Class 11: Three Dimensional Coordinate Geometry – Exercise 28.3 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: The vertices of the triangle are $\displaystyle A (5, 4, 5), B(1, -1,3) \text{ and } C(4, 3, 2).$ The internal bisector of angle $\displaystyle A$ meets $\displaystyle BC$ at $\displaystyle D.$ Find the coordinates of $\displaystyle D$ and the length $\displaystyle AD.$

Answer:

$\displaystyle AB = \sqrt{(5-1)^2 + (4+1)^2+(6-3)^2} = \sqrt{ 16+25+9 } = \sqrt{50} = 5\sqrt{2}$

$\displaystyle AC = \sqrt{(5-4)^2 + (4-3)^2+(6-2)^2} = \sqrt{ 1+1+16 } = \sqrt{18} = 3\sqrt{2}$

$\displaystyle AD$ is the internal bisector of $\displaystyle \angle BAC$

$\displaystyle \therefore \frac{BD}{DC} = \frac{AB}{AC} = \frac{5\sqrt{2}}{3\sqrt{2}} = \frac{5}{3}$

Therefore $\displaystyle D$ divides $\displaystyle BC$ in the ratio of $\displaystyle 5:3$

$\displaystyle \therefore D = \Bigg( \frac{5 \times 4 + 3 \times 1}{5+3}, \frac{5 \times 3 + 3 \times (-1)}{5+3}, \frac{5 \times 2 + 3 \times 3}{5+3} \Bigg)$

$\displaystyle \Rightarrow D = \Bigg( \frac{23}{8}, \frac{12}{8}, \frac{19}{8} \Bigg)$

$\displaystyle \Rightarrow D = \Bigg( \frac{23}{8}, \frac{3}{2}, \frac{19}{8} \Bigg)$

$\displaystyle \therefore AD = \sqrt{ \Bigg(5 - \frac{23}{8} \Bigg)^2 + \Bigg(5 - \frac{12}{8} \Bigg)^2 + \Bigg(6 - \frac{19}{8} \Bigg)^2 }$

$\displaystyle = \sqrt{ \frac{17^2+20^2 + 29^2}{8^2} }$

$\displaystyle = \sqrt{ \frac{289+400+841}{8^2} } = \frac{\sqrt{1530}}{8}$

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

Answer:

Let $\displaystyle C$ divides $\displaystyle AB$ in the ratio of $\displaystyle m:1$

$\displaystyle \text{Hence the coordinates of C are } \Bigg( \frac{8m+2}{m+1}, \frac{-3}{m+1}, \frac{10m+4}{m+1} \Bigg)$

The z coordinate of $\displaystyle C$ is $\displaystyle 8.$

$\displaystyle \therefore \frac{10m+4}{m+1} = 8$

$\displaystyle \Rightarrow 10m + 4 = 8m+8$

$\displaystyle \Rightarrow 2m = 4$

$\displaystyle \Rightarrow m = 2$

$\displaystyle \text{Hence the coordinates of C are } \Bigg( \frac{8\times 2+2}{2+1}, \frac{-3}{2+1}, 8 \Bigg) = ( 6, -1, 8)$

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

Answer:

Let $\displaystyle C$ divides $\displaystyle AB$ in the ratio of $\displaystyle m:1$

$\displaystyle \text{Hence the coordinates of C are } \Bigg( \frac{-m+2}{m+1}, \frac{2m+1}{m+1}, \frac{-3m+4}{m+1} \Bigg)$

The coordinates of $\displaystyle C$ is $\displaystyle (-4,1, - 10).$

$\displaystyle \therefore \frac{-m+2}{m+1} = -4, \frac{2m+3}{m+1}= 1, \frac{-3m+4}{m+1} =$

$\displaystyle \Rightarrow -m+2 = - 4 m-4, 2m+3 = m + 1 , -3m + 4 = - 10m-10$

$\displaystyle \Rightarrow -3m = - 6, m$

$\displaystyle \Rightarrow m = - 2, m = -2, m=-2$

Therefore $\displaystyle m = - 2$

Hence $\displaystyle C$ divides $\displaystyle AB$ in the ratio $\displaystyle 2:1$ ( externally)

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Question 4: Find the ratio in which the line joining $\displaystyle (2,4,5) \text{ and } (3,5,4)$ is divided by the yz-plane.

