Question

The incentre of the triangle formed by the points \[\widehat i + \widehat j + \widehat k,4\widehat i + \widehat j + \widehat k,4\widehat i + 5\widehat j + \widehat k\] is
A. \[\dfrac{{\widehat i + \widehat j + \widehat k}}{3}\]
B. \[\widehat i + 2\widehat j + 3\widehat k\]
C. \[3\widehat i + 2\widehat j + \widehat k\]
D. \[\widehat i + \widehat j + \widehat k\]

Answer 1

Hint: This problem deals with the incentre of the triangle. Given the position vectors of the triangle, we have to find the position vector of the incentre of the given triangle. Here the incentre of the triangle is the point where all of the angle bisectors meet in the triangle. If the vertices of the triangle are $\overrightarrow A ,\overrightarrow B ,\overrightarrow C $, then the formula of incentre of the triangle is given by:
$ \Rightarrow \dfrac{{\overrightarrow A \times \left| {\overrightarrow {BC} } \right| + \overrightarrow B \times \left| {\overrightarrow {CA} } \right| + \overrightarrow C \times \left| {\overrightarrow {AB} } \right|}}{{\left| {\overrightarrow {AB} } \right| + \left| {\overrightarrow {BC} } \right| + \left| {\overrightarrow {CA} } \right|}}$

