_1} + {x_3} = 8 \ldots \ldots (7)$

$\dfrac{{{y_1} + {y_3}}}{2} = - 1$

${y_1} + {y_3} = - 2 \ldots \ldots$(8)

$\dfrac{{{z_1} + {z_3}}}{z} = 7$

${z_1} + {z_3} = 14 \ldots \ldots$(9)

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

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

${x_1} + {x_2} + {x_3} = \dfrac{{16}}{3} = 8 \ldots ..(10)$

Similarly,${y_1} + {y_2} + {y_3} = 7 \ldots \ldots (11)$

${z_1} + {z_2} + {z_3} = 15 \ldots \ldots (12)$

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

${x_3} = - 4,\quad {x_1} = 12,\quad {x_2} = 0$

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

${y_3} = - 3,\quad {y_1} = 1,\quad {y_2} = 9$

Similarly,${z_3} = 9,\quad {z_1} = 5,\quad {z_2} = 1$

$\therefore$ 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 $R\& S$ be the points of trisection of the line  segment $QP$ . Then

$\Rightarrow 2PR = RQ$

$\Rightarrow \dfrac{{PQ}}{{RQ}} = \dfrac{1}{2}$

$S$divide $QP$in the ratio 1: 2

$\therefore$ Co-ordinates of point

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

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

Similarly,$PS = 2SQ$

$\Rightarrow \dfrac{{PS}}{{SQ}} = \dfrac{2}{1}$

$\therefore $$R divide the line segment QP in the ratio 2:1 \therefore co-ordinates of point R R\left[ {\dfrac{{2(10) + 1(4)}}{{1 + 2}},\dfrac{{2( - 16) + 1(2)}}{{1 + 2}},\dfrac{{2(6) + 1( - 6)}}{{1 + 2}}} \right] \therefore R(8, - 10,2) 14. Show that the point A(1,2,3),B( - 1, - 2, - 1),C(2,3,2) and D(4,7,6) taken in order form the vertices of a parallelogram. Do these form a rectangle? Ans: Mid-point of AC is \left( {\dfrac{{1 + 2}}{2},\dfrac{{2 + 3}}{2},\dfrac{{3 + 2}}{2}} \right) i.e. \left( {\dfrac{3}{2},\dfrac{5}{2},\dfrac{5}{2}} \right) also the mid-point of BD is \left( {\dfrac{{ - 1 + 4}}{2},\dfrac{{ - 2 + 7}}{2},\dfrac{{ - 1 + 6}}{2}} \right) i.e.\left( {\dfrac{3}{2},\dfrac{5}{2},\dfrac{5}{2}} \right) ThenACand BD have same mid-points \therefore$$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 ${\text{R}}$

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

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

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

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

$\left[ {4,\dfrac{{ - 3}}{{\dfrac{1}{2} + 1}} \cdot \dfrac{{10 \times \dfrac{1}{2} + 4}}{{\dfrac{1}{2} + 1}}} \right]$ 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 $A\left( {\dfrac{{1 - 5K}}{{K + 1}} \cdot \dfrac{{0 + 9K}}{{K + 1}},\dfrac{{ - 6 + 6K}}{{K + 1}}} \right)$

$B( - 3,P,q)$

$\dfrac{{1 - 5K}}{{K + 1}} = - 3$

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

$- 2K = - 4$

$K = \dfrac{{ - 4}}{{ - 2}}$

$K = 2$

$\therefore$ 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 $PR\left( {\dfrac{{3 - 1}}{2},\dfrac{{ - 1 + 1}}{2},\dfrac{{2 + 2}}{2}} \right)$

$= (1,0,2)$

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

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

$\dfrac{{x + 1}}{2} = 1 \Rightarrow x = 1$

$\dfrac{{y + 2}}{2} = 0 \Rightarrow y = - 2$

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

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$(x,y,z)$

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

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

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

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

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

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

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

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

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

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.

