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]$\left [ \frac{m.a_{2}\;+\;n.a_{1}}{m\;+\;n},\frac{m.b_{2}\;+\;n.b_{1}}{m\;+\;n},\frac{m.c_{2}\;+\;n.c_{1}}{m\;+\;n} \right ]$

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]$\left [ \frac{1\times 1\;+\;4\times -2}{1\;+\;4},\frac{1\times 3\;+\;4\times 1}{1\;+\;4},\frac{1\times 6\;+\;4\times 0}{1\;+\;4} \right ]\\$

(x, y, z) = [75,75,65]$\left [ \frac{-7}{5},\frac{7}{5},\frac{6}{5} \right ]$

Therefore, the coordinates of point P are: [75,75,65]$\left [ \frac{-7}{5},\frac{7}{5},\frac{6}{5} \right ]\\$

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]$\left [ \frac{m.a_{2}\;-\;n.a_{1}}{m\;-\;n},\frac{m.b_{2}\;-\;n.b_{1}}{m\;-\;n},\frac{m.c_{2}\;-\;n.c_{1}}{m\;-\;n} \right]$

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]$\left [ \frac{1\times 1\;-\;4\times -2}{1\;-\;4},\frac{1\times 3\;-\;4\times 1}{1\;-\;4},\frac{1\times 6\;-\;4\times 0}{1\;-\;4} \right ]\\$

(x, y, z) = [93,13,63]$\left [ \frac{9}{-3},\frac{-1}{-3},\frac{6}{-3} \right ]$

Therefore, the coordinates of point Q are: [3,13,2]$\left [ -3,\frac{1}{3},-2 \right ]$

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]$\Rightarrow \left [ \frac{(k\times 9)\;+\;(1\times 3)}{k\;+\;1},\frac{(k\times 8)\;+\;(1\times 2)}{k+1},\frac{(k\times -10)\;+\;(1\times -4)}{k\;+\;1} \right ]\\$

[9k+3k+1,8k+3k+1,10k4k+1]$\Rightarrow \left [ \frac{9k\;+\;3}{k\;+\;1},\frac{8k\;+\;3}{k\;+\;1},\frac{-10k\;-\;4}{k\;+\;1} \right ]$. . . . . . . . . . . . . . (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$\Rightarrow \frac{9k\;+\;3}{k\;+\;1}=5\\$

$\Rightarrow$ 9k + 3 = 5k + 5

Therefore, k = 12$\frac{1}{2}$

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]$\Rightarrow \left [ \frac{(k\times -2)\;+\;(1\times 3)}{k\;+\;1},\frac{(k\times 4)\;+\;(1\times -5)}{k\;+\;1},\frac{(k\times 7)\;+\;(1\times 8)}{k\;+\;1} \right ]\\$

$\\\Rightarrow$   [2k+3k+1,4k5k+1,7k+8k+1]$\left [ \frac{-2k\;+\;3}{k\;+\;1},\frac{4k\;-\;5}{k\;+\;1},\frac{7k\;+\;8}{k\;+\;1} \right ]$

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

4k5k+1=0$\Rightarrow \frac{4k\;-\;5}{k\;+\;1}=0\\$

Therefore, k=54$k = \frac{5}{4}$

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$\frac{1}{3}$, 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]$\Rightarrow \left [ \frac{(k\times 0)\;+\;(1\times -1)}{k\;+\;1},\frac{(k\times \frac{1}{3})\;+\;(1\times 2)}{k\;+\;1},\frac{(k\times 2)\;+\;(1\times 1)}{k\;+\;1} \right ]\\$ [1k+1,k3+2k+1,2k+1k+1]$\Rightarrow \left [ \frac{-1}{k\;+\;1},\frac{\frac{k}{3}\;+\;2}{k\;+\;1},\frac{2k\;+\;1}{k\;+\;1} \right ]\\$

Therefore, the coordinates of point A are: [1k+1,k+63k+3,2k+1k+1]$\left [ \frac{-1}{k\;+\;1},\frac{k\;+\;6}{3k\;+\;3},\frac{2k\;+\;1}{k\;+\;1} \right ]$

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

1k+1=2$\Rightarrow \frac{-1}{k+1}= 2\\$

1=2k+2$\Rightarrow -1= 2k+2\\$

Therefore, k = 32$\frac{-3}{2}\\$

Now, for k = 32$\frac{-3}{2}$, the coordinates of point A are:

$\\\Rightarrow$    [1(32)+1,(32)+63×(32)+3,2×(32)+1(32)+1]$\left [ \frac{-1}{\left ( \frac{-3}{2} \right )\;+\;1}\;,\;\frac{\left ( \frac{-3}{2} \right )\;+\;6}{3\times \left ( \frac{-3}{2} \right )\;+\;3}\;,\;\frac{2\times \left ( \frac{-3}{2} \right )\;+\;1}{\left ( \frac{-3}{2} \right )\;+\;1} \right ]\\$

$\\\Rightarrow$   [1×23+2,3+129+6,6+23+2]$\left [ \frac{-1\times 2}{-3\;+\;2}\;,\;\frac{-3\;+\;12}{-9\;+\;6}\;,\;\frac{-6\;+\;2}{-3\;+\;2} \right ]\\$

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

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

$\Rightarrow$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]$\Rightarrow \left [ \frac{(1\times 9)\;+\;(2\times 5)}{1\;+\;2},\frac{(1\times 15)\;+\;(2\times 3)}{1\;+\;2},\frac{(1\times 7)\;+\;(2\times -6)}{1\;+\;2} \right ]\\$ [193,213,53]$\Rightarrow \left [ \frac{19}{3},\frac{21}{3},\frac{-5}{3} \right ]\\$

Therefore, the coordinates of point P are:[193,7,53]$\left [ \frac{19}{3},7,\frac{-5}{3} \right ]$

For Point Q:

[(2×9)+(1×5)2+1,(2×15)+(1×3)2+1,(2×7)+(1×6)2+1]$\Rightarrow \left [ \frac{(2\times 9)\;+\;(1\times 5)}{2\;+\;1},\frac{(2\times 15)\;+\;(1\times 3)}{2\;+\;1},\frac{(2\times 7)\;+\;(1\times -6)}{2\;+\;1} \right ]\\$ [233,333,83]$\Rightarrow \left [ \frac{23}{3},\frac{33}{3},\frac{8}{3} \right ]\\$

Therefore, the coordinates of point Q are: [233,11,83]$\left [ \frac{23}{3},11,\frac{8}{3} \right ]$