A force vec F = alpha hat i + 3hat j + 6hat k | vedclass

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(C) For the angular momentum about the origin to be conserved,the net torque $\vec{	au}$ acting on the particle about the origin must be zero.The tor

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$-1$

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(C) For the angular momentum about the origin to be conserved,the net torque $\vec{	au}$ acting on the particle about the origin must be zero.The torque is given by the cross product: $\vec{	au} = \vec{r} 	imes \vec{F} = 0$.Calculating the cross product using the determinant method:$\vec{r} 	imes \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -6 & -12 \ \alpha & 3 & 6 \end{vmatrix} = 0$Expanding the determinant:$\hat{i} [(-6)(6) - (-12)(3)] - \hat{j} [(2)(6) - (-12)(\alpha)] + \hat{k} [(2)(3) - (-6)(\alpha)] = 0$$\hat{i} [-36 + 36] - \hat{j} [12 + 12\alpha] + \hat{k} [6 + 6\alpha] = 0$$0\hat{i} - (12 + 12\alpha)\hat{j} + (6 + 6\alpha)\hat{k} = 0$For this vector to be zero,each component must be zero:$12 + 12\alpha = 0 \Rightarrow 12\alpha = -12 \Rightarrow \alpha = -1$$6 + 6\alpha = 0 \Rightarrow 6\alpha = -6 \Rightarrow \alpha = -1$Thus,the value of $\alpha$ is $-1$.

$A$ force $\vec F = \alpha \hat i + 3\hat j + 6\hat k$ is acting at a point $\vec r = 2\hat i - 6\hat j - 12\hat k$. The value of $\alpha$ for which angular momentum about the origin is conserved is:

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