AP EAPCET 24 May 2025 Shift 1 Paper

The position vector of a is
a=−3

∧

i

+5

∧

j

+2

∧

k

Let, M be the mid-point of the hypotenuse BC.
So, the position vector of M is
m=6

∧

i

+2

∧

j

+5

∧

k

Let the centroid be G and its position vector be g.
So, g=‌

a+b+c

3

Since, the mid-point of the hypotenuse of a right triangle is equidistant from all three vertices
So, M is the circumcenter of a triangle and m=‌

b+c

2

⇒b+c=2m
Substituting this value in Eq. (i), we get
g‌=‌

a+2m

3

⇒g‌=‌

(−3 ∧ i +5 ∧ j +2 ∧ k )+2(6 ∧ i +2 ∧ j +5 ∧ k )

3

‌=‌

−3 ∧ i +5 ∧ j +2 ∧ k +12 ∧ i +4 ∧ j +10 ∧ k

3

‌=‌

9 ∧ i +9 ∧ j +12 ∧ k

3

‌=3

∧

i

+3

∧

j

+4

∧

k

So, the position vector of centroid is
3

∧

i

+3

∧

j

+4

∧

k