The vector from the origin to point \(A\) is given as \((6,-2,-4)\), and the unit vector directed from the origin toward point \(B\) is \((2,-2,1) / 3\). If points \(A\) and \(B\) are ten units apart, find the coordinates of point \(B\).

The coordinates of point A are (6, -2, -4) and it is mentioned that A is at the origin (0, 0, 0). Therefore, Vector OA is: \(OA = (-6, 2, 4)\)

The unit vector directed from O to B is given as (2, -2, 1) / 3. Therefore, the vector OB can be expressed as: \(OB = (2t, -2t, t)\) Where \(t\) is the scalar multiple. We know that the distance between A and B is 10 units. Let's represent the vector from A to B as Vector AB: \( AB= OB - OA = (2t - (-6), -2t - 2, t - 4)\) Since we know the distance between A and B which is 10 units, we can apply the distance formula on the AB vector: \( \sqrt{(2t + 6)^2 + (-2t - 2)^2 + (t - 4)^2} = 10\) Now, we need to solve the equation to find the value of \(t\).

To find the coordinates of point B, we need to solve the distance equation to find the value of \(t\): \( (2t + 6)^2 + (-2t - 2)^2 + (t - 4)^2 = 100\) Expanding and simplifying the equation, we get: \(10t^2 -8t - 48 = 0\) Factoring: \((5t - 4)(2t + 3) = 0\) So, t can have two possible values: - \(t_1 = \frac{4}{5}\) - \(t_2 = \frac{-3}{2}\) However, considering the given unit vector directed from the origin to point B is \((2, -2, 1) / 3\), we know that the magnitude of t should be positive. So, we should consider only the positive value of t, \(t_1 = \frac{4}{5}\). Now, we can find the coordinates of point B by substituting the value of \(t_1\) in OB: \(OB = (2(\frac{4}{5}), -2(\frac{4}{5}), \frac{4}{5})= (\frac{8}{5}, -\frac{8}{5}, \frac{4}{5})\) Therefore, the coordinates of point B are \((\frac{8}{5}, -\frac{8}{5}, \frac{4}{5})\).