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← JEE Advanced 2024 · Paper 2
MathematicsInteger+4Question 11 of 51

JEE Advanced 2024 · Paper 2, Mathematics question 11

Let p⃗=2i^+j^+3k^\vec p = 2\hat i + \hat j + 3\hat k and q⃗=i^−j^+k^\vec q = \hat i - \hat j + \hat k. If for some real numbers α\alpha, β\beta, and γ\gamma, we have

15i^+10j^+6k^=α(2p⃗+q⃗)+β(p⃗−2q⃗)+γ(p⃗×q⃗),15\hat i + 10\hat j + 6\hat k = \alpha\left(2\vec p + \vec q\right) + \beta\left(\vec p - 2\vec q\right) + \gamma\left(\vec p \times \vec q\right),

then the value of γ\gamma is _____.

Answer and worked solution

Official answer: 2

Compute the cross product:
p⃗×q⃗=∣i^j^k^2131−11∣=4i^+j^−3k^,∣p⃗×q⃗∣2=26.\vec p\times\vec q=\begin{vmatrix}\hat i&\hat j&\hat k\\2&1&3\\1&-1&1\end{vmatrix}=4\hat i+\hat j-3\hat k,\qquad|\vec p\times\vec q|^2=26.

Both 2p⃗+q⃗2\vec p+\vec q and p⃗−2q⃗\vec p-2\vec q are perpendicular to p⃗×q⃗\vec p\times\vec q. Dotting the given identity with p⃗×q⃗\vec p\times\vec q therefore leaves only the γ\gamma term:
(15,10,6)⋅(4,1,−3)=60+10−18=52=26γ  ⟹  γ=2.(15,10,6)\cdot(4,1,-3)=60+10-18=52=26\gamma\implies\gamma=2.

Topics:Cross productDot product
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