Let vec p =2 hat i +hat j +3 hat k and vec q =hat i -hat j +hat k . If for some real numbers α, β , and γ , we have 15 hat i +10 hat j +6 hat k =α(2 vec p +vec q )+β(vec p -2 vec q )+γ(vec p × vec q ), then the value of γ is \ \ \ \ \

Mathematics · Vector Algebra

JEE Advanced 2024 — Paper 2 — Question 11

Let $\vec{p}=2 \hat{i}+\hat{j}+3 \hat{k}$ and $\vec{q}=\hat{i}-\hat{j}+\hat{k}$ . If for some real numbers $\alpha, \beta$ , and $\gamma$ , we have

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 $\gamma$ is $\_\_\_\_\_$

Numerical answer — enter this value.

$\quad \vec{p}=2 \hat{i}+\hat{j}+3 \hat{k}, \vec{q}=\hat{i}-\hat{j}+\hat{k}$

$\vec{p} \times \overrightarrow{\mathbf{q}}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 1 & -1 & 1\end{array}\right|$

$\vec{p} \times \vec{q}=4 \hat{i}+\hat{j}-3 \hat{k}$ , $2 \vec{p}+\vec{q}=5 \hat{i}+\hat{j}+7 \hat{k}$

$\vec{p}-2 \vec{q}=0 \hat{i}+3 \hat{j}+\hat{k}$

Now, $15 \hat{i}+10 \hat{j}+6 \hat{k}=\alpha(2 \vec{p}+\vec{q})+\beta(\vec{p}-2 \vec{q})+\gamma(\vec{p} \times \vec{q})$

$15 \hat{i}+10 \hat{j}+6 \hat{k}=(5 \alpha+4 \gamma) \hat{i}+(\alpha+3 \beta+\gamma) \hat{j}+(7 \alpha+\beta-3 \gamma) \hat{k}$

$\Rightarrow 5 \alpha+4 \gamma=15 ; \alpha+3 \beta+\gamma=10 ; 7 \alpha+\beta-3 \gamma=6$

$\Rightarrow \gamma=2$

Answer key and solution verified before publishing.

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