Class 11 Maths Ncert Solutions Chapter 12 Ex 12.2 Introduction to Three Dimensional Geometry PDF

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

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 = (x2x1)2+(y2y1)2+(z2z1)2$\sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}+\left ( z_{2}- z_{1}\right )^{2}}\\$

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

AB=(42)2+(58)2+(89)2$AB = \sqrt{\left ( 4-2 \right )^{2}+\left ( 5-8 \right )^{2}+\left ( 8-9\right )^{2}}\\$ AB=4+9+1$\\\Rightarrow AB = \sqrt{4+9+1}$

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

Therefore, The distance between point A (2, 8, 9) and point B (4, 5, 8) is 14$\sqrt{14}$ 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$AB = \sqrt{\left [ 2-(-3) \right ]^{2}+\left ( 6-4 \right )^{2}+\left ( -1-5\right )^{2}}\\$

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

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

Therefore, The distance between point A (-3, 4, 5) and point B (2, 6, -1) is 65$\sqrt{65}$ 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$AB = \sqrt{\left [ 5-(-6) \right ]^{2}+ [-2-(-4)]^{2}+\left ( 6-1\right )^{2}}\\$

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

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

Therefore, The distance between point A (-6, -4, 1) and point B (5, -2, 6) is 56$5\sqrt{6}$ 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$AB = \sqrt{\left [ 6-(-1) \right ]^{2}+ (5-9)^{2}+\left ( -3-8\right )^{2}}\\$

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

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

Therefore, The distance between point A (-1, 9, 8) and point B (6, 5, -3) is 186$\sqrt{186}$ 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=(27)2+(30)2+[5(1)]2$AB=\sqrt{(-2-7)^{2}+(3-0)^{2}+[5-(-1)]^{2}}$

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

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

Therefore, AB = 314$3\sqrt{14}$ units

Now, BC=[1(2)]2+(23)2+(35)2$BC=\sqrt{[1-(-2)]^{2}+(2-3)^{2}+(3-5)^{2}}$

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

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

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

And, AC=(17)2+(20)2+[3(1)]2$AC=\sqrt{(1-7)^{2}+(2-0)^{2}+[3-(-1)]^{2}}$

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

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

Therefore, AC = 214$2\sqrt{14}$ units

Since, AC + BC = 214+14$2\sqrt{14}+\sqrt{14}$ = 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 = (x2x1)2+(y2y1)2+(z2z1)2$\sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}+\left ( z_{2}- z_{1}\right )^{2}}$

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

Now,  AB=(0+4)2+(79)2+(106)2$AB=\sqrt{(0+4)^{2}+(7-9)^{2}+(10-6)^{2}}\\$

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

Therefore, AB = 6 units.

Now,  BC=(10)2+(67)2+(610)2$BC=\sqrt{(-1-0)^{2}+(6-7)^{2}+(6-10)^{2}}\\$

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

Therefore, BC = 32$3\sqrt{2}\\$ units.

And,  AC=(1+4)2+(69)2+(66)2$AC=\sqrt{(-1+4)^{2}+(6-9)^{2}+(6-6)^{2}}\\$

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

Therefore, AC = 32$3\sqrt{2}$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=(04)2+(79)2+(10+6)2$AB=\sqrt{(0-4)^{2}+(7-9)^{2}+(-10+6)^{2}}\\$

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

Therefore, AB = 6 units.

Now,  BC=(10)2+(67)2+(6+10)2$BC=\sqrt{(1-0)^{2}+(6-7)^{2}+(-6+10)^{2}}\\$

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

Therefore, BC = 32$3\sqrt{2}$ units.

And,  AC=(14)2+(69)2+(6+6)2$AC=\sqrt{(1-4)^{2}+(6-9)^{2}+(-6+6)^{2}}\\$

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

Therefore, AC = 32$3\sqrt{2}$ 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=(41)2+(7+2)2+(85)2$AB=\sqrt{(4-1)^{2}+(-7+2)^{2}+(8-5)^{2}}\\$

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

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

Now,  BC=(24)2+(3+7)2+(48)2$BC=\sqrt{(2-4)^{2}+(-3+7)^{2}+(4-8)^{2}}\\$

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

Therefore, BC = 6 units.

