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Home » = \fracn+12 - \frac13n^3 \cdot \fracn^2(n+1)^24 - Baxtercollege

= \fracn+12 - \frac13n^3 \cdot \fracn^2(n+1)^24 - Baxtercollege

📅 March 11, 2026 👤 scraface

Mar 11, 2026

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Solution: Let $\mathbf{w} = \langle w_1, w_2, w_3
angle$. The cross product $\mathbf{w} imes \mathbf{u} = \mathbf{v}$ c...
egin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ w_1 & w_2 & w_3...
Compute the cross product:
\mathbf{w} imes \mathbf{u} = \langle w_2(1) - w_3(-2), w_3(3) - w_1(1...
angle = \langle w_2 + 2w_3, 3w_3 - w_1, -2w_1 - 3w_2
Set equal to $\mathbf{v} = \langle -1, 4, 2
angle$, leading to the system:
w_2 + 2w_3 = -1 \quad ext{(1)} \
3w_3 - w_1 = 4 \quad ext{(2)} \

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