Verified Solution • 2025–26 Session moderate

QJEEAdv24Maac4a: JEE Advanced 2024 Physics - Class 11 JEEAdv24Maac4a Solution: Let and . If for some real numbers , and , we have then the…

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,β\alpha, \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(2 \vec{p}+\vec{q})+\beta(\vec{p}-2 \vec{q})+\gamma(\vec{p} \times \vec{q}),

then the value of γ\gamma is ________.

Step-by-Step Verified Solution

15i^+10j^+6k^=α(2p→+q→)+β(p→−2q→)+γ(p→×q→)15 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}=\alpha(2 \overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}})+\beta(\overrightarrow{\mathrm{p}}-2 \overrightarrow{\mathrm{q}})+\gamma(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}})

taking dot with (p→×q→)(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}})

amp;∣15amp;10amp;62amp;1amp;31amp;−1amp;1∣=0+0+γ(p2q2−(p→⋅q→)2)[∵(p→×q→)2+(p→⋅q→)2=p2q2]amp;⇒52=26γamp;∴γ=2\begin{aligned} & \left|\begin{array}{ccc} 15 & 10 & 6 \\ 2 & 1 & 3 \\ 1 & -1 & 1 \end{array}\right|=0+0+{\gamma}\left(\mathrm{p}^2 \mathrm{q}^2-(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}})^2\right) \quad\left[\because(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}})^2+(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}})^2=\mathrm{p}^2 \mathrm{q}^2\right] \\ & \Rightarrow 52=26 \boldsymbol{\gamma} \\ & \therefore \gamma=2 \end{aligned}

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