Class 11 Maths Ncert Solutions Chapter 12 Ex 12.3 Introduction to Three Dimensional Geometry PDF

Class 11 Maths Ncert Solutions Ex 12.3

Class 11 Maths Ncert Solutions Chapter 12 Ex 12.3

Exercise 12.3

 Q.1: Find the coordinates of the point which divides the line segment joining the points (-2, 1, 0) and (1, 3, 6) internally in the ratio of 1:4 and externally in the ratio of 1:4.

Sol:

Since, the coordinates of point P, which divides the line segment joining two points A (a1, b1, c1) and B (a2, b2, c2) internally in the ratio m and n are:

[m.a2+n.a1m+n,m.b2+n.b1m+n,m.c2+n.c1m+n]

Therefore, the coordinates of point P, which divides the line segment joining two points A(-2, 1, 0) and B (1, 3, 6) internally in the ratio 1 and 4 are:

(x, y, z) = [1×1+4×21+4,1×3+4×11+4,1×6+4×01+4]

(x, y, z) = [75,75,65]

Therefore, the coordinates of point P are: [75,75,65]

Now, The coordinates of point Q, which divides the line segment joining two points A (a1, b1, c1) and B (a2, b2, c2) Externally in the ratio m and n are:

[m.a2n.a1mn,m.b2n.b1mn,m.c2n.c1mn]

Therefore, the coordinates of point Q, which divides the line segment joining two points A(-2, 1, 0) and B (1, 3, 6) externally in the ratio 1 and 4 are:

(x, y, z) = [1×14×214,1×34×114,1×64×014]

(x, y, z) = [93,13,63]

Therefore, the coordinates of point Q are: [3,13,2]

  

Q.2: Given that the points A (3, 2, -4) B (5, 4, -6) and C (9, 8, -10), are collinear. Find the ratio in which line AC is divided by B.

Sol.

Since, point A, point B and point C are collinear. Therefore, let us assume that point B divides the line segment joining point A and point C in the ratio k : 1.

Hence, by Section Formula, coordinates of point B are:

[(k×9)+(1×3)k+1,(k×8)+(1×2)k+1,(k×10)+(1×4)k+1]

[9k+3k+1,8k+3k+1,10k4k+1]. . . . . . . . . . . . . . (1)

And, the coordinates of point B are: (5, 4, -6) [Given]

Therefore, on comparing it with equation (1) we will get:

9k+3k+1=5

9k + 3 = 5k + 5

Therefore, k = 12

Hence, point B divides AB in the ratio of 1 : 2

 

Q.3: Find the ratio in which XZ – plane divides the line segment AB formed by joining the points (3, -5, 8) and (-2, 4, 7).

Sol.

Let the XZ plane divides the line segment PQ in the ratio k : 1

Hence, by Section Formula the coordinates of point of intersection are:

[(k×2)+(1×3)k+1,(k×4)+(1×5)k+1,(k×7)+(1×8)k+1]

   [2k+3k+1,4k5k+1,7k+8k+1]

Now, in XZ – plane, the y – coordinate of any point is zero.

4k5k+1=0

Therefore, k=54

Hence, the XZ plane divides the line segment formed by the joining of given points in the ratio of 2 : 3.

 

Q.4: By using the section formula; Show that the points P (-1, 2, 1), Q (0, 13, 2) and R (2, -3, 4) are collinear.

Sol.

Let, point A divides line PQ in the ratio k : 1

Hence, by Section Formula the coordinates of point A are:

[(k×0)+(1×1)k+1,(k×13)+(1×2)k+1,(k×2)+(1×1)k+1] [1k+1,k3+2k+1,2k+1k+1]

Therefore, the coordinates of point A are: [1k+1,k+63k+3,2k+1k+1]

Now, the value of ‘k’ for which point A coincides with point R:

1k+1=2

1=2k+2

Therefore, k = 32

Now, for k = 32, the coordinates of point A are:

    [1(32)+1,(32)+63×(32)+3,2×(32)+1(32)+1]

   [1×23+2,3+129+6,6+23+2]

Therefore, the coordinates of point A are:  (2, -3, 4)

Hence, the coordinates of point A coincides with the coordinates of point R.

R (2, -3, 4) is a point that divides PQ externally and is same as point A.

Hence, the points P, Q and R are colinear.

 

Q.5: The line segment joining points A (5, 3, -6) and B (9, 15, 7) is trisected by the points P and Q, Find the coordinates of points P and Q.

Sol. 

Point P divides the line segment AB in the ratio 1 : 2 and point Q divides the line segment AB in the ratio 2 : 1.

Therefore, by section formula the coordinates of point P and point Q are:

For Point P:

[(1×9)+(2×5)1+2,(1×15)+(2×3)1+2,(1×7)+(2×6)1+2] [193,213,53]

Therefore, the coordinates of point P are:[193,7,53]

For Point Q:

[(2×9)+(1×5)2+1,(2×15)+(1×3)2+1,(2×7)+(1×6)2+1] [233,333,83]

Therefore, the coordinates of point Q are: [233,11,83]

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