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

  • Last Updated : 16 May, 2021

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.

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