# 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:AC=5:3=m:n

Substitute values for m:n=5:3

(x

_{1},y_{1},z_{1})=(1,-1,3)(x

_{2},y_{2},z_{2})=(4,3,2)

### 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

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

Length of AB

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

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 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: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(x

_{1},y_{1},z_{1}) B(x_{2},y_{2},z_{2}) C(x_{3},y_{3},z_{3})From midpoint formula, we get following equations

x

_{1}+x_{2}=-4; x_{2}+x_{3}=8; x_{3}+x_{1}=12y

_{1}+y_{2}=6; y_{2}+y_{3}=-2; y_{3}+y_{1}=10z

_{1}+z_{2}=10; z_{2}+z_{3}=14; z_{3}+z_{1}=6Solving 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

Assume D divides BC

so

### 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 lies on sphere, so substitute in sphere equation

x

^{2}+y^{2}+z^{2}=5049(λ+4)

^{2}+(9λ+4)^{2}+4(9λ+4)^{2}=504(λ+1)^{2}729λ

^{2}+81λ^{2}+324λ^{2}+648λ+72λ+288λ+144+16+64=504λ^{2}+1008λ+504(1134-504)λ

^{2}+(1008-1008)λ+224-504=0630λ2=280

λ2=4/9

λ=2:3

### Question 10. Show that the plane ax + by + cz + d = 0 divides the line joining the points (x_{1},y_{1},z_{1}) and (x_{2},y_{2},z_{2}) in the ratio

**Solution:**

Assume ratio is λ:1

Plane is ax+by+cz+d=0

points (x

_{1},y_{1},z_{1}) and (x_{2},y_{2},z_{2})Assume point of intersection of line and plane is D

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

λ(ax

_{2}+by_{2}+cz_{2}+d)+ax_{1}+by_{1}+cz_{1}+d=0⇒

### 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

We know that

x

_{1}+x_{2}=2x

_{2}+x_{3}=6x

_{1}+x_{3}=-2Adding all gives ⇒2(x

_{1}+x_{2}+x_{3})=6so x

_{1}+x_{2}+x_{3}=3Similarly, y

_{1}+y_{2}+y_{3}=3, z_{1}+z_{2}+z_{3}=-6centroid =(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

(x

_{3},y_{3},z_{3})=(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

consider 2:! case, we get

(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

DR’s of AC are

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: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

m:n=-2:3

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

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