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Question

The incentre of the triangle formed by the points ^i+^j+^k, 4^i+^j+^k, and 4^i+5^j+^k is

A
^i+^j+^k3
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B
^i+2^j+3^k
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C
3^i+2^j+^k
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D
^i+^j+^k
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Solution

The correct option is D 3^i+2^j+^k

Position vector of A is ^i+^j+^k, gives coordinates as (1,1,1).

Position vector of B is 4^i+^j+^k, gives coordinates as (4,1,1).
Position vector of C is 4^i+5^j+^k, gives coordinates as (4,5,1).

AB=BA=3^i
BC=CB=4^j
AC=CA=3^i+4^j
Here, we can see that ABC is a right angled triangle with right angle at B.

|AB|=32=3
|BC|=42=4
|AC|=32+42=5

Point of Intersection of angle bisectors can be found using formula:
I=|BC|×A+|CA|×B+|AB|×C|AB|+|BC|+|CA|

I=4×(^i+^j+^k)+5×(4^i+^j+^k)+3×(4^i+5^j+^k)3+4+5

I=112(36^i+24^j+12^j)=3^i+2^j+^k

Hence, incenter will be 3^i+2^j+^k

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