Answer:

Given $\displaystyle A (2, 4, 5)$ and $\displaystyle B (3, 5, 4)$

Let yz-plane divide $\displaystyle AB$ at point $\displaystyle P$ in the ratio $\displaystyle (\lambda : 1)$

Therefore we have

$\displaystyle P = \Bigg( \frac{3\lambda+1}{\lambda+1} , \frac{5\lambda+4}{\lambda+1}, \frac{4\lambda+5}{\lambda+1} \Bigg)$

Since $\displaystyle P$ lies on yz-plane, the x-coordinate of $\displaystyle P$ will be zero.

$\displaystyle \therefore \frac{3\lambda+1}{\lambda+1} = 0$

$\displaystyle \Rightarrow 3\lambda+2 = 0$

$\displaystyle \therefore \lambda = \frac{-2}{3}$

Hence the yz-plane divides $\displaystyle AB$ externally in the ratio of $\displaystyle 2:3$

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

Answer:

Suppose the plane $\displaystyle x+y+z=5$ divides the line joining $\displaystyle A(2, -1. 3)$ and $\displaystyle B(-1, 2, 1)$ at a point $\displaystyle P$ in the ratio $\displaystyle (\lambda : 1)$

Therefore the coordinates of $\displaystyle P$ would be:

$\displaystyle P = \Bigg( \frac{-\lambda+2}{\lambda+1} , \frac{2\lambda-1}{\lambda+1}, \frac{\lambda+3}{\lambda+1} \Bigg)$

Since $\displaystyle P$ lies on the plane $\displaystyle x+y+z=5$ , the coordinates of $\displaystyle P$ must satisfy the equation of the plane.

$\displaystyle \frac{-\lambda+2}{\lambda+1} + \frac{2\lambda-1}{\lambda+1}+ \frac{\lambda+3}{\lambda+1} = 5$

$\displaystyle \Rightarrow -\lambda+2 + 2\lambda-1 +\lambda+3 = 5\lambda+5$

$\displaystyle \Rightarrow 2\lambda + 4 = 5 \lambda + 5$

$\displaystyle \Rightarrow -3\lambda = 1$

$\displaystyle \therefore \lambda = \frac{-1}{3}$

Hence the required ratio is $\displaystyle 1:3$ ( externally)

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

Answer:

Given points $\displaystyle A (3, 2, -4), B(9,8,-10) \text{ and } C (5,4,-6)$

Let $\displaystyle C$ divide $\displaystyle AB$ at point $\displaystyle P$ in the ratio $\displaystyle (\lambda : 1)$

Therefore we have

$\displaystyle C = \Bigg( \frac{9\lambda+3}{\lambda+1} , \frac{8\lambda+2}{\lambda+1}, \frac{-10\lambda-4}{\lambda+1} \Bigg)$

Given coordinates of $\displaystyle C$ are $\displaystyle ( 5, 4, -6)$

$\displaystyle \therefore \frac{9\lambda+3}{\lambda+1} = 5, \frac{8\lambda+2}{\lambda+1} = 4, \frac{-10\lambda-4}{\lambda+1} = -6$

$\displaystyle \text{These equations give } \lambda = \frac{1}{2}, \lambda = \frac{1}{2}, \lambda = \frac{1}{2}$

Hence, $\displaystyle C$ divides $\displaystyle AB$ in the ratio $\displaystyle \frac{1}{2}$

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

Answer:

Let $\displaystyle D(x_1, y_1, z_1) , E(x_2, y_2, z_2) ,$ and $\displaystyle F(x_3, y_3, z_3)$ be the vertices of the triangle.

And let $\displaystyle A(- 2, 3, 5), B(4, -1,7) \text{ and } C(6,5,3)$ be the midpoint of the slides $\displaystyle EF, FA$ and $\displaystyle DE$ respectively.

