Q34 / 54 · JEE Advanced 2020 (Paper 1) · Mathematics

In a triangle PQR , let a→=QR→, b→=RP→ and c→=PQ→ .

If a→ =3, b→ =4 and a→.c→−b→c→.a→−b→=a→a→+b→, then the value of a→×b→2 is _______

Correct answer: 108

Solution

a→+b→+c→=0

∵a→.c→−b→c→.a→−b→=a→a→+b→

⇒−b→+c→⋅c→−b→−a→+b→⋅a→−b→=a→a→+b→

⇒c→ 2−b→ 2a→ 2−b→ 2=33+4=37

⇒c→ 2=13

Now, a→+b→=−c→⇒a→+b→2=−c→2

⇒a→ 2+b→ 2+2a→.b→=c→ 2

⇒9+16+2a→.b→=13

⇒a→·b→=−6

Now, a→×b→2+a→.b→2=a→2.b→2

⇒a→×b→2+36=916⇒a→×b→2=108

Hence, the correct answer is 108.

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