Question

If I is the center of a circle inscribed in a triangle ABC, then
\[\left| \overrightarrow{BC} \right|\overrightarrow{IA}+\left| \overrightarrow{CA} \right|\overrightarrow{IB}+\left| \overrightarrow{AB} \right|\overrightarrow{IC}\] is
(A) \[\overline{0}\]
(B) \[\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}\]
(C) \[\dfrac{\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}}{3}\]
(D) \[\dfrac{\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}}{2}\]

Answer 1

Hint: Get the position of the vertices A, B, and C of \[\Delta ABC\] with respect to the incenter I. Use the formula for the position vector of incenter of \[\Delta ABC\] , \[\dfrac{BC.\overrightarrow{a}+CA.\overrightarrow{b}+AB.\overrightarrow{c}}{BC+CA+AB}\] where \[\overrightarrow{a}\] , \[\overrightarrow{b}\] , and \[\overrightarrow{c}\] are the affixes of vertices of \[\Delta ABC\] and get the position vector of I. Since the calculated position vector is with respect to I so, the position vector of I with respect to I must be equal to zero. Now, solve it further and calculate the value of \[\left| \overrightarrow{BC} \right|\overrightarrow{IA}+\left| \overrightarrow{CA} \right|\overrightarrow{IB}+\left| \overrightarrow{AB} \right|\overrightarrow{IC}\] .

Complete step by step solution:
According to the question, we are given that I is the center of the circle inscribed in the triangle ABC and we are asked to find the possible value of the expression, \[\left| \overrightarrow{BC} \right|\overrightarrow{IA}+\left| \overrightarrow{CA} \right|\overrightarrow{IB}+\left| \overrightarrow{AB} \right|\overrightarrow{IC}\] .
First of all, let us assume that I is the incenter of \[\Delta ABC\] .

From the above diagram, we can observe that
The position vector of vertex A with respect to the incenter I = \[\overrightarrow{IA}\] …………………………………………….(1)
The position vector of vertex B with respect to the incenter I = \[\overrightarrow{IB}\] ……………………………………….……(2)
The position vector of vertex C with respect to the incenter I = \[\overrightarrow{IC}\] …………………………………………….(3)
We know the formula for the incenter of \[\Delta ABC\] , \[\dfrac{BC.\overrightarrow{a}+CA.\overrightarrow{b}+AB.\overrightarrow{c}}{BC+CA+AB}\] where \[\overrightarrow{a}\] , \[\overrightarrow{b}\] , and \[\overrightarrow{c}\] are the affixes of the vertices of \[\Delta ABC\] ………………………………………..(4)
Now, from equation (1), equation (2), equation (3), and equation (4), we get
The affix of the incenter of \[\Delta ABC\] = \[\dfrac{BC\left| \overrightarrow{IA} \right|+CA\left| \overrightarrow{IB} \right|+AB\left| \overrightarrow{IC} \right|}{BC+CA+AB}\] ………………………………..(5)
But the position vector of I with respect to I must be equal to zero ……………………………………..(6)
Now, from equation (5) and equation (6), we get
\[\Rightarrow \overrightarrow{0}=\dfrac{BC.\overrightarrow{IA}+CA.\overrightarrow{IB}+AB.\overrightarrow{IC}}{BC+CA+AB}\]
\[\Rightarrow \overrightarrow{0}=BC.\overrightarrow{IA}+CA.\overrightarrow{IB}+AB.\overrightarrow{IC}\] ……………………………………………….(7)
BC, CA, and AB can also be written as \[\left| \overrightarrow{BC} \right|\] , \[\left| \overrightarrow{CA} \right|\] , and \[\left| \overrightarrow{AB} \right|\] …………………………………………..(8)
Now, from equation (7) and equation (8), we get
\[\Rightarrow \overrightarrow{0}=\left| \overrightarrow{BC} \right|\overrightarrow{IA}+\left| \overrightarrow{CA} \right|\overrightarrow{IB}+\left| \overrightarrow{AB} \right|\overrightarrow{IC}\]
So, the correct answer is “Option A”.

Note: For this type of question, one must remember the formula for the position vector of the incenter of any triangle. That is the position vector of incenter of \[\Delta ABC\] , \[\dfrac{BC.\overrightarrow{a}+CA.\overrightarrow{b}+AB.\overrightarrow{c}}{BC+CA+AB}\] where \[\overrightarrow{a}\] , \[\overrightarrow{b}\] , and \[\overrightarrow{c}\] are the affixes of vertices of \[\Delta ABC\] .