Now, $\displaystyle A$ is the mid-point of $\displaystyle EF.$

$\displaystyle \therefore \frac{x_2+x_3}{2} = - 2, \frac{y_2+y_3}{2} = 3, \frac{z_2+z_3}{2} = 5$

$\displaystyle \Rightarrow x_2+x_3 = -4, y_2+y_3= 6, z_2+z_3 = 10 \ \ \ \text{ ... ... ... ... ... i)}$

Also, $\displaystyle B$ is the mid-point of $\displaystyle FD.$

$\displaystyle \therefore \frac{x_1+x_3}{2} = 4, \frac{y_1+y_3}{2} = -1, \frac{z_1+z_3}{2} = 7$

$\displaystyle \Rightarrow x_1+x_3 = 8, y_1+y_3= -2, z_1+z_3 = 14 \ \ \ \text{ ... ... ... ... ... ii)}$

And, $\displaystyle C$ is the mid-point of $\displaystyle DE.$

$\displaystyle \therefore \frac{x_1+x_2}{2} = 6, \frac{y_1+y_2}{2} = 5, \frac{z_1+z_2}{2} = 3$

$\displaystyle \Rightarrow x_1+x_2 = 12, y_1+y_2= 10, z_1+z_2 = 6 \ \ \ \text{ ... ... ... ... ... iii)}$

Adding first three equations in i), ii) and iii) we get

$\displaystyle 2 ( x_1 + x_2+x_3) = -4 + 8 + 12 \Rightarrow x_1 + x_2+x_3 = 8$

Solving the first three equations, we get $\displaystyle x_1 = 12, x_2 = 0 , x_3 = -4$

Adding the next three equation in i), ii) and iii) we get

$\displaystyle 2(y_1+y_2+y_3) = 6-2+10 \Rightarrow y_1+y_2+y_3 = 7$

Solving the next three equations in i) , ii) and iii) we get

$\displaystyle y_1 = 1, y_2 = 9 , y_3 = -3$

Similarly, Adding the next three equation in i), ii) and iii) we get

$\displaystyle 2(z_1+z_2+z_3) = 10+14+6 \Rightarrow z_1+z_2+z_3 = 7$

Solving the next three equations in i) , ii) and iii) we get

$\displaystyle z_1 = 5, z_2 = 1 , z_3 = 9$

Therefore the vertices of the triangle are $\displaystyle (12, 1, 5), (0, 9, 1), (-4, -3, 9)$

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

Answer:

$\displaystyle AB = \sqrt{ (0-1)^2+(4-2)^2+(1-3)^2 } = \sqrt{ 1+4+4 } = \sqrt{9} = 3$

$\displaystyle AC = \sqrt{ (-1-1)^2+(-1-2)^2+(-3-3)^2 } = \sqrt{ 4+9+36 } = \sqrt{49} = 7$

$\displaystyle AD$ is the internal bisector of $\displaystyle \angle BAC$

$\displaystyle \therefore \frac{BD}{DC} = \frac{AB}{AC} = \frac{3}{7}$

Therefore $\displaystyle D$ divides $\displaystyle BC$ in the ratio of $\displaystyle 3:7$

$\displaystyle \therefore D = \Bigg( \frac{3(-1) + 3(0)}{3+7} , \frac{3(-1) + 7(4)}{3+7}, \frac{3(-3) + 7(1)}{3+7} \Bigg)$

$\displaystyle \Rightarrow D = \Bigg( \frac{-3}{10} , \frac{25}{10}, \frac{-2}{10} \Bigg)$

$\displaystyle \Rightarrow D = \Bigg( \frac{-3}{10} , \frac{5}{2}, \frac{-1}{5} \Bigg)$

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Question 9: Find the ratio in which the sphere $\displaystyle x^2+y^2+z^2=504$ divides the line joining the points $\displaystyle (12, - 4,8) \text{ and } (27, - 9,18).$

Answer:

Let the sphere $\displaystyle x^2+y^2+z^2=504$ meet the line joining the points $\displaystyle (12, - 4,8) \text{ and } (27, - 9,18)$ at $\displaystyle (x_1, y_1, z_1)$

Therefore we have $\displaystyle {x_1}^2+{y_1}^2+{z_1}^2=504 \ \ \ \text{ ... ... ... ... ... i) }$

