NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.2 - Introduction to Three Dimensional Geometry

Question 1: (i) Find the distance between the following pairs of points: (2, 3, 5) and (4, 3, 1)

Answer:

The distance between two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by

$d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$

So, distance between (2, 3, 5) and (4, 3, 1) is given by

$d=\sqrt{(4-2{)}^2+(3-3{)}^2+(1-5{)}^2}$

$d=\sqrt{4+0+16}$

$d=\sqrt{20}$

$d=2\sqrt{5}$

Question 1: (ii) Find the distance between the following pairs of points: (–3, 7, 2) and (2, 4, –1)

Answer:

The distance between two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by

$d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$

So, distance between (–3, 7, 2) and (2, 4, –1) is given by

$d=\sqrt{(2-(-3){)}^2+(4-7{)}^2+(-1-2{)}^2}$

$d=\sqrt{25+9+9}$

$d=\sqrt{43}$

Question 1: (iii) Find the distance between the following pairs of points:(–1, 3, – 4) and (1, –3, 4)

Answer:

The distance between two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by

$d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$

So, distance between (–1, 3, – 4) and (1, –3, 4) is given by

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

$d=\sqrt{4+36+64}$

$d=\sqrt{104}$

$d=2\sqrt{26}$

Question 1: (iv) Find the distance between the following pairs of points: (2, –1, 3) and (–2, 1, 3).

Answer:

The distance between two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by

$d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$

So, distance between (2, –1, 3) and (–2, 1, 3).) is given by

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

$d=\sqrt{16+4+0}$

$d=\sqrt{20}$

$d=2\sqrt{5}$

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

Answer:

Given,theree points A=(–2, 3, 5), B=(1, 2, 3) and C=(7, 0, –1)

The distance between two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by

$d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$

The distance AB :

$AB=\sqrt{(1-(-2){)}^2+(2-3{)}^2+(3-5{)}^2}$

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

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

$AB=\sqrt{14}$

The distance BC:

$BC=\sqrt{(7-1{)}^2+(0-2{)}^2+(-1-3{)}^2}$

$BC=\sqrt{36+4+16}$

$BC=\sqrt{56}$

$BC=2\sqrt{14}$

The distance CA

$CA=\sqrt{(7-(-2){)}^2+(0-3{)}^2+(-1-5{)}^2}$

$CA=\sqrt{81+9+36}$

$CA=\sqrt{126}$

$CA=3\sqrt{14}$

As we can see here,

$AB+BC=\sqrt{14}+2\sqrt{14}=3\sqrt{14}=AC$

Hence we can say that point A,B and C are colinear.

Question 3: (i) Verify the following: (0, 7, –10), (1, 6, – 6) and (4, 9, – 6) are the vertices of an isosceles triangle.

Answer:

Given Three points A=(0, 7, –10), B=(1, 6, – 6) and C=(4, 9, – 6)

The distance between two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by

$d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$

The distance AB

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

$AB=\sqrt{1+1+16}$

$AB=\sqrt{18}$

The distance BC

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

$BC=\sqrt{9+9+0}$

$BC=\sqrt{18}$

The distance CA

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

$CA=\sqrt{16+4+16}$

$CA=\sqrt{36}$

$CA=6$

As we can see $AB=BC\ne CA$

Hence we can say that ABC is an isosceles triangle.

Question 3:(ii) Verify the following

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

Answer:

Given Three points A=(0, 7, 10), B=(-1, 6, 6) and C=(-4, 9, 6)

The distance between two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by

$d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$

The distance AB

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

$AB=\sqrt{1+1+16}$

$AB=\sqrt{18}$

The distance BC

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

$BC=\sqrt{9+9+0}$

$BC=\sqrt{18}$

The distance CA

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

$CA=\sqrt{16+4+16}$

$CA=\sqrt{36}$

$CA=6$

As we can see

$(AB{)}^2+(BC{)}^2=18+18=36=(CA{)}^2$

Since this follows pythagorus theorem, we can say that ABC is a right angle triangle.

