Class 11 Maths Ncert Solutions Ex 12.2

Class 11 Maths Ncert Solutions Chapter 12 Ex 12.2

 Distance Formula:

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

AB = (x2x1)2+(y2y1)2+(z2z1)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 = (x2x1)2+(y2y1)2+(z2z1)2

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

AB=(42)2+(58)2+(89)2 AB=4+9+1

AB=14 units

Therefore, The distance between point A (2, 8, 9) and point B (4, 5, 8) is 14 units

 

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

AB=[2(3)]2+(64)2+(15)2

   AB=25+4+36

 AB=65 units

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

 

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

AB=[5(6)]2+[2(4)]2+(61)2

AB=121+4+25

   AB=150 units

Therefore, The distance between point A (-6, -4, 1) and point B (5, -2, 6) is 56 units

 

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

AB=[6(1)]2+(59)2+(38)2

   AB=49+16+121

  AB=186 units

Therefore, The distance between point A (-1, 9, 8) and point B (6, 5, -3) is 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=(27)2+(30)2+[5(1)]2

   AB=81+9+36

AB=126 units

Therefore, AB = 314 units

Now, BC=[1(2)]2+(23)2+(35)2

   BC=9+1+4

   BC=14 units

Therefore, BC = 14 units

And, AC=(17)2+(20)2+[3(1)]2

  AC=36+4+16

AC=56 units

Therefore, AC = 214 units

Since, AC + BC = 214+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 = (x2x1)2+(y2y1)2+(z2z1)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=(0+4)2+(79)2+(106)2

   AB=16+4+16=36 units.

Therefore, AB = 6 units.

Now,  BC=(10)2+(67)2+(610)2

   BC=1+1+16=18 units.

Therefore, BC = 32 units.

And,  AC=(1+4)2+(69)2+(66)2

   AC=9+9=18 units.

Therefore, AC = 32units.

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=(04)2+(79)2+(10+6)2

   AB=36 units.

Therefore, AB = 6 units.

Now,  BC=(10)2+(67)2+(6+10)2

   BC=1+1+16=18 units.

Therefore, BC = 32 units.

And,  AC=(14)2+(69)2+(6+6)2

AC=9+9=18 units.

Therefore, AC = 32 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=(41)2+(7+2)2+(85)2

   AB=9+25+9=43 units.

Therefore, AB =43 units.

Now,  BC=(24)2+(3+7)2+(48)2

  BC=4+16+16=36 units.

Therefore, BC = 6 units.

Now,  CD=(12)2+(2+3)2+(14)2

   AC=9+25+9=433 units.

Therefore, CD = 43 units.

And,  AD=(11)2+(2+2)2+(15)2

   AD=4+16+16=36 units.

Therefore, AD = 6 units.

Since, BC = AD = 6 units and AB = CD = 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

  [(x3)2+(y2)2+(z1)2]2=[(x+1)2+(y2)2+(z3)2]2

  (x3)2+(y2)2+(z1)2=(x+1)2+(y2)2+(z3)2

  x2+96x+y2+44y+z2+12z=x2+1+2x+y2+44y+z2+96z

– 6x – 4y – 2z = +2x – 4y – 6z

– 6x – 4y – 2z = + 2x – 4y – 6z

   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

  (x3)2+y2+z2+[x(3)]2+y2+z2=12

  (x3)2+y2+z2=12(x+3)2+y2+z2

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

  ((x3)2+y2+z2)2=(12(x+3)2+y2+z2)2

  (x3)2+y2+z2=144+[(x+3)2+y2+z2]24(x+3)2+y2+z2

  x2+96x+y2+z2=144+x2+9+6x+y2+z224(x+3)2+y2+z2

   24(x+3)2+y2+z2=144+12x

  2(x+3)2+y2+z2=(x+12)

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

  4[(x+3)2+y2+z2]=(x+12)2

  4x2+4y2+4z2+36+24x=x2+24x+144

  3x2+4y2+4z2108=0

Therefore, the required equation is: 3x2+4y2+4z2108=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:

[m.a2+n.a1m+n,m.b2+n.b1m+n,m.c2+n.c1m+n]

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:

[m.a2n.a1mn,m.b2n.b1mn,m.c2n.c1mn]