Let the point $\displaystyle (x_1, y_1, z_1)$ divide the line joining $\displaystyle (12, - 4,8) \text{ and } (27, - 9,18)$ in the ratio $\displaystyle (\lambda : 1 )$

$\displaystyle \therefore x_1 = \frac{27 \lambda + 12}{\lambda + 1} , y_1 = \frac{-9 \lambda -4}{\lambda + 1} , z_1 = \frac{18 \lambda + 8}{\lambda + 1}$

Substituting these values in equation i) we get

$\displaystyle \Bigg( {\frac{27 \lambda + 12}{\lambda + 1}} \Bigg) ^2+\Bigg( {\frac{-9 \lambda -4}{\lambda + 1}} \Bigg)^2+\Bigg( {\frac{18 \lambda + 8}{\lambda + 1}} \Bigg)^2=504$

$\displaystyle \Rightarrow 9( 9\lambda + 4)^2 + ( 9\lambda + 4)^2 + 4( 9\lambda+4)^2 = 504(\lambda+1)^2$

$\displaystyle \Rightarrow 9\lambda + 4 = \pm 6 ( \lambda + 1)$

$\displaystyle \Rightarrow \lambda = \pm \frac{2}{3}$

Therefore the sphere divides the line joining $\displaystyle (12, - 4,8) \text{ and } (27, - 9,18)$ internally in the ratio of $\displaystyle 2:3$ and externally in the ratio of $\displaystyle -2:3$

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Question 10: Show that the plane $\displaystyle ax + by + cz + d = 0$ divides the line joining the points $\displaystyle (x_1, y_1, z_1) \text{ and } ( x_2, y_2, z_2)$ in the ratio $\displaystyle -\frac{ ax_1 + by_1 + cz_1 + d }{ ax_2 + by_2 + cz_3 + d }.$

Answer:

Let $\displaystyle A(x_1, y_1, z_1) \text{ and } B( x_2, y_2, z_2)$

Let the line joining $\displaystyle A$ and $\displaystyle B$ be divided by the plane $\displaystyle ax + by + cz + d = 0$ at point $\displaystyle P$ in the ratio $\displaystyle (\lambda : 1)$

$\displaystyle \therefore P = \Bigg( \frac{\lambda x_2 + x_1}{\lambda + 1}, \frac{\lambda y_2 + y_1}{\lambda + 1} , \frac{\lambda z_2 + z_1}{\lambda + 1} \Bigg)$

Since $\displaystyle P$ lies on the given plane $\displaystyle ax + by + cz + d = 0$ , it should satisfy it. Hence,

$\displaystyle a \Bigg(\frac{\lambda x_2 + x_1}{\lambda + 1} \Bigg) + b \Bigg(\frac{\lambda y_2 + y_1}{\lambda + 1} \Bigg) + c \Bigg(\frac{\lambda z_2 + z_1}{\lambda + 1} \Bigg) + d = 0$

$\displaystyle \Rightarrow a(\lambda x_2 + x_1) + b ( \lambda y_2 + y_1) + c(\lambda z_2 + z_1) + d(\lambda + 1) = 0$

$\displaystyle \Rightarrow \lambda (ax_2 + by_2 + cz_2 + d) + (ax_1 + by_1 + cz_1 + d) = 0$

$\displaystyle \Rightarrow \lambda = - \frac{ax_1 + by_1 + cz_1 + d}{ax_2 + by_2 + cz_2 + d}$

Therefore the plane $\displaystyle ax + by + cz + d = 0$ divides the line joining the points $\displaystyle (x_1, y_1, z_1) \text{ and } ( x_2, y_2, z_2) \text{ in the ratio } -\frac{ ax_1 + by_1 + cz_1 + d }{ ax_2 + by_2 + cz_3 + d }.$

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Question 11: Find the centroid of a triangle, mid-points of whose sides are $\displaystyle (1, 2, -3), (3, 0,1) \text{ and } (- 1, 1, - 4).$

Answer:

Let $\displaystyle A(x_1, y_1, z_1) , B(x_2, y_2, z_2) ,$ and $\displaystyle C(x_3, y_3, z_3)$ be the vertices of the given triangle.