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

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

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

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

Now, coordinates of point $G$ dividing $S{G_1}$ in the ratio 3: 1are

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

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

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

$A\left(x_{1}, y_{1}, z_{1}\right), B\left(x_{2}, y_{2}, z_{2}\right)$ and $C\left(x_{3}, y_{3}, z_{3}\right)$

Mid-point of $A C$ is $E$

$\therefore \quad\left(\dfrac{x_{1}+x_{3}}{2}, \dfrac{y_{1}+y_{3}}{2}, \dfrac{z_{1}+z_{3}}{2}\right) \equiv(0,4,-2)$

So, $C\left(x_{3}, y_{3}, z_{3}\right) \equiv C\left(-x_{1}, 8-y_{1},-4-z_{1}\right)$

(i)

Mid-point of $A B$ is $F$

$\therefore \quad\left(\dfrac{x_{1}+x_{2}}{2}, \dfrac{y_{1}+y_{2}}{2}, \dfrac{z_{1}+z_{2}}{2}\right) \equiv(2,3,4)$

So, $B\left(x_{2}, y_{2}, z_{2}\right) \equiv B\left(4-x_{1}, 6-y_{1}, 8-z_{1}\right)$

(ii) Mid-point of BC is

$\therefore \quad \dfrac{-x_{1}+4-x_{1}}{2} =1,$$\dfrac{8-y_{1}+6-y_{1}}{2}=5$

$\dfrac{-4-z_{1}+8-z_{1}}{2}=-1$

$\Rightarrow \quad x_{1}=1, y_{1}=2 \text { and } z_{1}=3$

$\therefore \quad A \equiv(1,2,3)$

So, $B \equiv(3,4,5) \quad$ [Using (ii)]

and $C \equiv(-1,6,-7) \quad$ [Using (i)]

Centroid, $G \equiv\left(\dfrac{1+3-1}{3}, \dfrac{2+4+6}{3}, \dfrac{3+5-7}{3}\right)$

$\equiv\left(1,4, \dfrac{1}{3}\right)$

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.

Ans: Let co-ordinate of Point $R$ be $(x,y,z)$ which divide line segment joining the point  $P$and $Q$ in the ratio ${m_1}:{m_2}$

Clearly , $\vartriangle PR{L^\prime }~\Delta QR{M^\prime }\quad [By - AA$similarity]

$\therefore \dfrac{{P{L^\prime}}}{{M{Q^\prime }}} = \dfrac{{PR}}{{RQ}}$

$\Rightarrow \dfrac{{L{L^\prime } - LP}}{{MQ - M{M^\prime }}} = \dfrac{{{m_1}}}{{{m_2}}}$

$\Rightarrow \dfrac{{NR - LP}}{{MQ - NR}} = \dfrac{{{m_1}}}{{{m_2}}}$

${\because L{L^\prime } = NR}$

${{\text{ and }}M{M^\prime } = NR}$

$\Rightarrow \dfrac{{z - {z_1}}}{{{z_2} - z}} = \dfrac{{{m_1}}}{{{m_2}}}$

$\Rightarrow z = \dfrac{{{m_1}{z_2} + {m_2}{z_1}}}{{{m_1} + {m_2}}}$

Similarly , we get

$y = \dfrac{{{m_1}{y_2} + {m_2}{y_1}}}{{{m_1} + {m_2}}}$

4. Show that the plane $px + qy + rz + s = 0$ divides the line joining the points $\left( {{x_1},{y_1},{z_1}} \right)$and $\left( {{x_2},{y_2},{z_2}} \right)$ in the ratio $\dfrac{{p{x_1} + q{y_1} + r{z_1} + s}}{{p{x_2} + q{y_2} + r{z_2} + s}}$.

Ans: Let the plane $px + qy + rz + s = 0$ divide the line joining the points $\left( {{x_1},{y_1},{z_1}} \right)$ and$\left( {{x_2},{y_2},{z_2}} \right)$ in the ratio $\lambda = 1$.

