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JEE Advanced 2024Paper 2
Exam
JEE Advanced
Year
2024
Paper
Paper 2
Subject
Mathematics
Chapter
Vector Algebra

JEE Advanced 2024 · Paper 2 · Mathematics · Vector Algebra

Let p =2 i + j +3 k and q = i - j + k . If for some real numbers α,...

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

Correct answer

0

Step-by-step solution

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}})takingdotwithtaking dot with(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}) \left|151062131−11\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]\begin{aligned} & \Rightarrow 52=26 \gamma \\ & \therefore \gamma=2\end{aligned}$$$
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