And $\displaystyle D(1, 2, -3), E(3, 0,1) \text{ and } F(- 1, 1, - 4)$ be the mid points of the sides $\displaystyle BC, CA \text{ and } AB$ respectively.

$\displaystyle D$ is the midpoint of $\displaystyle BC$

$\displaystyle \therefore \frac{x_2+x_3}{2} = 1, \ \ \frac{y_2+y_3}{2} = 2, \ \ \frac{z_2+z_3}{2} = -3$

$\displaystyle \Rightarrow x_2+x_3 = 2, \ \ y_2+y_3 = 4, \ \ z_2+z_3 = -6 \ \ \ \text{ ... ... ... ... ... i)}$

$\displaystyle E$ is the midpoint of $\displaystyle CA$

$\displaystyle \therefore \frac{x_1+x_3}{2} = 3, \ \ \frac{y_1+y_3}{2} = 0, \ \ \frac{z_1+z_3}{2} = 1$

$\displaystyle \Rightarrow x_1+x_3 = 6, \ \ y_1+y_3 = 0, \ \ z_1+z_3 = 2 \ \ \ \text{ ... ... ... ... ... ii)}$

$\displaystyle F$ is the midpoint of $\displaystyle AB$

$\displaystyle \therefore \frac{x_1+x_2}{2} = -1, \ \ \frac{y_1+y_2}{2} = 1, \ \ \frac{z_1+z_2}{2} = -4$

$\displaystyle \Rightarrow x_1+x_2 = -2, \ \ y_1+y_2 = 2, \ \ z_1+z_2 = -8 \ \ \ \text{ ... ... ... ... ... iii)}$

Adding equation i) , ii) and iii) we get

$\displaystyle 2(x_1+x_2+x_3) = 6 \Rightarrow x_1+x_2+x_3 = 3$

$\displaystyle 2(y_1+y_2+y_3) = 6 \Rightarrow y_1+y_2+y_3 = 3$

$\displaystyle 2(z_1+z_2+z_3) = -12 \Rightarrow z_1+z_2+z_3 = -6$

Therefore the coordinate of the centroid is given by

$\displaystyle \Bigg( \frac{x_1+x_2+x_3}{3} , \frac{y_1+y_2+y_3}{3} , \frac{z_1+z_2+z_3}{3} \Bigg) = (1, 1, -2)$

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Question 12: The centroid of a triangle $\displaystyle ABC$ is at the point $\displaystyle (1, 1, 1).$ If the coordinates of $\displaystyle A \text{ and } B$ are $\displaystyle (3, -5,7) \text{ and } (-1,7, -6)$ respectively, find the coordinates of the point $\displaystyle C$ .

Answer:

Let $\displaystyle G$ be the centroid of the $\displaystyle \triangle ABC$

Given $\displaystyle G = ( 1, 1, 1)$

Given $\displaystyle A(3, -5,7) \text{ and } B(-1,7, -6)$

Let $\displaystyle C= ( x, y, z)$

$\displaystyle \text{Therefore } 1 = \frac{3-1+x}{3} \Rightarrow 2+x = 3 \Rightarrow x = 1$

$\displaystyle \text{Similarly, } 1 = \frac{-5+7+y}{3} \Rightarrow 2+y = 3 \Rightarrow y = 1$

$\displaystyle \text{and } 1 = \frac{7-6+z}{3} \Rightarrow 1+z = 3 \Rightarrow x = 2$

$\displaystyle \text{Hence, centroid } G = ( 1, 1, 2)$

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Question 13: Find the coordinates of the points which trisect the line segment joining the points $\displaystyle P (4,2, - 6) \text{ and } Q(-10, -16, 6).$

Answer:

Given $\displaystyle P (4,2, - 6) \text{ and } Q(-10, -16, 6)$

Let $\displaystyle A$ and $\displaystyle B$ be the points that trisect $\displaystyle PQ.$

Therefore $\displaystyle PA = AB = BQ$

Therefore $\displaystyle PA : AQ = 1:2$

Thus $\displaystyle A$ divides $\displaystyle PQ$ internally in the ratio $\displaystyle 1:2$