Complete step-by-step solution:
Given the position vectors of the vertices of the triangle.
Let the vertices of the triangle be A, B and C.
The position vectors of A is \[\widehat i + \widehat j + \widehat k\], which is mathematically expressed below:
$\overrightarrow A = \widehat i + \widehat j + \widehat k$
The position vectors of B is \[4\widehat i + \widehat j + \widehat k\], which is mathematically expressed below:
$\overrightarrow B = 4\widehat i + \widehat j + \widehat k$
The position vectors of C is \[4\widehat i + 5\widehat j + \widehat k\], which is mathematically expressed below:
$\overrightarrow C = 4\widehat i + 5\widehat j + \widehat k$
Now finding the vectors $\overrightarrow {AB} ,\overrightarrow {BC} ,\overrightarrow {CA} $,
First finding the vector $\overrightarrow {AB} $, as given below:
\[ \Rightarrow \overrightarrow {AB} = \overrightarrow B - \overrightarrow A \]
\[ \Rightarrow \overrightarrow {AB} = \left( {4\widehat i + \widehat j + \widehat k} \right) - \left( {\widehat i + \widehat j + \widehat k} \right)\]
Simplifying the above expression as shown below:
\[ \Rightarrow \overrightarrow {AB} = 4\widehat i + \widehat j + \widehat k - \widehat i - \widehat j - \widehat k\]
\[\therefore \overrightarrow {AB} = 3\widehat i\]
Now finding the vector $\overrightarrow {BC} $, as given below:
\[ \Rightarrow \overrightarrow {BC} = \overrightarrow C - \overrightarrow B \]
\[ \Rightarrow \overrightarrow {BC} = \left( {4\widehat i + 5\widehat j + \widehat k} \right) - \left( {4\widehat i + \widehat j + \widehat k} \right)\]
Simplifying the above expression as shown below:
\[ \Rightarrow \overrightarrow {BC} = 4\widehat i + 5\widehat j + \widehat k - 4\widehat i - \widehat j - \widehat k\]
\[\therefore \overrightarrow {BC} = 4\widehat j\]
First finding the vector $\overrightarrow {CA} $, as given below:
\[ \Rightarrow \overrightarrow {CA} = \overrightarrow A - \overrightarrow C \]
\[ \Rightarrow \overrightarrow {CA} = \left( {\widehat i + \widehat j + \widehat k} \right) - \left( {4\widehat i + 5\widehat j + \widehat k} \right)\]
Simplifying the above expression as shown below:
\[ \Rightarrow \overrightarrow {CA} = \widehat i + \widehat j + \widehat k - 4\widehat i - 5\widehat j - \widehat k\]
\[\therefore \overrightarrow {CA} = - 3\widehat i - 4\widehat j\]
Now finding the values of \[\left| {\overrightarrow {AB} } \right|,\left| {\overrightarrow {BC} } \right|,\left| {\overrightarrow {CA} } \right|\].
First finding \[\left| {\overrightarrow {AB} } \right|\], as given below:
$ \Rightarrow $\[\left| {\overrightarrow {AB} } \right| = \sqrt {{{\left( {3\widehat i} \right)}^2}} \]
\[ \Rightarrow \left| {\overrightarrow {AB} } \right| = \sqrt {{{\left( 3 \right)}^2}} \]
\[\therefore \left| {\overrightarrow {AB} } \right| = 3\]
Now finding \[\left| {\overrightarrow {BC} } \right|\], as given below:
$ \Rightarrow $\[\left| {\overrightarrow {BC} } \right| = \sqrt {{{\left( {4\widehat j} \right)}^2}} \]
\[ \Rightarrow \left| {\overrightarrow {BC} } \right| = \sqrt {{{\left( 4 \right)}^2}} \]
\[\therefore \left| {\overrightarrow {BC} } \right| = 4\]
Now finding \[\left| {\overrightarrow {CA} } \right|\], as given below:
$ \Rightarrow $\[\left| {\overrightarrow {CA} } \right| = \sqrt {{{\left( { - 3\widehat i} \right)}^2} + {{\left( { - 4\widehat j} \right)}^2}} \]
\[ \Rightarrow \left| {\overrightarrow {CA} } \right| = \sqrt {{{\left( 3 \right)}^2} + {{\left( 4 \right)}^2}} \]
\[\therefore \left| {\overrightarrow {CA} } \right| = 5\]
Let the incentre of the triangle be $I$, which is given below.
Now the incentre of the triangle is given by:
$ \Rightarrow I = \dfrac{{\overrightarrow A \times \left| {\overrightarrow {BC} } \right| + \overrightarrow B \times \left| {\overrightarrow {CA} } \right| + \overrightarrow C \times \left| {\overrightarrow {AB} } \right|}}{{\left| {\overrightarrow {AB} } \right| + \left| {\overrightarrow {BC} } \right| + \left| {\overrightarrow {CA} } \right|}}$
Substituting the position vectors $\overrightarrow A ,\overrightarrow B ,\overrightarrow C $, and the values of $\left| {\overrightarrow {AB} } \right|,\left| {\overrightarrow {BC} } \right|,\left| {\overrightarrow {CA} } \right|$ in the above expression:
$ \Rightarrow I = \dfrac{{\left( {\widehat i + \widehat j + \widehat k} \right) \times 4 + \left( {4\widehat i + \widehat j + \widehat k} \right) \times 5 + \left( {4\widehat i + 5\widehat j + \widehat k} \right) \times 3}}{{3 + 4 + 5}}$
$ \Rightarrow I = \dfrac{{\left( {4\widehat i + \widehat {4j} + 4\widehat k} \right) + \left( {20\widehat i + 5\widehat j + 5\widehat k} \right) + \left( {12\widehat i + 15\widehat j + 3\widehat k} \right)}}{{12}}$
Grouping the like terms and unlike terms together and simplifying as given below:
$ \Rightarrow I = \dfrac{{36\widehat i + 2\widehat {4j} + 12\widehat k}}{{12}}$
$ \Rightarrow I = 3\widehat i + 2\widehat j + \widehat k$
$\therefore $The position vector of the incentre of the triangle is $3\widehat i + 2\widehat j + \widehat k$.
The incentre of the given triangle is $3\widehat i + 2\widehat j + \widehat k$.

Option C is the correct answer.

Note: Please note that while solving the incentre of the triangle, we have to be clear in one thing that the incentre is the point where all of the angle bisectors meet in the triangle, it need not necessarily be the center of the triangle. Remember here that do not confuse that the incentre of the triangle is the same as the centroid of the triangle. It need not be true. But it can be in a few cases.