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Vector Algebra question

2024 · Shift 2 · Q28
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Vector Algebra question

2024 · Shift 2 · Q28

JEE AdvancedMathematicsVector AlgebraNumerical+4 / −1
Let p⃗=2i^+j^+3k^\vec{p}=2 \hat{i}+\hat{j}+3 \hat{k}p​=2i^+j^​+3k^ and q⃗=i^−j^+k^\vec{q}=\hat{i}-\hat{j}+\hat{k}q​=i^−j^​+k^. If for some real numbers α,β\alpha, \betaα,β, and γ\gammaγ, 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}),15i^+10j^​+6k^=α(2p​+q​)+β(p​−2q​)+γ(p​×q​), then the value of γ\gammaγ is ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 2

  1. Given vectors

p⃗=2i^+j^+3k^,q⃗=i^−j^+k^\vec p = 2\hat i + \hat j + 3\hat k, \qquad \vec q = \hat i - \hat j + \hat kp​=2i^+j^​+3k^,q​=i^−j^​+k^

We need to use

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).15i^+10j^​+6k^=α(2p​+q​)+β(p​−2q​)+γ(p​×q​).

Our goal is to find γ\gammaγ.

  1. Compute the required vectors

(i) 2p⃗+q⃗2\vec p + \vec q2p​+q​

2p⃗=2(2,1,3)=(4,2,6)2\vec p = 2(2,1,3) = (4,2,6)2p​=2(2,1,3)=(4,2,6)

So,

2p⃗+q⃗=(4,2,6)+(1,−1,1)=(5,1,7).2\vec p + \vec q = (4,2,6) + (1,-1,1) = (5,1,7).2p​+q​=(4,2,6)+(1,−1,1)=(5,1,7).

Thus,

2p⃗+q⃗=5i^+j^+7k^.2\vec p + \vec q = 5\hat i + \hat j + 7\hat k.2p​+q​=5i^+j^​+7k^.

(ii) p⃗−2q⃗\vec p - 2\vec qp​−2q​

2q⃗=2(1,−1,1)=(2,−2,2)2\vec q = 2(1,-1,1) = (2,-2,2)2q​=2(1,−1,1)=(2,−2,2)

Hence,

p⃗−2q⃗=(2,1,3)−(2,−2,2)=(0,3,1).\vec p - 2\vec q = (2,1,3) - (2,-2,2) = (0,3,1).p​−2q​=(2,1,3)−(2,−2,2)=(0,3,1).

Thus,

p⃗−2q⃗=3j^+k^.\vec p - 2\vec q = 3\hat j + \hat k.p​−2q​=3j^​+k^.

(iii) p⃗×q⃗\vec p \times \vec qp​×q​

\hat i & \hat j & \hat k \\ 2 & 1 & 3 \\ 1 & -1 & 1 \end{vmatrix}$$ Expanding: $$\vec p \times \vec q = \hat i\begin{vmatrix}1 & 3 \\-1 & 1\end{vmatrix} - \hat j\begin{vmatrix}2 & 3 \\1 & 1\end{vmatrix} + \hat k\begin{vmatrix}2 & 1 \\1 & -1\end{vmatrix}$$ $$= \hat i(1\cdot 1 - 3\cdot(-1)) - \hat j(2\cdot 1 - 3\cdot 1) + \hat k(2\cdot(-1) - 1\cdot 1)$$ $$= \hat i(1+3) - \hat j(2-3) + \hat k(-2-1)$$ $$= 4\hat i + \hat j - 3\hat k.$$ So, $$\vec p \times \vec q = (4,1,-3).$$ --- 3. **Substitute into the given equation** Now, $$ (15,10,6) = \alpha(5,1,7) + \beta(0,3,1) + \gamma(4,1,-3). $$ Equating components: $$15 = 5\alpha + 4\gamma \qquad ...(1)$$ $$10 = \alpha + 3\beta + \gamma \qquad ...(2)$$ $$6 = 7\alpha + \beta - 3\gamma \qquad ...(3)$$ --- 4. **Solve for $\gamma$** From (1): $$5\alpha + 4\gamma = 15$$ $$\alpha = \frac{15 - 4\gamma}{5}. $$ Substitute into (2): $$10 = \frac{15 - 4\gamma}{5} + 3\beta + \gamma$$ Multiply by 5: $$50 = 15 - 4\gamma + 15\beta + 5\gamma$$ $$50 = 15 + \gamma + 15\beta$$ $$15\beta = 35 - \gamma$$ $$\beta = \frac{35 - \gamma}{15}. $$ Now substitute $\alpha$ and $\beta$ into (3): $$6 = 7\left(\frac{15 - 4\gamma}{5}\right) + \frac{35 - \gamma}{15} - 3\gamma$$ Take LCM $15$: $$6 = \frac{21(15 - 4\gamma) + (35 - \gamma) - 45\gamma}{15}$$ $$6 = \frac{315 - 84\gamma + 35 - \gamma - 45\gamma}{15}$$ $$6 = \frac{350 - 130\gamma}{15}$$ $$90 = 350 - 130\gamma$$ $$130\gamma = 260$$ $$\gamma = 2.$$ --- 5. **Final answer** $$\boxed{2}$$ The derived value of $\gamma$ matches the stored correct answer.
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