$\displaystyle \therefore A = \Bigg( \frac{1 \times 10 + 2 \times 4}{1+2}, \frac{1 \times (-16) + 2 \times 2}{1+2}, \frac{1 \times 6 + 2 \times (-6)}{1+2} \Bigg)$

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

$\displaystyle = \Bigg( \frac{18}{3}, \frac{-12}{3}, \frac{-6}{3} \Bigg)$

$\displaystyle = ( 6, -4, -2)$

Similarly,

Thus $\displaystyle B$ divides $\displaystyle AQ$ internally in the ratio $\displaystyle 1:1$

$\displaystyle \therefore B= \Bigg( \frac{6+10}{2}, \frac{-4-16}{2}, \frac{-2+6}{2} \Bigg)$

$\displaystyle = \Bigg( \frac{16}{2}, \frac{-20}{2}, \frac{4}{2} \Bigg)$

$\displaystyle = ( 8, -10, 2)$

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Question 14: Using section formula, show that the points $\displaystyle A (2,-3,4),B(-1,2,1) \text{ and } C(0,1/3,2)$ are collinear.

Answer:

$\displaystyle \text{Given } A (2,-3,4),B(-1,2,1) \text{ and } C(0,\frac{1}{3},2)$

Let $\displaystyle C$ divides $\displaystyle AB$ in the ratio of $\displaystyle \lambda :1$

$\displaystyle \text{Therefore the coordinates of C are } \Bigg( \frac{-\lambda+2}{\lambda+1}, \frac{2\lambda-3}{\lambda+1}, \frac{\lambda+4}{\lambda+1} \Bigg)$

$\displaystyle \text{Comparing with } C(0,\frac{1}{3}, 2) \text{ we get, }$

$\displaystyle \frac{-\lambda+2}{\lambda+1} = 0 \ \ \Rightarrow -\lambda + 2 = 0 \ \ \Rightarrow \lambda = 2$

$\displaystyle \frac{2\lambda-3}{\lambda+1} = \frac{1}{3} \ \ \Rightarrow 6\lambda -9 = \lambda + 1 \ \ \Rightarrow \lambda = 2$

$\displaystyle \frac{\lambda+4}{\lambda+1} = 2 \ \ \Rightarrow \lambda + 4 = 2\lambda +2 \ \ \Rightarrow \lambda = 2$

$\displaystyle \lambda$ in all the three cases is the same. Therefore we can say that the given points are collinear.

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

Answer:

Given $\displaystyle P (3, 2, - 4), Q (5, 4, -6) \text{ and } R (9, 8, - 10)$ are collinear.

Let $\displaystyle Q$ divides $\displaystyle PR$ in the ratio of $\displaystyle \lambda :1$

$\displaystyle \text{Therefore the coordinates of Q are } \Bigg( \frac{9\lambda+23}{\lambda+1}, \frac{8\lambda+2}{\lambda+1}, \frac{-10\lambda-4}{\lambda+1} \Bigg)$

$\displaystyle \text{Comparing with } Q (5, 4, -6) \text{ we get, }$

$\displaystyle \frac{9\lambda+3}{\lambda+1} =5 \ \ \Rightarrow 9\lambda + 3 = 5\lambda +5 \ \ \Rightarrow \lambda = \frac{1}{2}$

$\displaystyle \frac{8\lambda+2}{\lambda+1} = 4 \ \ \Rightarrow 8\lambda +2 = 4\lambda + 4 \ \ \Rightarrow \lambda = \frac{1}{2}$

$\displaystyle \frac{-10\lambda-4}{\lambda+1} = -6 \ \ \Rightarrow -10\lambda - 4 = -6\lambda -6 \ \ \Rightarrow \lambda = \frac{1}{2}$

Hence we can say, $\displaystyle Q$ divides $\displaystyle PR$ in the ratio of $\displaystyle 1:2$

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Question 16: Find the ratio in which the line segment joining the points $\displaystyle (4, 8, 10) \text{ and } (6, 10, - 8)$ is divided by the $\displaystyle yz-plane.$