Question

Show that the points $O\left( {0,0,0} \right),A\left( {2, - 3,3} \right),B\left( { - 2,3, - 3} \right)$ are collinear. Find the ratio in which each point divides the segment joining the other two.

Answer 1

First, we will try to prove that the area of the triangle is equal to 0 so that the given points are collinear. Hence, we will find the ratio through which each point divides the segment joining the other two. Eventually arriving at the final answer.

Complete step-by-step answer:
Let's find the area of $\vartriangle OAB$ which is given by,
$\Delta = \dfrac{1}{2}|\mathop {OA}\limits^ \to | \times |\mathop {OB}\limits^ \to |\sin \theta$
Where $\theta$ is the angle between $\mathop {OA}\limits^ \to = (2 - 0)\mathop i\limits^ \wedge + ( - 3 - 0)\mathop j\limits^ \wedge + (3 - 0)\mathop z\limits^ \wedge$ and $\mathop {OB}\limits^ \to = ( - 2 - 0)\mathop i\limits^ \wedge + ( - 3 - 0)\mathop j\limits^ \wedge + ( - 3 - 0)\mathop z\limits^ \wedge$
Now, $\mathop {OB}\limits^ \to$
Rearranging the terms, we get
$\Rightarrow \cos (\theta ) = \dfrac{{\mathop {OA}\limits^ \to \times \mathop {OB}\limits^ \to }}{{|\mathop {OB}\limits^ \to ||\mathop {OA}\limits^ \to |}}$
Substituting the values of $\mathop {OA}\limits^ \to$, $\mathop {OB}\limits^ \to$, $|\mathop {OA}\limits^ \to |$ and $|\mathop {OB}\limits^ \to |$
$\Rightarrow \cos (\theta ) = \dfrac{{\left( {2\mathop i\limits^ \wedge - 3\mathop j\limits^ \wedge + 3\mathop z\limits^ \wedge } \right) \times \left( { - 2\mathop i\limits^ \wedge + 3\mathop j\limits^ \wedge - 3\mathop z\limits^ \wedge } \right)}}{{\sqrt {{2^2} + ( - 3){}^2 + 3{}^2} \times \sqrt {{{( - 2)}^2} + 3{}^2 + ( - 3){}^2} }}$
Now as we know in dot product $\mathop i\limits^ \wedge \times \mathop i\limits^ \wedge = 1,\mathop j\limits^ \wedge \times \mathop j\limits^ \wedge = 1,\mathop k\limits^ \wedge \times \mathop k\limits^ \wedge = 1$ and $\mathop i\limits^ \wedge \times \mathop j\limits^ \wedge \,or\mathop i\limits^ \wedge \times \mathop k\limits^ \wedge = 0,\mathop j\limits^ \wedge \times \mathop i\limits^ \wedge \,or\,\mathop j\limits^ \wedge \times \mathop k\limits^ \wedge = 0\,or\,\mathop k\limits^ \wedge \times \mathop j\limits^ \wedge = 0$, we get
$\Rightarrow \cos (\theta ) = \dfrac{{2\mathop i\limits^ \wedge \times \left( { - 2\mathop i\limits^ \wedge } \right) + \left( { - 3} \right)\mathop j\limits^ \wedge \times 3\mathop j\limits^ \wedge + 3\mathop i\limits^ \wedge \times \left( { - 3\mathop i\limits^ \wedge } \right)}}{{\sqrt {{2^2} + ( - 3){}^2 + 3{}^2} \times \sqrt {{{( - 2)}^2} + 3{}^2 + ( - 3){}^2} }}$