$\therefore x = \dfrac{{\widehat {2{x_2}} + {x_1}}}{{\lambda + 1}} = y = \dfrac{{\lambda {y_2} + {y_1}}}{{\lambda + 1}} = z = \dfrac{{\lambda {z_2} + {z_1}}}{{\lambda + 1}}$

$\because$ Plane $px + qy + rz + s = 0$Passing through $\left( {{x_0}y,z} \right)$

$\therefore p\dfrac{{\left( {\lambda {x_2} + {x_1}} \right)}}{{\lambda + 1}} + q\dfrac{{\left( {\lambda {y_2} + {y_1}} \right)}}{{\lambda + 1}} + r\dfrac{{\left( {\lambda {z_2} + {z_1}} \right)}}{{2 + 1}} + s = 0$

$p\left( {\lambda {x_2} + {x_1}} \right) + q\left( {\lambda {y_2} + {y_1}} \right) + r\left( {\lambda {z_2} + {z_1}} \right) + s(\lambda + 1) = 0$

$\lambda \left( {p{x_2} + q{y_2} + r{z_2} + s} \right) + \left( {p{x_1} + q{y_1} + r{z_1} + s} \right) = 0$

$\lambda = - \dfrac{{\left( {p{x_1} + q{y_1} + r{z_1} + s} \right)}}{{\left( {p{x_2} + q{y_2} + r{z_2} + s} \right)}}$

Hence Proved.

5.Prove that the points$0(0,0,0),P(2,0,0),Q(1,\sqrt 3 ,0)$, and $R\left( {1,\dfrac{1}{{\sqrt 3 }},\dfrac{{2\sqrt 2 }}{{\sqrt 3 }}} \right)$ are

The vertices of a regular tetrahedron.

Ans: To prove $O,P,Q,R$ are vertices of regular tetrahedron.

${{\text{ We have to show that }}}$

${|OP| = |OQ| = |OR| = |PQ| = |QR| = |RP|}$

${|OP| = \sqrt {{{(0 - 2)}^2} + {0^2} + {0^2}} = 2{\text{ unit }}}$

${|{\text{OQ}}| = \sqrt {{{(0 - 1)}^2} + {{(0 - \sqrt 3 )}^2} + {0^2}} = \sqrt {1 + 3} = \sqrt 4 = 2{\text{ unit }}}$

${|OR| = \sqrt {{{(0 - 1)}^2} + \left( {0 - \dfrac{1}{{\sqrt 3 }}} \right) + {{\left( {0 - \dfrac{{2\sqrt 2 }}{3}} \right)}^2}} }$

${ = \sqrt {1 + \dfrac{1}{3} + \dfrac{8}{3}} }$

${ = \sqrt {\dfrac{{12}}{3}} = \sqrt 4 = 2{\text{ unit }}}$

${|AB| = \sqrt {{{(2 - 1)}^2} + {{(0 - \sqrt 3 )}^2} + {{(10 - 0)}^2}} = \sqrt {1 + 3 + 0} }$

${ = \sqrt 4 = 2{\text{ unit }}}$

${|BC| = \sqrt {{{(1 - 1)}^2} + {{\left( {\sqrt 3 - \dfrac{1}{{\sqrt 3 }}} \right)}^2} + {{\left( {0 - \dfrac{{2\sqrt 2 }}{{\sqrt 3 }}} \right)}^2}} }$

${ = \sqrt {0 + {{\left( {\dfrac{2}{{\sqrt 3 }}} \right)}^2} + \dfrac{8}{3}} }$

$= \sqrt {\dfrac{{12}}{3}} = 2{\text{unit}}$

$|{\text{CA}}| = \sqrt {{{(1 - 2)}^2} + {{\left( {\dfrac{1}{{\sqrt 3 }} - 0} \right)}^2} + {{\left( {\dfrac{{2\sqrt 2 }}{{\sqrt 3 }} - 0} \right)}^2}}$

$= \sqrt {1 + \dfrac{1}{3} + \dfrac{8}{3}}$

$= \sqrt {\dfrac{{12}}{3}} = 2{\text{unit}}$

$\therefore |{\text{AB}}| = |{\text{BC}}| = |{\text{CA}}| = |{\text{OA}}| = |{\text{OB}}| = |{\text{OC}}| = 2$ unit

${\text{O,P,Q,R}}$ are vertices of a regular tetrahedron.

