Class 11 Maths NCERT Solutions Chapter 11 Ex 11.2

The distance between two points (x_1,y_1,z_1) and (x_2,y_2,z_2) in space is:

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

(i) (2, 3, 5) and (4, 3, 1)

d=\sqrt{(4-2)^2+(3-3)^2+(1-5)^2}=\sqrt{4+0+16}=\sqrt{20}=2\sqrt{5}

(ii) (−3, 7, 2) and (2, 4, −1)

d=\sqrt{(2-(-3))^2+(4-7)^2+(-1-2)^2}=\sqrt{25+9+9}=\sqrt{43}

(iii) (−1, 3, −4) and (1, −3, 4)

d=\sqrt{(1-(-1))^2+(-3-3)^2+(4-(-4))^2}=\sqrt{4+36+64}=\sqrt{104}=2\sqrt{26}

(iv) (2, −1, 3) and (−2, 1, 3)

d=\sqrt{(-2-2)^2+(1-(-1))^2+(3-3)^2}=\sqrt{16+4+0}=\sqrt{20}=2\sqrt{5}

Let P(-2,3,5), Q(1,2,3) and R(7,0,-1). Three points are collinear if the sum of the distances between two pairs of points equals the distance between the remaining pair.

Distance PQ:

PQ=\sqrt{(1-(-2))^2+(2-3)^2+(3-5)^2}=\sqrt{9+1+4}=\sqrt{14}

Distance QR:

QR=\sqrt{(7-1)^2+(0-2)^2+(-1-3)^2}=\sqrt{36+4+16}=\sqrt{56}=2\sqrt{14}

Distance PR:

PR=\sqrt{(7-(-2))^2+(0-3)^2+(-1-5)^2}=\sqrt{81+9+36}=\sqrt{126}=3\sqrt{14}

Now, PQ+QR=\sqrt{14}+2\sqrt{14}=3\sqrt{14}=PR.

(i) Isosceles triangle

Let A(0,7,-10), B(1,6,-6) and C(4,9,-6).

AB=\sqrt{(1-0)^2+(6-7)^2+(-6+10)^2}=\sqrt{1+1+16}=\sqrt{18}=3\sqrt{2}

BC=\sqrt{(4-1)^2+(9-6)^2+(-6+6)^2}=\sqrt{9+9+0}=\sqrt{18}=3\sqrt{2}

CA=\sqrt{(0-4)^2+(7-9)^2+(-10+6)^2}=\sqrt{16+4+16}=\sqrt{36}=6

Since AB=BC=3\sqrt{2}, two sides are equal.

(ii) Right angled triangle

Let A(0,7,10), B(-1,6,6) and C(-4,9,6).

AB^2=(-1-0)^2+(6-7)^2+(6-10)^2=1+1+16=18

BC^2=(-4+1)^2+(9-6)^2+(6-6)^2=9+9+0=18

CA^2=(0+4)^2+(7-9)^2+(10-6)^2=16+4+16=36

Checking: AB^2+BC^2=18+18=36=CA^2.

(iii) Parallelogram

Let A(-1,2,1), B(1,-2,5), C(4,-7,8) and D(2,-3,4).

AB=\sqrt{(1+1)^2+(-2-2)^2+(5-1)^2}=\sqrt{4+16+16}=6

CD=\sqrt{(2-4)^2+(-3+7)^2+(4-8)^2}=\sqrt{4+16+16}=6

BC=\sqrt{(4-1)^2+(-7+2)^2+(8-5)^2}=\sqrt{9+25+9}=\sqrt{43}

DA=\sqrt{(-1-2)^2+(2+3)^2+(1-4)^2}=\sqrt{9+25+9}=\sqrt{43}

Since AB=CD and BC=DA, opposite sides are equal — as an additional check, the diagonals AC and BD also bisect each other, since their mid-points both equal \left(\dfrac{3}{2},-\dfrac{5}{2},\dfrac{9}{2}\right).

Let P(x,y,z) be any point equidistant from A(1,2,3) and B(3,2,-1), so PA=PB, i.e., PA^2=PB^2:

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

The (y-2)^2 terms cancel. Expanding the rest:

(x^2-2x+1)+(z^2-6z+9)=(x^2-6x+9)+(z^2+2z+1)

-2x+1-6z+9=-6x+9+2z+1

-2x-6z+10=-6x+2z+10

4x-8z=0

Let P(x,y,z) be any point with PA+PB=10, where A(4,0,0) and B(-4,0,0):

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

Rearranging and squaring:

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

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

The y^2+z^2 terms cancel. Expanding the x-terms:

(x^2-8x+16)=100-20\sqrt{(x+4)^2+y^2+z^2}+(x^2+8x+16)

-8x=100-20\sqrt{(x+4)^2+y^2+z^2}+8x

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

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

Squaring again:

25\left[(x+4)^2+y^2+z^2\right]=(25+4x)^2

25(x^2+8x+16)+25y^2+25z^2=625+200x+16x^2

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

9x^2+25y^2+25z^2=225

Dividing throughout by 225:

\dfrac{x^2}{25}+\dfrac{y^2}{9}+\dfrac{z^2}{9}=1