Write the vector \( \mathbf{v}=\langle-6,-3,-6\rangle \) as a product of its magnitude and a unit vector with the same direction as \( \mathbf{v} \). \( \langle-6,-3,-6\rangle=\square\langle(\square,(\square),(\square)\rangle \) (Simplify your answers. Use integers or fractions for any numbers in the expression.)
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The Deep Dive
To express the vector \( \mathbf{v}=\langle-6,-3,-6\rangle \) as a product of its magnitude and a unit vector in the same direction, we first find the magnitude of \( \mathbf{v} \). The magnitude \( |\mathbf{v}| \) is calculated using the formula: \[ |\mathbf{v}| = \sqrt{(-6)^2 + (-3)^2 + (-6)^2} = \sqrt{36 + 9 + 36} = \sqrt{81} = 9. \] Next, we find the unit vector in the direction of \( \mathbf{v} \). The unit vector \( \mathbf{u} \) is given by: \[ \mathbf{u} = \frac{\mathbf{v}}{|\mathbf{v}|} = \frac{\langle-6, -3, -6\rangle}{9} = \left\langle \frac{-6}{9}, \frac{-3}{9}, \frac{-6}{9} \right\rangle = \langle -\frac{2}{3}, -\frac{1}{3}, -\frac{2}{3} \rangle. \] Now we can express \( \mathbf{v} \) as: \[ \mathbf{v} = 9 \left\langle -\frac{2}{3}, -\frac{1}{3}, -\frac{2}{3} \right\rangle. \] Therefore, we can write: \[ \langle-6,-3,-6\rangle = 9 \langle -\frac{2}{3}, -\frac{1}{3}, -\frac{2}{3} \rangle. \]