Q.

Between the following two statements:

Statement-I: Let a→=i^+2j^–3k^ and b→=2i^+j^–k^. Then the vector r→ satisfying a→×r→=a→×b→ and a→·r→=0 is of magnitude 10.

Statement-II: In a triangle ABC, cos 2A + cos 2B + cos 2C ≥–32.          [2024]

1 Both Statement-I and Statement-II are incorrect.  
2 Statement-I is incorrect but Statement-II is correct.  
3 Statement-I is correct but Statement-II is incorrect.  
4 Both Statement-I and Statement-II are correct.  

Ans.

(2)

We have, a→×r→=a→×b→

⇒  a→×(r→–b→)=0  ⇒  a→=λ(r→–b→)

⇒  a→·a→=λ(a→·r→–a→·b→)

⇒  |a→|2=λ(0–7)  ⇒  14=–7λ

⇒  λ=–2  ⇒  r→=b→–a→2=2b→–a→2=3i^+k^2

So, |r→|=102

So, Statement-I is incorrect.

Let △ABC be given triangle and O be its circumcentre

Now, OA = a→, OB = b→ and OC = c→

Now, |OA| = |OB| = |OC|

(distance from circumcentre will be same for all vertices.)

Consider (OA + OB + OC)

=(a→+b→+c→)2≥0

⇒  |a→|2+|b→|2+|c→|2+2a→·b→+2b→·c→+2c→·a→≥0

⇒  3|a→|2+2|a→||b→|cos2C+2|b→||c→|cos2A+2|c→||a→|cos2B≥0

⇒  3|a→|2+2|a→|2[cos2A+cos2B+cos2C]≥0

⇒  2[cos2A+cos2B+cos2C]≥–3      [∵ |a→| >0]

⇒  cos 2A + cos 2B + cos 2C ≥–32

Hence, Statement-II is correct.