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JEE Advanced 2020 · Paper 1, Mathematics question 16

In a triangle PQRPQR, let a⃗=QR→\vec a = \overrightarrow{QR}, b⃗=RP→\vec b = \overrightarrow{RP} and c⃗=PQ→\vec c = \overrightarrow{PQ}. If

∣a⃗∣=3,∣b⃗∣=4anda⃗⋅(c⃗−b⃗)c⃗⋅(a⃗−b⃗)=∣a⃗∣∣a⃗∣+∣b⃗∣,|\vec a| = 3,\qquad |\vec b| = 4\qquad\text{and}\qquad \frac{\vec a\cdot(\vec c - \vec b)}{\vec c\cdot(\vec a - \vec b)} = \frac{|\vec a|}{|\vec a| + |\vec b|},

then the value of ∣a⃗×b⃗∣2|\vec a\times\vec b|^2 is ______

Answer and worked solution

Official answer: 108

Around the triangle a⃗+b⃗+c⃗=0⃗\vec a+\vec b+\vec c=\vec 0, so c⃗=−(a⃗+b⃗)\vec c=-(\vec a+\vec b).

Numerator and denominator.
a⃗⋅(c⃗−b⃗)=−∣a⃗∣2−2a⃗⋅b⃗=−9−2a⃗⋅b⃗,\vec a\cdot(\vec c-\vec b)=-|\vec a|^2-2\vec a\cdot\vec b=-9-2\vec a\cdot\vec b,
c⃗⋅(a⃗−b⃗)=−(∣a⃗∣2−∣b⃗∣2)=7.\vec c\cdot(\vec a-\vec b)=-(|\vec a|^2-|\vec b|^2)=7.

The condition.
−9−2a⃗⋅b⃗7=37  ⟹  a⃗⋅b⃗=−6.\frac{-9-2\vec a\cdot\vec b}7=\frac37\implies\vec a\cdot\vec b=-6.

Cross product.
∣a⃗×b⃗∣2=∣a⃗∣2∣b⃗∣2−(a⃗⋅b⃗)2=144−36=108.|\vec a\times\vec b|^2=|\vec a|^2|\vec b|^2-(\vec a\cdot\vec b)^2=144-36=108.
Answer: 108

Topics:Cross productDot product
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