Let vec{v}=alpha hat{i}+2 hat{j}-3 hat{k},vec | vedclass

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(C) The scalar triple product is given by $[\vec{u} \vec{v} \vec{w}] = \vec{u} \cdot (\vec{v} 	imes \vec{w})$.The minimum value of the scalar triple

Vedclass · Accepted Answer

$3501$

Vedclass · Answer

(C) The scalar triple product is given by $[\vec{u} \vec{v} \vec{w}] = \vec{u} \cdot (\vec{v} 	imes \vec{w})$.The minimum value of the scalar triple product is $-|\vec{u}| |\vec{v} 	imes \vec{w}| = -\alpha \sqrt{3401}$.Given $|\vec{u}| = \alpha$,we have $|\vec{v} 	imes \vec{w}| = \sqrt{3401}$.Calculating the cross product $\vec{v} 	imes \vec{w}$:$\vec{v} 	imes \vec{w} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ \alpha & 2 & -3 \ 2\alpha & 1 & -1 \end{vmatrix} = \hat{i}(-2 + 3) - \hat{j}(-\alpha + 6\alpha) + \hat{k}(\alpha - 4\alpha) = \hat{i} - 5\alpha \hat{j} - 3\alpha \hat{k}$.Now,$|\vec{v} 	imes \vec{w}|^2 = 1^2 + (-5\alpha)^2 + (-3\alpha)^2 = 1 + 25\alpha^2 + 9\alpha^2 = 1 + 34\alpha^2$.Equating to $3401$: $1 + 34\alpha^2 = 3401 \implies 34\alpha^2 = 3400 \implies \alpha^2 = 100 \implies \alpha = 10$.Since the minimum value occurs when $\vec{u}$ is in the opposite direction of $\vec{v} 	imes \vec{w}$,we have $\vec{u} = -k(\hat{i} - 5\alpha \hat{j} - 3\alpha \hat{k})$ for some $k > 0$.$|\vec{u}| = k \sqrt{1 + 34\alpha^2} = k \sqrt{3401} = \alpha = 10 \implies k = \frac{10}{\sqrt{3401}}$.Then $\vec{u} = -\frac{10}{\sqrt{3401}}(\hat{i} - 50\hat{j} - 30\hat{k})$.$|\vec{u} \cdot \hat{i}|^2 = |-\frac{10}{\sqrt{3401}}|^2 = \frac{100}{3401}$.Thus $m = 100$ and $n = 3401$. Since $m$ and $n$ are coprime,$m + n = 100 + 3401 = 3501$.

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