Class 11 RD Sharma Solutions - Chapter 28 Introduction to 3D Coordinate Geometry - Exercise 28.3

Question 1. The vertices of the triangle are A(5, 4, 6), B(1, -1, 3), and C(4, 3, 2). The internal bisector of angle A meets BC at D. Find the coordinates of D and the Length AD. 

Solution:

We know that the angle bisector divides opposite side in ratio of other two sides 

⇒D divides BC in ratio of AB:AC

A(5, 4, 6), B(1, -1, 3) and C(4, 3, 2)

AB=\sqrt{16+25+9}=\sqrt{50}=5\sqrt{2}\\ AC=\sqrt{1+1+16}=\sqrt{18}=3\sqrt{2}

AB:AC=5:3=m:n

D(x,y)=(\frac{mx_2+mx_1}{m+n},\frac{my_2+my_1}{m+n},\frac{mz_2+zn_1}{m+n})

Substitute values for m:n=5:3

(x₁,y₁,z₁)=(1,-1,3)

(x₂,y₂,z₂)=(4,3,2)

D=(\frac{22}{8},\frac{3}{2},\frac{19}{8})

Question 2. A point C with z-coordinate 8 lies on the line segment joining the points A(2, -3, 4) and B(8, 0, 10). Find its coordinates.

Solution:

Z-coordinate 8

A(2, -3, 4) and B(8, 0, 10)

DR's of AB=(6,3,6)

DR's of BC=(x-8,y-0,8-10)

Given A,B,C lie on same line 

So values of DR's should be proportional

\frac{x-8}{6}=\frac{y}{3}=\frac{8-10}{6}   

So x=6, y=-1

point is (6,-1,8)

Question 3. Show that three points A(2, 3, 4), B(-1, 2, -3) and C(-4, 1, -10) are collinear and find the ratio in which C divides AB.

Solution:

If points are collinear then all points lie on same line

and DR's should be proportional 

A(2, 3, 4), B(-1, 2, -3) and C(-4, 1, -10)

DR's of AB=(3,1,7)

DR's of BC=(3,1,7)

So A,B,C are collinear 

Length of AC=\sqrt{36+4+196}=\sqrt{236}

Length of AB=\sqrt{9+1+49}=\sqrt{59}

Ratio is AC:AB=2:1

So C divides AB in ratio 2:1 externally

Question 4. Find the ratio in which the line joining (2, 4, 5) and (3, 5, 4) is divided by the yz-plane.

Solution:

yz plane means x=0

Given (2,4,5) and (3,5,4)

Assume ratio to be m:n

Lets equate x term

0=\frac{3m+2n}{m+n}

3m=-2n

m:n=-2:3

Which means yz plane divides the line in 2:3 ratio externally.

Question 5. Find the ratio in which the line segment joining the points (2, -1, 3) and (-1, 2, 1) is divided by the plane x+ y + z = 5. 

Solution:

(2, -1, 3) and (-1, 2, 1)

x+y+z=5

Assume plane divides line in ratio λ:1

so point P which is dividing line in λ:1 ratio is

P=(\frac{-λ+2}{λ+1},\frac{2λ-1}{λ+1},\frac{λ+3}{λ+1})

P lies on plane  x+y+z=5

-λ+2+2λ-1+λ+3=5λ5

3λ=-1⇒λ=-1:3

Question 6. If the points A(3, 2, -4), B(9, 8, -10), and C(5, 4, -6) are collinear, find the ratio in which C divides AB. 

Solution:

A(3, 2, -4), B(9, 8, -10) and C(5, 4, -6)

AC=\sqrt{4+4+4}=2\sqrt{3}\\ AB=\sqrt{36+36+36}=6\sqrt{3}\\ BC=\sqrt{16+16+16}=4\sqrt{3}

AC:BC=1:2

Question 7. The mid-points of the sides of a triangle ABC are given by (-2, 3, 5), (4, -1, 7), and (6, 5, 3). Find the coordinates of A, B, and C. 

Solution:

Given midpoints (-2, 3, 5), (4, -1, 7) and (6, 5, 3)

Assume D is midpoint of AB, E is midpoint  of BC

F is midpoint of CA

A(x₁,y₁,z₁) B(x₂,y₂,z₂) C(x₃,y₃,z₃)

From midpoint formula, we get following equations 

x₁+x₂=-4;   x₂+x₃=8;   x₃+x₁=12

y₁+y₂=6;    y₂+y₃=-2;  y₃+y₁=10

z₁+z₂=10;  z₂+z₃=14;  z₃+z₁=6

Solving above set of equations we get

A=(0,9,1)

B=(-4,-3,9)

C=(12,1,5)

Question 8. A(1, 2, 3), B(0, 4, 1), C(-1, -1, -3) are the vertices of a triangle ABC. Find the point in which the bisector of the angle ∠BAC meets BC. 

Solution:

A(1, 2, 3), B(0, 4, 1), C(-1, -1, -3)

Angle bisector at A divides  BC in ratio of AB:AC

AB=\sqrt{1+4+4}=3 AC=\sqrt{4+9+36}=7

Assume D divides  BC

so D=(\frac{-3}{10},\frac{25}{10},\frac{-2}{10})

Question 9. Find the ratio in which the sphere x2+y2 +z2 = 504 divides the line joining the points (12, -4, 8) and (27, -9, 18). 

