TG EAMCET 2 May 2025 Shift 1 Paper

Given, A=

∧

i

+2

∧

j

+3

∧

k

,B=7

∧

i

−

∧

k

P=−2

∧

i

+3

∧

j

+5

∧

k

If P divides AB in the ratio m:n, then its harmonic conjugate Q divides AB in the ratio −m:n.
Now, AB=B−A
‌=(7−1)

∧

i

+(0−2)

∧

j

+(−1−3)

∧

k

‌=6

∧

i

−2

∧

j

−4

∧

k

Let P divides AB in the ratio m:n, so
‌P=‌

nA+mB

m+n

‌⇒‌‌−2

∧

i

+3

∧

j

+5

∧

k

‌‌‌=‌

n( ∧ i +2 ∧ j +3 ∧ k )+m(7 ∧ i − ∧ k )

m+n

‌⇒‌‌(m+n)(−2

∧

i

+3

∧

j

+5

∧

k

)
‌‌‌=n(

∧

i

+2

∧

j

+3

∧

k

)+m(7

∧

i

−

∧

k

)
Comparing the coefficients of

∧

i

,

∧

j

and

∧

k

on both sides
⇒‌−2(m+n)‌=n(1)+m(7)
⇒‌−2m−2n‌=n+7m
⇒‌−9m‌=3n
⇒‌‌

m

n

‌=−3m
∴‌‌

−m

n

‌=‌

1

−3

⇒‌‌

−1

−3

=‌

1

3

Now using section formula,
Q‌=‌

3A+1B

3+1

‌=‌

3( ∧ i +2 ∧ j +3 ∧ k )+(7 ∧ i − ∧ k )

4

‌=‌

10 ∧ i +6 ∧ j +8 ∧ k

4

=‌

5

2

∧

i

+‌

3

2

∧

j

+2

∧

k

So, sum of scalar components of
Q‌=‌

5

2

+‌

3

2

+2
‌=‌

8

2

+2=4+2=6