Class 11 Maths NCERT Solutions Chapter 11 Miscellaneous Exercise

In a parallelogram ABCD, the diagonals AC and BD bisect each other, so they share the same mid-point. Mid-point of AC, using A(3,-1,2) and C(-1,1,2):

M=\left(\dfrac{3+(-1)}{2},\dfrac{-1+1}{2},\dfrac{2+2}{2}\right)=(1,0,2)

Let the fourth vertex be D(x,y,z). Since M is also the mid-point of BD, using B(1,2,-4):

\left(\dfrac{1+x}{2},\dfrac{2+y}{2},\dfrac{-4+z}{2}\right)=(1,0,2)

Equating each coordinate:

\dfrac{1+x}{2}=1\;\Rightarrow\;x=1,\qquad\dfrac{2+y}{2}=0\;\Rightarrow\;y=-2,\qquad\dfrac{-4+z}{2}=2\;\Rightarrow\;z=8

A median joins a vertex to the mid-point of the opposite side.

Median from A

Mid-point of BC: M_A=\left(\dfrac{0+6}{2},\dfrac{4+0}{2},\dfrac{0+0}{2}\right)=(3,2,0).

AM_A=\sqrt{(3-0)^2+(2-0)^2+(0-6)^2}=\sqrt{9+4+36}=\sqrt{49}=7

Median from B

Mid-point of AC: M_B=\left(\dfrac{0+6}{2},\dfrac{0+0}{2},\dfrac{6+0}{2}\right)=(3,0,3).

BM_B=\sqrt{(3-0)^2+(0-4)^2+(3-0)^2}=\sqrt{9+16+9}=\sqrt{34}

Median from C

Mid-point of AB: M_C=\left(\dfrac{0+0}{2},\dfrac{0+4}{2},\dfrac{6+0}{2}\right)=(0,2,3).

CM_C=\sqrt{(0-6)^2+(2-0)^2+(3-0)^2}=\sqrt{36+4+9}=\sqrt{49}=7

The centroid of a triangle with vertices (x_1,y_1,z_1), (x_2,y_2,z_2), (x_3,y_3,z_3) is:

\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)

Since the centroid is the origin (0,0,0), each coordinate sum must be zero.

x-coordinates:

\dfrac{2a+(-4)+8}{3}=0\;\Rightarrow\;2a+4=0\;\Rightarrow\;a=-2

y-coordinates:

\dfrac{2+3b+14}{3}=0\;\Rightarrow\;3b+16=0\;\Rightarrow\;b=-\dfrac{16}{3}

z-coordinates:

\dfrac{6+(-10)+2c}{3}=0\;\Rightarrow\;2c-4=0\;\Rightarrow\;c=2

Let P(x,y,z) be any point in the set. Then:

PA^2=(x-3)^2+(y-4)^2+(z-5)^2

PB^2=(x+1)^2+(y-3)^2+(z+7)^2

By the given condition PA^2+PB^2=k^2:

(x-3)^2+(y-4)^2+(z-5)^2+(x+1)^2+(y-3)^2+(z+7)^2=k^2

Expanding each term:

(x^2-6x+9)+(x^2+2x+1)+(y^2-8y+16)+(y^2-6y+9)+(z^2-10z+25)+(z^2+14z+49)=k^2

Collecting like terms:

2x^2+2y^2+2z^2-4x-14y+4z+109=k^2