Question

Prove that the points $(4,5, - 5),(0, - 11,3)$ and $(2, - 3, - 1)$ are collinear.

Answer 1

Hint: According to given in the question we have to prove that the given points $(4,5, - 5),(0, - 11,3)$ and $(2, - 3, - 1)$ are collinear so, first of all we have to understand when the given three points are collinear which is explained below:
First of all we have to find the distance between the first and the second point let the points are A and B and after that we have to find the distance between the second and the third point let that the third point s C and then we have to find the distance between third and the first point and if the sum of distances between the two lines are equal to the distance between any two of the points which forming a line can be defined as a collinear.
Now, as mentioned above we have to find the distance between points $(4,5, - 5)$ and $(0, - 11,3)$ with the help of the formula as mentioned below:

Formula used: $ \Rightarrow PQ = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} - {z_1})}^2}} ...................(A)$
Where PQ is the distance between the points $P({x_1},{y_1},{z_1})$ and $Q({x_2},{y_2},{z_2})$
Same as we have to find the distance between the points $(0, - 11,3)$ and $(2, - 3, - 1)$ with the help of the formula (A) above.
Now, we have to find the distance between the points $(2, - 3, - 1)$ and $(4,5, - 5)$ with the help of the formula (A) above.

Complete step-by-step solution:
Step 1: First of all we have to find the distance between the first A $(4,5, - 5)$ and the second point B $(0, - 11,3)$ with the help of the formula (A) as mentioned in the solution hint. Hence,
 $
   \Rightarrow AB = \sqrt {{{(0 - 4)}^2} + {{( - 11 - 5)}^2} + {{(3 - ( - 5))}^2}} \\
   \Rightarrow AB = \sqrt {{{(4)}^2} + {{( - 16)}^2} + {{(8)}^2}}
 $
On solving the expression obtained just above,
$
   \Rightarrow AB = \sqrt {16 + 256 + 64} \\
   \Rightarrow AB = \sqrt {336} \\
   \Rightarrow AB = 4\sqrt {21} ..............(1)
 $
Step 2: Now, same as the step 1 we have to find the distance between the second point B $(0, - 11,3)$ and the third point C $(2, - 3, - 1)$ with the help of the formula (A) as mentioned in the solution hint. Hence,
$
   \Rightarrow BC = \sqrt {{{(2 - 0)}^2} + {{( - 3 - ( - 11))}^2} + {{( - 1 - 3)}^2}} \\
   \Rightarrow BC = \sqrt {{{(2)}^2} + {{(8)}^2} + {{( - 4)}^2}}
 $
On solving the expression obtained just above,
$
   \Rightarrow BC = \sqrt {4 + 64 + 16} \\
   \Rightarrow BC = \sqrt {84} \\
   \Rightarrow BC = 2\sqrt {21} ..............(2)
 $
Step 3: Now, same as the step 2 we have to find the distance between the second point C $(2, - 3, - 1)$ and the third point A $(4,5, - 5)$ with the help of the formula (A) as mentioned in the solution hint. Hence,
$
   \Rightarrow AC = \sqrt {{{(2 - 4)}^2} + {{( - 3 - 5)}^2} + {{( - 1 - ( - 5))}^2}} \\
   \Rightarrow AC = \sqrt {{{( - 2)}^2} + {{( - 8)}^2} + {{(4)}^2}}
 $
On solving the expression obtained just above,
$
   \Rightarrow AC = \sqrt {4 + 64 + 16} \\
   \Rightarrow AC = \sqrt {84} \\
   \Rightarrow AC = 2\sqrt {21} ..............(3)
 $
Step 4: Now, from the expressions (1), (2), and (3) as obtained above, we have to check the condition that the points A, B, and C are collinear as mentioned in the solution hint.
$ \Rightarrow BC + AC = AB$
On substituting all the values in the expression just above,
$
   \Rightarrow 2\sqrt {21} + 2\sqrt {21} = 4\sqrt {21} \\
   \Rightarrow 4\sqrt {21} = 4\sqrt {21}
 $
Hence, points A, B, and C are collinear.

Hence, with the help of the formula (A) we have proved that the three points are collinear.

Note: Another method:
Step 1: First of all we have to arrange the all three points $(4,5, - 5),(0, - 11,3)$ and $(2, - 3, - 1)$ in terms of determinant to check if the points are collinear or not. Hence,
$ = \left| {\begin{array}{*{20}{c}}
  4&5&{ - 5} \\
  0&{ - 11}&3 \\
  2&{ - 3}&{ - 1}
\end{array}} \right|$
Step 2: Now, on solving the determinant as obtained in the solution step 1,
$
   = 4(11 + 9) + 0 + 2(15 - 55) \\
   = 80 - 80 \\
   = 0
 $
Hence, as we can see that the value of determinant is 0 so all the three points are collinear.