TG EAMCET 4 May 2025 Shift 1 Paper

Let position vector of point A be a.
Let position yector of point B be b.
Let position vector of point C be c.
Let position vector of point D be d.
Position vector of centroid of
△BCD=‌−

∧

i

+2

∧

j

−3

∧

k

‌

b+c+d

3

=‌−

∧

i

+2

∧

j

−3

∧

k

d=‌−3

∧

i

+6

∧

j

−9

∧

k

−b−c
d=‌−3

∧

i

+6

∧

j

−9

∧

k

‌−(−2

∧

i

+

∧

j

+3

∧

k

+3

∧

i

+2

∧

j

−

∧

k

)
=‌−3

∧

i

+6

∧

j

−9

∧

k

−

∧

i

−3

∧

j

−2

∧

k

=−4

∧

i

+3

∧

j

−11

∧

k

Let position vector of centroid G of tetrahedron ABCD be g.
g‌=‌

a+b+c+d

4

GD‌=d−(‌

a+b+c+d

4

)
‌=‌

3d−|a+b+c|

4

‌=‌

−12 ∧ i +9 ∧ j −33 ∧ k −(2 ∧ i + ∧ j +5 ∧ k )

4

‌=‌

−14 ∧ i +8 ∧ j −38 ∧ k

4

‌=‌

−7 ∧ i +4 ∧ j −19 ∧ k

2

|GD|‌=‌

1

2

×√72+42+192
‌=‌

√2×√213

√2×√2

=√‌

213

2