The points A, B and C with position vectors, $\overrightarrow a = 3\widehat i - 4\widehat j - 4\widehat k,\overrightarrow b = 2\widehat i - \widehat j + \widehat k,$ and $\overrightarrow c = \widehat i - 3\widehat j - 5\widehat k$ respectively form the vertices of a?
1. Right angled triangle
2. Isosceles triangle
3. Equilateral triangle
4. None of these

Hint: In this question, we are given the position vectors $\overrightarrow a = 3\widehat i - 4\widehat j - 4\widehat k,\overrightarrow b = 2\widehat i - \widehat j + \widehat k,$ and $\overrightarrow c = \widehat i - 3\widehat j - 5\widehat k$ of the points A, B, and C and we have to determine these position vectors are of which triangle. Start the solution by finding the magnitude of each side. And then put the values in Pythagoras theorem to check whether the triangle is right angled or not.

Formula Used:
Pythagoras theorem –
${\left( {hypotenuse} \right)^2} = {\left( {base} \right)^2} + {\left( {perpendicular} \right)^2}$
Magnitude of three-dimensional vector –
$Magnitude = \sqrt {{a^2} + {b^2} + {c^2}} $ where the vector is $a\widehat i + b\widehat j + c\widehat k$

Complete step by step Solution:
Given that,
Position vectors of given points A, B, C are
$\overrightarrow a = 3\widehat i - 4\widehat j - 4\widehat k$
$\overrightarrow b = 2\widehat i - \widehat j + \widehat k$
$\overrightarrow c = \widehat i - 3\widehat j - 5\widehat k$
Here, we’ll find the sides of the triangle using position vectors
$AB = \left| {\overrightarrow b - \overrightarrow a } \right|$
$ = \left| {2\widehat i - \widehat j + \widehat k - \left( {3\widehat i - 4\widehat j - 4\widehat k} \right)} \right|$
$ = \left| { - \widehat i + 3\widehat j + 5\widehat k} \right|$
$ = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( 3 \right)}^2} + {{\left( 5 \right)}^2}} $
$ = \sqrt {35} $
$BC = \left| {\overrightarrow c - \overrightarrow b } \right|$
$ = \left| {\widehat i - 3\widehat j - 5\widehat k - \left( {2\widehat i - \widehat j + \widehat k} \right)} \right|$
$ = \left| { - \widehat i - 2\widehat j - 6\widehat k} \right|$
$ = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( { - 2} \right)}^2} + {{\left( { - 6} \right)}^2}} $
$ = \sqrt {41} $
$CA = \left| {\overrightarrow a - \overrightarrow c } \right|$
$ = \left| {3\widehat i - 4\widehat j - 4\widehat k - \left( {\widehat i - 3\widehat j - 5\widehat k} \right)} \right|$
$ = \left| {2\widehat i - \widehat j + \widehat k} \right|$
$ = \sqrt {{{\left( 2 \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( 1 \right)}^2}} $
$ = \sqrt 6 $
From the above required values,
${\left( {BC} \right)^2} = {\left( {AB} \right)^2} + {\left( {CA} \right)^2}$
${\left( {\sqrt {41} } \right)^2} = {\left( {\sqrt {35} } \right)^2} + {\left( {\sqrt 6 } \right)^2}$
$41 = 35 + 6$
$41 = 41$
Here, the left-hand side is equal to the right-hand side..
It implies that the side of triangle satisfies the Pythagoras theorem i.e., ${\left( {hypotenuse} \right)^2} = {\left( {base} \right)^2} + {\left( {perpendicular} \right)^2}$
As we know, Right angle triangle is the triangle that follows Pythagoras theorem
Therefore, the points A, B and C with position vectors, $\overrightarrow a = 3\widehat i - 4\widehat j - 4\widehat k,\overrightarrow b = 2\widehat i - \widehat j + \widehat k,$ and $\overrightarrow c = \widehat i - 3\widehat j - 5\widehat k$ are the vertices of the Right-angled triangle.

Hence, the correct option is 1.

Note: The key concept involved in solving this problem is the good knowledge of Triangle. Students must know that to check the triangle is right- angled always use Pythagoras theorem if the sides satisfy the theorem it means the triangle is right-angled. For equilateral no need to do anything all sides will be equal and for isosceles any two sides will be equal.