Now as we know in the dot product $\mathop i\limits^ \wedge \times \mathop i\limits^ \wedge = 1,\mathop j\limits^ \wedge \times \mathop j\limits^ \wedge = 1,\mathop k\limits^ \wedge \times \mathop k\limits^ \wedge = 1$
$\Rightarrow \cos (\theta ) = \dfrac{{ - 4 - 9 - 9}}{{\sqrt {4 + 9 + 9} \times \sqrt {4 + 9 + 9} }}$
On simplification we get,
$\Rightarrow \cos (\theta ) = \dfrac{{ - (4 + 9 + 9)}}{{4 + 9 + 9}}$
On cancelling common terms, we get
$\Rightarrow \cos (\theta ) = - 1$
On Substituting $- 1 = \cos (180^\circ )$, we get
$\Rightarrow \cos (\theta ) = \cos (180^\circ )$
Comparing the above angles.
$\Rightarrow \theta = 180^\circ$
$\Rightarrow \Delta = 0$ as $\sin (180^\circ ) = 0$
$\Rightarrow$ O, A, B are collinear
Now, let A divides OB in $k:1$
We know that the section formula for 3d geometry is $p(x,y,z) = \left( {\dfrac{{m{x_1} + n{x_2}}}{{m + n}},\dfrac{{m{y_1} + n{y_2}}}{{m + n}},\dfrac{{m{z_1} + n{z_2}}}{{m + n}}} \right)$ Where $m = k,n = 1,x = 2,{x_1} = - 2,{x_2} = 0$
Now, comparing only the x coordinate of the section formula we get
$x = \dfrac{{m{x_1} + n{x_2}}}{{m + n}}$
Dividing numerator and denominator by n on the RHS, we get
$\Rightarrow x = \dfrac{{\dfrac{{m{x_1} + n{x_2}}}{n}}}{{\dfrac{{m + n}}{n}}}$
On simplification we get
$\Rightarrow x = \dfrac{{\dfrac{m}{n}{x_1} + {x_2}}}{{\dfrac{m}{n} + 1}}$
Since we have taken $\dfrac{m}{n} = k$
$\Rightarrow 2 = \dfrac{{ - 2k + 0}}{{k + 1}}$
On cross multiplication we get
$\Rightarrow 2(k + 1) = - 2k$
In simplification we get
$\Rightarrow 2k + 2 = - 2k$
On rearranging we get
$\Rightarrow 2k + 2k = - 2$
On simplification we get
$\Rightarrow 4k = - 2$
On dividing the equation by 4, we get,
$\Rightarrow k = \dfrac{{ - 1}}{2}$
$\Rightarrow$ A divides OB in $1:2$ externally
Now for B, Where $\dfrac{m}{n} = k$, $x = - 2$, ${x_1} = - 2$, ${x_2} = 0$
$x = \dfrac{{m{x_1} + n{x_2}}}{{m + n}}$
Dividing numerator and denominator by n on the RHS
$\Rightarrow x = \dfrac{{\dfrac{{m{x_1} + n{x_2}}}{n}}}{{\dfrac{{m + n}}{n}}}$
On simplification we get
$\Rightarrow x = \dfrac{{\dfrac{m}{n}{x_1} + {x_2}}}{{\dfrac{m}{n} + 1}}$
Since we have taken $\dfrac{m}{n} = k$
$\Rightarrow - 2 = \dfrac{{2k + 0}}{{k + 1}}$
On cross multiplication we get
$\Rightarrow - 2(k + 1) = 2k$
On simplification we get
$\Rightarrow - 2k - 2 = 2k$
On rearranging we get
$\Rightarrow - 2k - 2k = 2$
On simplification we get
$\Rightarrow - 4k = 2$
On dividing the equation by -4 we get
$\Rightarrow k = \dfrac{{ - 1}}{2}$
$\Rightarrow$ B divides OA in $1:2$ externally for O

Now for O, Where $\dfrac{m}{n} = k$, $x = 0$, ${x_1} = - 2$, ${x_2} = 2$
$x = \dfrac{{m{x_1} + n{x_2}}}{{m + n}}$
Dividing numerator and denominator by n on the RHS
$\Rightarrow x = \dfrac{{\dfrac{{m{x_1} + n{x_2}}}{n}}}{{\dfrac{{m + n}}{n}}}$
On simplification we get
$\Rightarrow x = \dfrac{{\dfrac{m}{n}{x_1} + {x_2}}}{{\dfrac{m}{n} + 1}}$
Since we have taken $\dfrac{m}{n} = k$
$\Rightarrow 0 = \dfrac{{ - 2k + 2}}{{k + 1}}$
On cross multiplication we get
$\Rightarrow 0 = - 2k + 2$
On rearranging we get
$\Rightarrow 2k = 2$
On dividing the equation by 2 we get,
$\Rightarrow k = \dfrac{2}{2}$
On simplification we get
$\Rightarrow k = \dfrac{1}{1}$
$\Rightarrow$ O divides AB in $1:1$ internally.

Note: In these types of questions, we need to remember that we cannot solve an equation with two variables for the unique solution so we need to convert the equation of section formula’s ratio from $\dfrac{m}{n}$ to $\dfrac{k}{1}$ by taking $k = \dfrac{m}{n}$ hence we get one variable and we can solve the equation.