Question

If $(6,10,10)$ , $(1,0, - 5)$, $(6, - 10,0)$ are vertices of a triangle, Find the direction ratios of its sides. Determine whether it is right angled or isosceles?

Answer 1

Hint : In this problem, we have to find direction ratio of the side of the triangle when the vertices are given $(6,10,10)$ , $(1,0, - 5)$, $(6, - 10,0)$. Let us assume the direction ratios of a line are the three sides as $a,b,c$ which are proportional to direction cosines.
Direction cosines is the angle made by the lines with the axis’s.
If $ABC$ be a line joining the $A({x_1},{y_1},{z_1})$,$B({x_1},{y_1},{z_1})$ and $C({x_1},{y_1},{z_1})$ then the direction ratios of the line $ABC$ are $({x_2} - {x_1})$ , $({y_2} - {y_1})$ and $({z_2} - {z_1})$.
Direction ratio of $AB$ is negative of the direction ratio of $BA$.
Angle between two lines is given by $\cos \theta = \dfrac{{({x_1}{x_2} + {y_1}{y_2} + {z_1}{z_2})}}{{\sqrt {{x_1}^2 + {y_1}^2 + {z_1}^2} \sqrt {{x_2}^2 + {y_2}^2 + {z_2}^2} }}$, where $\theta$ is the angle between the lines.
To find the direction ratio, we simply subtract the coordinates of the vertices of the triangle.

Complete step-by-step answer :
Let $ABC$ be the given triangle with the vertices $A(6,10,10)$ , $B(1,0, - 5)$ and $C(6, - 10,0)$.on comparing with the general form,
If $ABC$ be a line joining the $A({x_1},{y_1},{z_1})$, $B({x_1},{y_1},{z_1})$ and $C({x_1},{y_1},{z_1})$ then the direction ratios of the line $ABC$ are $({x_2} - {x_1})$ , $({y_2} - {y_1})$ and $({z_2} - {z_1})$.
So, we have to find the direction ratio of $AB$, $BC$ and $AC$.
Formula for finding the vertices of the direction ratios of the line $ABC$ are $({x_2} - {x_1})$ , $({y_2} - {y_1})$ and $({z_2} - {z_1})$.
To Determine the direction ratios of $AB$:
Let us assume $A({x_1},{y_1},{z_1}) = A(6,10,10)$ and $B({x_2},{y_2},{z_2}) = B(1,0, - 5)$.
The direction ratios of $AB$ are $(1 - 6)$ , $(0 - 10)$ , $(( - 5) - 10)$
i.e . $AB( - 5, - 10, - 15)$
Similarly, To Determine the direction ratios of $BC$:
Let us assume $B({x_1},{y_1},{z_1}) = B(1,0, - 5)$ and $C({x_2},{y_2},{z_2}) = C(6, - 10,0)$.
The direction ratios of $BC$ are $(6 - 1)$ , $( - 10 - 0)$ , $(0 - ( - 5))$ .
i.e . $BC(5, - 10,5)$.
Similarly, To Determine the direction ratios of $AC$:
Let us assume $A({x_1},{y_1},{z_1}) = A(6,10,10)$ and $C({x_2},{y_2},{z_2}) = C(6, - 10,0)$.
The direction ratios of $AC$ are $(6 - 6)$ , $( - 10 - 10)$ , $(0 - 10)$ .
i.e . $AC(0, - 20, - 10)$
Therefore, $AB( - 5, - 10, - 15)$,$BC(5, - 10,5)$,and $AC(0, - 20, - 10)$.
We needs to Finding the angle between the two lines by the formula is,
$\cos \theta = \dfrac{{({x_1}{x_2} + {y_1}{y_2} + {z_1}{z_2})}}{{\sqrt {{x_1}^2 + {y_1}^2 + {z_1}^2} \sqrt {{x_2}^2 + {y_2}^2 + {z_2}^2} }}$ --------- (1)
Now, Finding the angle $(\theta = A)$ between the two lines $AB( - 5, - 10, - 15)$ and $AC(0, - 20, - 10)$, then
Substitute all the values in the equation (1), then
(1)$\Rightarrow \cos A = \dfrac{{(( - 5)(0) + ( - 10)( - 20) + ( - 15)( - 10))}}{{\sqrt {{{( - 5)}^2} + {{( - 10)}^2} + {{( - 15)}^2}} \sqrt {{{(0)}^2} + {{( - 20)}^2} + {{( - 10)}^2}} }}$
On simplifying, we get
$\cos A = \dfrac{{200 + 150}}{{\sqrt {350} \sqrt {500} }} = \dfrac{{350}}{{\sqrt {350} \sqrt {500} }} = \sqrt {\dfrac{7}{{10}}}$
Therefore, $\angle A = {\cos ^{ - 1}}\sqrt {\dfrac{7}{{10}}}$ ----------(2)
Similarly, Finding the angle $(\theta = B)$ between the two lines $AB( - 5, - 10, - 15)$ and $BC(5, - 10,5)$, then
Substitute all the values in the equation (1), then
(1) $\Rightarrow \cos B = \dfrac{{(( - 5)(5) + ( - 10)( - 10) + ( - 15)(5))}}{{\sqrt {{{(5)}^2} + {{(10)}^2} + {{(15)}^2}} \sqrt {{{(5)}^2} + {{( - 10)}^2} + {{(5)}^2}} }}$
$\cos B = \dfrac{{ - 25 + 100 - 75}}{{\sqrt {350} \sqrt {150} }} = \dfrac{0}{{\sqrt {350} \sqrt {150} }} = 0$
Therefore, $\angle B = {\cos ^{ - 1}}0 = \dfrac{\pi }{2}$ ----------(3)
Now, Now, Finding the angle $(\theta = C)$ between the two lines $BC(5, - 10,5)$ and $AC(0, - 20, - 10)$, then
Substitute all the values in the equation (1), then
Similarly, $\cos C = \dfrac{{(( - 5)(0) + (10)( - 20) + ( - 5)( - 10))}}{{\sqrt {{{( - 5)}^2} + {{(10)}^2} + {{( - 5)}^2}} \sqrt {{{(0)}^2} + {{( - 20)}^2} + {{( - 10)}^2}} }}$
$\cos C = \dfrac{{ - 200 + 50}}{{\sqrt {150} \sqrt {500} }} = \dfrac{{ - 150}}{{\sqrt {150} \sqrt {500} }} = - \sqrt {\dfrac{3}{{10}}}$ (negative sign of cosine is become positive)
Therefore, $\angle C = {\cos ^{ - 1}}\sqrt {\dfrac{3}{{10}}}$ ---------(4)
Hence, we see that $\angle B = \dfrac{\pi }{2}$, and $\angle A \ne \angle C$.
Therefore $ABC$is a right angle triangle, not an isosceles triangle.

Note : For this we will first find the direction ratio of each side. Then we will find the angle between these lines to check whether the given triangle is right angle or isosceles.
We find the given triangle is a right angled triangle by the cosine angle formula.