Question 3:(iii) Verify the following: (–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are the vertices of a parallelogram.

Answer:

Given A=(–1, 2, 1), B=(1, –2, 5), C=(4, –7, 8) and D=(2, –3, 4)

Given Three points A=(0, 7, –10), B=(1, 6, – 6) and C=(4, 9, – 6)

The distance between two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by

$d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$

The distance AB

$AB=\sqrt{(1-(-1){)}^2+(-2-2{)}^2+(5-1{)}^2}$

$AB=\sqrt{4+16+16}$

$AB=\sqrt{36}$

$AB=6$

The distance BC

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

$BC=\sqrt{9+25+9}$

$BC=\sqrt{43}$

The distance CD

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

$CA=\sqrt{4+16+16}$

$CA=\sqrt{36}$

$CA=6$

The distance DA

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

$DA=\sqrt{9+25+9}$

$DA=\sqrt{43}$

Here As we can see

$AB=6=CA$ And $BC=\sqrt{43}=DA$

As the opposite sides of quadrilateral are equal, we can say that ABCD is a parallelogram.

Question 4: Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1).

Answer:

Given, two points A=(1, 2, 3) and B=(3, 2, –1).

Let the point P= (x,y,z) be a point which is equidistance from the points A and B.

so,

The distance PA= The distance PB

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

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

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

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

$[(x-1{)}^2-(x-3{)}^2]+[(y-2{)}^2-(y-2{)}^2]+[(z-3{)}^2-(z+1{)}^2]=0$

Now lets apply the simplification property,

$a^2-b^2=(a+b)(a-b)$

$[(2)(2x-4)]+0+[(-4)(2z-2)]=0$

$4x-8-8z+8=0$

$4x-8z=0$

$x-2z=0$

Hence locus of the point which is equidistant from A and B is $x-2z=0$.

Question 5: Find the equation of the set of points P, the sum of whose distances from A (4, 0, 0) and B (– 4, 0, 0) is equal to 10

Answer:

Given,

Two points A (4, 0, 0) and B (– 4, 0, 0)

Let the point P(x,y,z) be a point whose distance from A and B is 10.

So,

The distance PA+The distance PB=10

$\sqrt{(x-4{)}^2+(y-0{)}^2+(z-0{)}^2}+\sqrt{(x-(-4){)}^2+(y{)}^2+(z{)}^2}$

$=10$

$\sqrt{(x-4{)}^2+(y{)}^2+(z{)}^2}+\sqrt{(x+4{)}^2+(y{)}^2+(z{)}^2}=10$

$\sqrt{(x-4{)}^2+(y{)}^2+(z{)}^2}=10-\sqrt{(x+4{)}^2+(y{)}^2+(z{)}^2}$

Squaring on both sides:

$(x-4{)}^2+(y{)}^2+(z{)}^2=100-20\sqrt{(x+4{)}^2+(y{)}^2+(z{)}^2}$$+(x+4{)}^2+(y{)}^2+(z{)}^2$

$\begin{array}{rl}& (x-4{)}^2-(x+4{)}^2=100-20\sqrt{(x+4{)}^2+(y{)}^2+(z{)}^2}\\ & -16x=100-20\sqrt{(x+4{)}^2+(y{)}^2+(z{)}^2}\end{array}$

$20\sqrt{(x+4{)}^2+(y{)}^2+(z{)}^2}=100+16x$

$5\sqrt{(x+4{)}^2+(y{)}^2+(z{)}^2}=25+4x$

Now again squaring both sides,

$25\left((x+4{)}^2+(y{)}^2+(z{)}^2\right)=625+200x+16x^2$

$25x^2+200x+400+25y^2+25z^2=625+200x+16x^2$

$9x^2+25y^2+25z^2-225=0$

Hence, the equation of the set of points P, the sum of whose distances from A and B is equal to 10 is $9x^2+25y^2+25z^2-225=0$.