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 ____.

JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Vector Algebra

Let

p = 2 \hat{i} + \hat{j} + 3 \hat{k}

and

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 p + q) + β (p - 2 q) + γ (p \times q),

then the value of

γ

is ____.

Answer: 2

Taking the dot product of both sides with

(p \times q)

:

Since

(p \times q) \cdot p = 0

and

(p \times q) \cdot q = 0

, the left side becomes:

(15 \hat{i} + 10 \hat{j} + 6 \hat{k}) \cdot (p \times q)

Compute

p \times q

p \times q = \hat{i} 21 \hat{j} 1 - 1 \hat{k} 31 = \hat{i} (1 + 3) - \hat{j} (2 - 3) + \hat{k} (- 2 - 1) = 4 \hat{i} + \hat{j} - 3 \hat{k}

So:

(15 \hat{i} + 10 \hat{j} + 6 \hat{k}) \cdot (4 \hat{i} + \hat{j} - 3 \hat{k}) = 60 + 10 - 18 = 52

The right side gives:

γ ∣ p \times q ∣^{2} = γ (16 + 1 + 9) = 26 γ

Therefore:

26 γ = 52 \Rightarrow γ = 2

JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Integer answer