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

Question 1: Three vertices of a parallelogram ABCD are A(3, – 1, 2), B (1, 2, – 4) and C (– 1, 1, 2). Find the coordinates of the fourth vertex.

Answer:

Given

The three vertices of the parallelogram ABCD are

A(3, – 1, 2), B (1, 2, – 4) and C (– 1, 1, 2).

Let the fourth vertex be (a, b, c )

Now, as we know, the concept that the diagonals of the parallelogram bisect each other,

Hence, here in parallelogram ABCD, the midpoint of the line segment AC will be equal to the midpoint of the line segment BD.

So,

$\left ( \frac{3-1}{2},\frac{-1+1}{2},\frac{2+2}{2} \right )=\left ( \frac{1+a}{2},\frac{2+b}{2},\frac{-4+c}{2} \right )$

On comparing both points, we get

$a=1,b=-2\:and\:c=8$

Hence the fourth vertex of the is (1,-2,8)

Question 2: Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0,4, 0) and (6, 0, 0).

Answer:

Given,

Three vertices of the triangle are A (0, 0, 6), B (0,4, 0), and C(6, 0, 0).

Now,

Let D be the midpoint of AB, E be the midpoint of BC, and F be the midpoint of AC.

Vertex of the D =

$\left ( \frac{0+0}{2},\frac{0+4}{2},\frac{6+0}{2} \right )=(0,2,3)$

Vertices of E =

$\left ( \frac{0+6}{2},\frac{4+0}{2},\frac{0+0}{2} \right )=(3,2,0)$

Vertices of the F =

$\left ( \frac{0+6}{2},\frac{0+0}{2},\frac{6+0}{2} \right )=(3,0,3)$

Now, the Medians of the triangle are CD, AE, and BF

So the lengths of the medians are

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

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

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

Hence, the median lengths are $7,7 \:and\:\sqrt{34}$.

Question 3: If the origin is the centroid of the triangle PQR with vertices P (2a, 2, 6), Q (– 4, 3b, –10) and R(8, 14, 2c), then find the values of a, b and c.

Answer:

Given,

Triangle PQR with vertices P (2a, 2, 6), Q (– 4, 3b, –10) and R(8, 14, 2c),

Now, as we know,

The centroid of a triangle is given by

$\left (\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3} ,\frac{z_1 +z_2+z_3}{3}\right )$

Where coordinates of vertices of the triangle are $(x_1,x_2,x_3),(y_1,y_2,y_3)\:and\:(z_1,z_2,z_3)$

Since the Centroid of the triangle, PQR, is the origin (0,0,0),

$\left (\frac{2a-4+8}{3},\frac{2+3b+14}{3},\frac{6-10+2c}{3} \right )=(0,0,0)$

On equating both coordinates, we get

$a=-2,b=\frac{-16}{3}\:and\:c=2$

Question 4: If A and B are the points (3, 4, 5) and (–1, 3, –7), respectively, find the equation of the set of points P such that $PA ^2 + PB ^ 2 = k^ 2$ , where k is a constant.

Answer:

Given Points,

A (3, 4, 5) and B (–1, 3, –7),

Let the coordinates of point P be (x,y,z)

Now,

Given condition :

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

$PB^2=x^2+2x+1+y^2-6y+9+z^2+14z+49$

$PB^2=x^2+y^2+z^2+2x-6y+14z+59$

And

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

$PA^2=x^2+y^2+z^2-6x-8y-10z+50$

Now, Given the Condition

$PA ^2 + PB ^ 2 = k^ 2$

$x^2+y^2+z^2+2x-6y+14z+59+x^2+y^2+z^2-6x-8y-10z+50=k^2$

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

Hence Equation of the set of points P is

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