6. If ${\text{P}}$ and  ${\text{Q}}$ are the points $( - 2,2,3)$ and $( - 1,4, - 3)$ respectively, then find the locus of

${\text{A}}$ such that $3|AP| = 2|\;AQ|$.

Ans: The Given points $P( - 2,2,3)$ and $Q( - 1,4, - 3)$

Suppose co-ordinates of point $A(x,y,z)$

$|AP| = \sqrt {{{(x + 2)}^2} + {{(y - 2)}^2} + {{(2 - 3)}^2}}$

$|{\text{AP}}| = \sqrt {{x^2} + {y^2} + {z^2} + 4x - 4y - 6z + 17}$

$|AQ| = \sqrt {{{(x + 1)}^2} + {{(y - 4)}^2} + {{(z + 3)}^2}}$

$|AQ| = \sqrt {{x^2} + {y^2} + {z^2} + 2x - 8y + 6z + 26}$

$9A{P^2} = 4\;A{Q^2}$

$9\left( {{x^2} + {y^2} + {z^2} + 4x - 4y - 6z + 17} \right) = 4\left( {{x^2} + {y^2} + {z^2} + 2x - 8y + 6z + 26} \right)$

$\left( {5{x^2} + 5{y^2} + 5{z^2} + 28x - 4y - 30z + 49 = 0} \right)$

## FAQs on Important Questions for CBSE Class 11 Maths Chapter 11 - Introduction to Three Dimensional Geometry

1. What is the main concept of 3D Geometry covered in Chapter 12 of Class 11 Maths?

Ans: This chapter is an introduction to the students for 3D geometry. This is a branch of mathematics dealing with the study of points, lines and solid shapes in three-dimensional coordinate systems. This chapter should be learnt with full focus so it becomes easier for the students to have a flow of the chapter so they are not left with any confusion. It also teaches the Z-coordinate. For further details, kindly visit the Vedantu website and refer to the Important Questions PDF. The notes and solutions are present on Vedantu's official website (vedantu.com) and mobile app for free of cost.

2. Where will Important Questions for Chapter 11 of Class 11 Maths be available?

Ans: The Important Questions of Chapter 11 of Class 11 Maths are available on Vedantu’s website. These questions will help the students to practice what they have learnt in Chapter 11. Students will also get an idea about the questions that might be asked in the exam, helping them prepare. Students should have complete focus on this chapter since it is an introduction and there will be a lot of unfamiliar terms. To know and get the concepts clear, focus and attention are required.

3. How many exercises are there in Chapter 11 of Class 11 Maths?

Ans: There are three exercises (11.1, 11.2, and 11.3) and miscellaneous exercises in this chapter. Practising these numerical problems will help the students to be perfect with the concept and also help them to think logically. The solutions for all these are systematically given in  Vedantu from NCERT solutions. These will come in handy if students have any confusion and they can rely upon these solutions with full confidence.

4. What are the x and y-axis taken in the plane known as and also what is the form of coordinates of points XY in a plane?

Ans: The X-axis and the Y-axis together are known as the XY plane and the form in which the coordinates of points XY plane is (x, y, 0). Students should know this and implement it while doing the problems or the questions might be asked in the form of fill in blanks or true or false. Understanding the basic concepts in the chapter is very important. Practising the exercise numericals in the proper step will help the students to understand in a comprehend manner.

5. How will you find the distance between the following points (2,3,5) and (4,3,1)?

Ans: Let us first take the point P as (2,3,5) and Q as (4,3,1).

Now, using the formula for distance PQ= [(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2] and given values x1=2, y1=3, z1=5 and x2=4, y2=3, z2=1,

We will substitute the values in this equation -

Distance PQ = [(4 – 2)2 + (3 – 3)2 + (1 – 5)2]

= √[(2)2 + (0)(2) + (-4)(2)]

= [4 + 0 + 8]

= √12

= 23

∴ The distance is 2√3 units.