Solution:

(12, -4, 8) and (27, -9, 18)

Assume point P is dividing line λ:1 ratio, we get

P=(\frac{27λ+12}{λ+1},\frac{-9λ-4}{λ+1},\frac{8λ+8}{λ+1})

P lies on sphere, so substitute in sphere equation

x²+y²+z²=504

9(λ+4)²+(9λ+4)²+4(9λ+4)²=504(λ+1)²

729λ²+81λ²+324λ²+648λ+72λ+288λ+144+16+64=504λ²+1008λ+504

(1134-504)λ²+(1008-1008)λ+224-504=0

630λ2=280

λ2=4/9

λ=2:3

Question 10. Show that the plane ax + by + cz + d = 0 divides the line joining the points (x₁,y₁,z₁) and (x₂,y₂,z₂) in the ratio \frac{ax_1+by_1+cz+d}{-ax_2+by_2+cz_2+d}

Solution:

Assume ratio is λ:1

Plane is ax+by+cz+d=0

points (x₁,y₁,z₁) and  (x₂,y₂,z₂)

Assume point of intersection of line and plane is D

D=(\frac{λx_2+x_1}{λ+1},\frac{λy_2+y_1}{λ+1},\frac{λz_2+z_1}{λ+1})

As D lies on plane, substitute D in plane equation, we get

λ(ax₂+by₂+cz₂+d)+ax₁+by₁+cz₁+d=0

 ⇒λ=-\frac{ax_1+by_1+cz_1+d}{ax_2+by_2+cz_2+d}

Question 11. Find the centroid of a triangle, mid-points of whose sides are (1, 2, -3), (3, 0, 1), and (-1, 1, -4). 

Solution:

(1, 2, -3), (3, 0, 1) and (-1, 1, -4)

Centroid of triangle is given by

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

We know that 

x₁+x₂=2

x₂+x₃=6

x₁+x₃=-2

Adding all gives ⇒2(x₁+x₂+x₃)=6

so x₁+x₂+x₃=3

Similarly, y₁+y₂+y₃=3,  z₁+z₂+z₃=-6

centroid =(1,1,-2)

Question 12. The centroid of a triangle ABC is at the point (1, 1, 1). If the coordinate of A and B are (3, -5, 7) and (-1, 7, -6) respectively, find the coordinates of the point C. 

Solution:

Given centroid (1,1,1)

A(3,-5,7) and B(-1,7,-6)

Equating terms, we get

1=\frac{3-1+x_3}{3}\\ 1=\frac{-5+7+y_3}{3}\\ 1=\frac{7-6+z_3}{3}

(x₃,y₃,z₃)=(1,1,2)

Question 13. Find the coordinates of the points which trisect the line segment joining the points P(4, 2, -6) and Q(10, -16, 6). 

Solution:

Trisection points are those which divide line in ratio 1:2 or 2:1

P(4, 2, -6) and Q(10, -16, 6)

Consider 1:2 case, we get

(\frac{10+8}{3},\frac{-16+4}{3},\frac{6-12}{3})=(6,-4,-2)

consider 2:! case, we get

(\frac{20+4}{3},\frac{-32+2}{3},\frac{12-6}{3})=(8,-10,2)

(6,-4,-2) and (8,-10,2) are trisection  points

Question 14. Using section formula, show that the points A(2, -3, 4), B(-1, 2, 1) and C(0, 1/3, 2) are collinear. 

Solution:

A(2, -3, 4), B(-1, 2, 1) and C(0, 1/3, 2)

DR's of BC are (-1,\frac{5}{3},-1)

DR's of AC are(2,\frac{-10}{3},2)

Its clear that all DR's are proportional

Question 15. Given that P(3, 2, -4), Q(5, 4, -6), and R(9, 8, -10) are collinear. Find the ratio in which Q divides PR. 

Solution:

 P(3, 2, -4), Q(5, 4, -6) and R(9, 8, -10)

PQ=\sqrt{4+4+4}=2\sqrt{3}\\ QR=\sqrt{16+16+16}=4\sqrt{3}

PQ:QR=1:2

Question 16. Find the ratio in which the line segment joining the points (4, 8, 10) and (6, 10, -8) is divided by the yz-plane. 

Solution:

(4, 8, 10) and (6, 10, -8) is divided by the yz-plane

Equation of yz-plane is x=0

Assume ratio is m:n

Equating x-term, we get

0=\frac{6m+4n}{m+n}  

m:n=-2:3

So, XY plane divides the line segment in ratio 2:3 externally.

Summary

Exercise 28.3 deals with the distance formula and section formula in 3D coordinate geometry. Students will learn to find the distance between two points, coordinates of the centroid of a triangle, and coordinates of a point which divides a line segment in a given ratio. They will also learn to apply the distance formula and section formula to solve problems in 3D geometry. This exercise helps students understand the concepts of distance and division of line segments in 3D space, which is crucial for understanding advanced topics in calculus, physics, and engineering.