Question

answer the question ​

Answer 1

Question

Show that the points A(5,4,2) ,B(6,2,-1) and C(8,-2,-7) are collinear.

Solution

To show that all three points are collinear we have prove the distance sum of any sides is equal to the third one.

E.g,=>AB+BC=CA

$x_{1}=5,x_{2}=6 \&\: x_{3}=8$

$y_{1}=4,y_{2}=2\&\: y_{3}=-2$

$z_{1}=2,z_{2}=-1 \&\: z_{3}=-7$

$\sf AB= \sqrt{ {(x_{2} - x_{1})}^{2} + {(y_{2 }- y_{1})}^{2} + {(z_{2} - z_{1})}^{2} }$

$\longmapsto \sqrt{ {(6 - 5)}^{2} + {(2 - 4)}^{2} + {( - 1 - 2)}^{2} }$

$\longmapsto \sqrt{ {(1)}^{2} + {( - 2)}^{2} + {( - 3)}^{2} }$

$\longmapsto \sqrt{1 + 4 + 9} = \sqrt{14}$

$\sf \bold{AB= \sqrt{14}}$

$\sf BC= \sqrt{ {(x_{3} - x_{2})}^{2} + {(y_{3 }- y_{2})}^{2} + {(z_{3} - z_{2})}^{2} }$

$\longmapsto \sqrt{ {(8 - 6)}^{2} + {( - 2 - 2)}^{2} + {( - 7 + 1)}^{2} }$

$\longmapsto \sqrt{ {(2)}^{2} + {( - 4)}^{2} + {( - 6)}^{2} }$

$\longmapsto \sqrt{4 + 16 + 36} = \sqrt{56} = 2 \sqrt{14}$

$\sf \bold{BC = 2 \sqrt{14}}$

$\sf CA= \sqrt{ {(x_{3} - x_{1})}^{2} + {(y_{3 }- y_{1})}^{2} + {(z_{3} - z_{1})}^{2} }$

$\longmapsto \sqrt{ {(8 - 5)}^{2} + {( - 2 - 4)}^{2} + {( - 7 - 2)}^{2} }$

$\longmapsto \sqrt{ {(3)}^{2} + {( - 6)}^{2} + {( - 9)}^{2} }$

$\longmapsto \sqrt{9 + 36 + 81} = \sqrt{126} = 3 \sqrt{14}$

$\sf \bold{CA = 3 \sqrt{14}}$

$\sf\bold{AB + BC = CA}$

Hence ,the sum of two sides is equal to the third side.

$\therefore$The above points are collinear.