Question

Show that the points 0(0,0,0), A(2,-3, 3), B(-2,3,-3) are collinear. Find the ratio in which each point divides the segment joining the other two ​

Answer 1

$\large\underline{\sf{Solution-}}$

Given coordinates are

Coordinates of 0 is (0,0,0)

Coordinates of A is (2,-3, 3)

Coordinates of C is B(-2,3,-3)

[Remark :- To Prove that these three points are collinear, we assume that let any point divide the other two point in the ratio k : 1 and if on simplifying, we get same value of k, then points are collinear. ]

Let assume that

0(0,0,0) divides the line segment joining the points A(2,-3, 3) and B(-2,3,-3) in the ratio k : 1

We know,

Section formula

Let A(x₁, y₁, z₁) and B(x₂, y₂, z₂) be two points in the plane and C(x, y, z) be the point which divides AB internally in the ratio m₁ : m₂, then the coordinates of C is given by

$\begin{gathered} \boxed{\tt{ (x, y,z) = \bigg(\dfrac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}}, \dfrac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}},\dfrac{m_{1}z_{2}+m_{2}z_{1}}{m_{1}+m_{2}}\bigg)}} \\ \end{gathered}$

So, on substituting the values, we get

$\rm \: (0,0,0) = \bigg(\dfrac{2k - 2}{k + 1} ,\dfrac{3k - 3}{k + 1}, \dfrac{ - 3k + 3}{k + 1} \bigg)$

So, on comparing, we get

$\rm \: \dfrac{2k - 2}{k + 1} = 0$

$\rm \: 2k - 2 = 0$

$\rm \: 2k = 2$

$\rm\implies \:k = 1$

On comparing y - coordinate, we get

$\rm \: \frac{3k - 3}{k + 1} = 0$

$\rm \: 3k - 3 = 0$

$\rm \: 3k = 3$

$\rm\implies \:k = 1$

On comparing z - coordinate, we get

$\rm \: \frac{ - 3k + 3}{k + 1} = 0$

$\rm \: - 3k + 3 = 0$

$\rm \: - 3k = - 3$

$\rm\implies \:k = 1$

So,

$\rm\implies \:O \: divides \: AB \: in \: the \: ratio \: 1 : 1$

$\rm\implies \:O \: is \: the \: midpoint \: of \: AB \:$

$\rm\implies \:O,A,B \: are \: collinear.$

Hence,

O divides AB in the ratio 1 : 1 internally.

A divides OB in the ratio 1 : 2 externally

B divides OA in the ratio 1 : 2 externally

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Additional Information :-

Let A(x₁, y₁, z₁) and B(x₂, y₂, z₂) be two points in the plane and C(x, y, z) be midpoint, then the coordinates of C is given by

$\begin{gathered} \boxed{\tt{ (x, y,z) = \bigg(\dfrac{x_{2}+x_{1}}{2}, \dfrac{y_{2} + y_{1}}{2},\dfrac{z_{2}+z_{1}}{2}\bigg)}} \\ \end{gathered}$

Answer 2

Step-by-step explanation:

$\colorbox{violet} {}\color{red} \underline{ \begin{array}{ || |l| || } \hline \color{magenta} \\ \hline \boxed{ \text{ \tt \: Solution:-} } \end{array}}$

$\color{green}\text{Let us assume that point B divides the line \( \sf A C \) in the ratio \( \sf k: 1 .\)}$

$\color{purple} \[ \begin{array}{l} \text { Let } \tt\left(x_{1}, y_{1}, z_{1}\right)=(0,0,0) \\ \\ \tt(x, y, z)=(2,-3,3) \\ \\ \tt \left(x_{2}, y_{2}, z_{2}\right)=(-2,3,-3) \end{array} \]$

$\color{olive} \[ \begin{array}{l} \text{By section formula:} \\ \tt x=\dfrac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}} \\ \\ \tt \Rightarrow 2=\dfrac{k \cdot(-2)+1 \cdot(0)}{k+1} \\ \\ \tt \Rightarrow k=\dfrac{-1}{4} \end{array} \]$

on neglecting the negative sign we get, therefore ratio is 1:4.