Now,  CD=(12)2+(2+3)2+(14)2$CD =\sqrt{(-1-2)^{2}+(2+3)^{2}+(1-4)^{2}}$

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

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

And,  AD=(11)2+(2+2)2+(15)2$AD =\sqrt{(-1-1)^{2}+(2+2)^{2}+(1-5)^{2}}\\$

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

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

$\Rightarrow$   [(x3)2+(y2)2+(z1)2]2=[(x+1)2+(y2)2+(z3)2]2$\left [ \sqrt{(x-3)^{2}+(y-2)^{2}+(z-1)^{2}} \;\right ]^{2}=\left [ \sqrt{(x+1)^{2}+(y-2)^{2}+(z-3)^{2}} \;\right ]^{2}$

$\\\Rightarrow$   (x3)2+(y2)2+(z1)2=(x+1)2+(y2)2+(z3)2$(x-3)^{2}+(y-2)^{2}+(z-1)^{2}=(x+1)^{2}+(y-2)^{2}+(z-3)^{2}$

$\\\Rightarrow$   x2+96x+y2+44y+z2+12z=x2+1+2x+y2+44y+z2+96z$x^{2}+9-6x+y^{2}+4-4y+z^{2}+1-2z=x^{2}+1+2x+y^{2}+4-4y+z^{2}+9-6z$

$\\\Rightarrow$ – 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

$\Rightarrow$   (x3)2+y2+z2+[x(3)]2+y2+z2=12$\sqrt{(x-3)^{2}+y^{2}+z^{2}}+\sqrt{[x-(-3)]^{2}+y^{2}+z^{2}}=12\\$

$\Rightarrow$   (x3)2+y2+z2=12(x+3)2+y2+z2$\sqrt{(x-3)^{2}+y^{2}+z^{2}}=12-\sqrt{(x+3)^{2}+y^{2}+z^{2}}$

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

$\Rightarrow$   ((x3)2+y2+z2)2=(12(x+3)2+y2+z2)2$\left ( \sqrt{(x-3)^{2}+y^{2}+z^{2}} \right )^{2}= \left ( 12-\sqrt{(x+3)^{2}+y^{2}+z^{2}} \right )^{2}\\$

$\Rightarrow$   (x3)2+y2+z2=144+[(x+3)2+y2+z2]24(x+3)2+y2+z2$(x-3)^{2}+y^{2}+z^{2}= 144+[(x+3)^{2}+y^{2}+z^{2}]-24\sqrt{(x+3)^{2}+y^{2}+z^{2}}\\$

$\Rightarrow$   x2+96x+y2+z2=144+x2+9+6x+y2+z224(x+3)2+y2+z2$x^{2}+9-6x+y^{2}+z^{2}=144+x^{2}+9+6x+y^{2}+z^{2}-24\sqrt{(x+3)^{2}+y^{2}+z^{2}}\\$

$\Rightarrow$   24(x+3)2+y2+z2=144+12x$24\sqrt{(x+3)^{2}+y^{2}+z^{2}}=144+12x\\$

$\Rightarrow$   2(x+3)2+y2+z2=(x+12)$2\sqrt{(x+3)^{2}+y^{2}+z^{2}}=(x+12)$

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

$\Rightarrow$   4[(x+3)2+y2+z2]=(x+12)2$4[(x+3)^{2}+y^{2}+z^{2}]=(x+12)^{2}\\$

$\Rightarrow$   4x2+4y2+4z2+36+24x=x2+24x+144$4x^{2}+4y^{2}+4z^{2}+36+24x=x^{2}+24x+144\\$

$\Rightarrow$   3x2+4y2+4z2108=0$3x^{2}+4y^{2}+4z^{2}-108=0$

Therefore, the required equation is: 3x2+4y2+4z2108=0$3x^{2}+4y^{2}+4z^{2}-108=0$

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:

[m.a2+n.a1m+n,m.b2+n.b1m+n,m.c2+n.c1m+n]$\left [ \frac{m.a_{2}\;+\;n.a_{1}}{m\;+\;n},\frac{m.b_{2}\;+\;n.b_{1}}{m\;+\;n},\frac{m.c_{2}\;+\;n.c_{1}}{m\;+\;n} \right ]$

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:

[m.a2n.a1mn,m.b2n.b1mn,m.c2n.c1mn]$\left [ \frac{m.a_{2}\;-\;n.a_{1}}{m\;-\;n},\frac{m.b_{2}\;-\;n.b_{1}}{m\;-\;n},\frac{m.c_{2}\;-\;n.c_{1}}{m\;-\;n} \right]$